Functional Analysis Notes

Functional Analysis Notes

Hang, Spring 2009

Table of contents

Hahn-Banach Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Separation of Convex Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Generalization of Hahn Banach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Hahn-Banach for C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Hölder’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Duality Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6Generalized Minkowski . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6Completeness of Lp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Examples of Banach Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Completion of a Normed Linear Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

A Variational Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Strict and Uniform Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Uniformly Convex Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Orthogonal Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Riesz Representation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18Adjoints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Orthonormal Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Existence of nontrivial elements in X ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Duality Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25Application of Duality to Density Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25Dual Variational Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Weak Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Reflexive Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Application to Galerkin’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Uniform Boundedness Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Lower Semicontinuity under Weak Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33Weak* Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Weak and Weak* Convergence in L1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Interpolation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Locally Convex Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

1

Weak Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49Topological Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50Locally Convex Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51Frechet Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52Separation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54Double Polar Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55Krein Milman Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

Where to get Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Application of Krein Milman . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Proof of Krein Milman . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

Spectrum of Bounded Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

Holomorphic Functions from C→X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61Facts about σ(A) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63Functional Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Fredholm Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

Index and Pseudoinverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68Fredholm Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72Compact Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

Spectrum of Compact Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Spectrum of Compact Operators in Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Texts:

1. Lax, Functional Analysis

2. Reed-Simon, Methods of Modern Mathematical Physics, Vol. 1

3. Yosida, Functional Analysis

4. Rudin, Functional Analysis

5. Conway, (GTM) A Course in Functional Analysis

Some motivational remarks about Functional Analysis? Linear spaces, linear operators in infinite dimen-sions requires much more machinery. A large focus on linear function spaces and various applications.“Functional” refers to a linear function from X → K (X is linear space, K is some scalar field) and suchfunctionals will play a central role in studying function spaces.

Hahn-Banach Theorems

Given a real vector space X/R (notation), a subspace X0 ⊂X and a linear functional l0 defined on X0, itis easy to extend l0 to a linear functional l on X (for instance, if we decompose X = X0 + Y and definel(x0 + y)= l0(x0)). Usually l0 has an additional property that we want to preserve in the extension.

Even though we have restricted to real vector spaces, the results carry through to complex vector spaceswith only a slight alteration.

2

Theorem 1. Let X/R be a real vector space and X0 ⊂ X a subspace, with l0 a linear functional definedon X0. Suppose that l0(x)≤ p(x) for all x∈X0, where p:X→R satisfying

• Positive Homogeneity. p(λx)=λp(x) for λ> 0

• Subadditivity. p(x+ y)≤ p(x)+ p(y)

Then there exists a linear functional l defined on X such that l |X0= l0 and l≤ p. (i.e. l is an extension of

l0 that is also dominated by p)

Remark: p can be relaxed to be just convex as well.

Proof. First show you can extend by one dimension while preserving the domination property, e.g. fromX0 to X0 + spany for y ∈X0. Then turn the set of all possible extensions into a poset and apply Zorn’slemma (existence of maximal linearly ordered chain), and finally show that there is a largest element inthe maximal linearly ordered chain.

Example 2. (Extension of Positive Functionals) Let E be a set and

X = f :E→R s.t. supx∈E

|f(x)|<∞

Assume Y ⊂ X is a linear subspace that contains an element h such that h(x) ≥ c0 > 0 (bounded awayfrom zero), then if l0: Y → R is a linear functional that is positive (i.e. for all g ∈ Y such that g ≥ 0, wehave that l0(g)≥ 0), then l0 can be extended to a linear functional l defined on X that is also positive.

The proof involves finding an appropriate p to apply Hahn Banach. Use

p(f)4 infg≥fg∈Y

l0(g)

and prove that p is positive homogeneous, subadditive. Then Hahn Banach shows that for any negativef ∈X (f ≤ 0), l(f)≤ p(f)≤ 0. For the last inequality note that 0∈ Y so that p(f)≤ p(0)= 0.

Separation of Convex Sets

Theorem 3. Let X/R be a real vector space, and A, B ⊂X convex with A ∩B = ∅, A has an absorbingpoint. Then there exists a nonzero linear functional l: X →R such that l(x) ≤ l(y) for all x ∈ A and y ∈B.

We say that x0 is an absorbing point of A if for any x, we can find a scalar λ such that x ∈ x0 + λ(A −x0). That is, when A is scaled relative to x0, for sufficiently large (or small) scalars every point is included(absorbed).

Remark 4. When 0 is an absorbing point of a convex set A⊂X, we can define the Minkowski functional(gauge) pA by

pA(x) = infλ>0x∈λA

λ

which is a useful positive homogeneous, subadditive function.

Example 5. If A=B1, then pA(x) = |x|.

3

The theorem can be reduced to the following lemma:

Lemma 6. Let A ⊂ X be a convex subset with an absorbing point, and let x0 ∈ A. Then there exists anonzero linear functional l:X→R such that l |A≤ l(x0).

Proof. (of Theorem given Lemma) Use A−B, which still has an absorbing point, and since A, B aredisjoint, 0∈A−B, and thus for some linear functional l we have that l |A−B ≤ l(0) = 0, or in other words,l(x)≤ l(y) for x∈A and y ∈B.

Proof. (of Lemma) Without loss of generality assume 0 is an absorbing point of A (otherwise shift it).Then we can take the Minkowski functional of A. Then define a linear functional on the subspacespanned by x0, Rx0, by l0(tx0) = pA(tx0). This means in particular that l0(tx0) ≤ pA(tx0), and we canextend to a linear functional on X such that l(x)≤ pA(x). Then for any x∈A, we have that

l(x)≤ pA(x)≤ 1 = pA(x0)= l(x0)

Generalization of Hahn Banach

Theorem 7. Let X/R be a real vector space and let X0 be a subspace. Let l0: X0 → R be a linear func-tional on X0 and p:X→R a positive homogeneous and subadditive function with l0≤ p|X0

as before. Addi-tionally, let A be a collection of linear maps from X→X such that

1. For all A∈A, AX0⊂X0 (i.e. X0 is invariant under A)

2. For all A∈A, l0(Ax) = l0(x) for all x∈X0.

3. For all A,B ∈A, AB=BA (commutative linear maps)

4. For all A∈A, p(Ax)= p(x) for all x∈X

Then there exists an extension to a linear functional l: X →R such that l|X0= l0, l ≤ p and l(Ax) = l(x)

for x∈X.

Example 8. From real analysis (Vitali) we know that there does not exist a nontrivial finite measure on

the power set of the unit circle S1 which is rotation invariant. However, if we relax the condition of count-able additivity to just finite additivity, we can use the theorem above to show that there does exist afinitely-additive rotation-invariant set function µ:P (S1)→ [0,∞).

Let X = u:S1→R , supz∈S1 |u(z)|<∞, A= Rθ: θ ∈R where Rθu(z) = u(eiθz). Let Y be the subset ofX consisting of the Lebesgue-measurable functions. Then on Y we can define l0(u) =

∫

S1 udθ, where we

note that for Lebesgue measurable sets A, the Lebesgue measure is recovered by λ(A) = l0(1A). Also l0 isrotation-invariant l0(Rθu) = l0(u) by the rotation-invariance of the Lebesgue measure. Now to extend toarbitrary subsets, we simply extend l0 to a rotation-invariant functional l defined on all of X and defineµ(A)= l(1A), and this gives us a finitely additive (linear) rotation-invariant set function as desired.

The only thing that remains is to find a positive-homogeneous and subadditive function p that is alsorotation-invariant and that dominates l. Since l0(u)≤

∫

S1 udθ ≤ 2π |u|∞, we can use p(u) = 2π |u|∞, which

is subadditive and positive homogeneous and dominates l0.

Proof. First turn A into an algebra by using A1 = A1A2Am: Ai ∈ A ⊃ A. Then examine the convexhull C = co(A) (set of convex combinations of A), and use the functional

q(x)= infA∈C

p(Ax)

4

noting that q(x)≤ p(x). The reason q is useful is because q(x−Ax)≤ 0 and q(Ax− x)≤ 0:

q(x−Ax)≤ 1

np[(I +A+ +An−1)(I −A)x] =

1

np((I −An)x)≤ p(x)+ p(− x)

n→ 0

Note we need q to be positive homogeneous and subadditive. This follows from

p

(

A+B

2x+

A+B

2y

)

≤ p(Ax)+ p(Bx)

2+p(Ay) + p(By)

2

and taking infimum over A,B ∈C gives the result.

Got A so that p(Ax) is close to inf, p(By) is close to inf.... bound below by p(Ax+By)

the same computation works for q(Ax − x) as well. This additional property shows us that the extensionto l≤ q satisfies l(x−Ax)≤ 0 and l(Ax− x)≤ 0 so that l(x)= l(Ax) as desired.

Hahn-Banach for CGiven a real-valued linear functional l1: X → R there exists a unique complex-valued functional l: X → C

such that l1 = Rel, and also l(x) = l1(x) − i l1(ix). Then we can prove the following version of Hahn-Banach for complex scalar field:

Theorem 9. Let X/C be a vector space over C, and X0 ⊂ X a subspace. Let l0: X0 → C be a complexvalued linear functional over X0, and p: X → R be positive homogeneous (p(αx) = |α|p(x)) and subaddi-tive, with |l0| ≤ p on X0. Then there exists an extension to a linear functional l:X→C such that l |X0

= l0and |l(x)| ≤ p(x).

Proof. Apply Hahn-Banach to Re l0 to get an extension to the real functional l1 over X. Then there is aunique functional l with real part l1, satisfying the requirements.

Banach Spaces

Preliminaries: norm, seminorm, Cauchy sequence, complete.

A Banach space is a complete normed linear space. Examples are C(X, R) the space of continuous func-tions from a compact, Hausdorff space X to R, and Lp(X, µ) for 1≤ p≤∞ (to show below).

Hölder’s Inequality

Quick proof: |fg |1 ≤ |f |p|g |q for 1

p+

1

q= 1, 1 ≤ p, q ≤∞. The case p= 1, q = ∞ is immediate. In general,

without loss of generality |f |p= |g |q= 1 by scaling. By concavity of lnx, we note that

|f(x)| |g(x)| ≤ 1

p|f(x)|p+

1

q|g(x)|q

so that integrating gives |fg |1≤ 1 as desired.

This inequality can be generalized slightly as well. First note that |fα|p = |f |αpα by definition. Then sup-

pose that1

r=

1

p+

1

q. Then

|fg |Lr = ||f |r |g |r |L11/r≤

∣

∣|f |r∣

∣

Lp/r

1/r ∣∣|g |r

∣

∣

Lq/r

1/r= |f |Lp|g |Lq

5

and inductively, we have that if1

r=

1

p1+ +

1

pm, then

|f1 fm|Lr ≤ |f1|Lp1 |fm|Lpm

Duality Principle

Let p′ denote the conjugate exponent satisfying1

p+

1

p′= 1. Assume (X, µ) is a σ-finite measure space.

Then for any f :X→C,

|f |p= sup|g|p′=1

fg integrable

∣

∣

∣

∣

∫

X

fg dµ

∣

∣

∣

∣

Proof. An upper bound on∫

fg is obtained by Hölder, the other constructs an explicit g ∈ Lp′

that

achieves the upper bound, g = ϕfp/p′

|f |pp/p′

where |ϕ| = 1 and ϕ |f | = f . The case where |f |p = ∞ uses σ-

finiteness for X =⋃

kXk, Yn=

⋃

k=1n

Xk, and fn=min (f , n)1Yn. Then RHS≥ |fn|p→∞

Generalized Minkowski

Theorem 10. Let 1≤ p≤∞ and (X, µ), (Y , ν) be σ-finite. Then if F :X ×Y → [0,∞], then

( ∫

X

( ∫

Y

F (x, y)dν(y)

)p

dµ(x)

)1/p

≤∫

Y

[ ∫

X

F (x, y)p dµ(x)

]1/p

dν(y)

Or in other words,∥

∥

∥

∥

∫

Y

F ( · , y)dν(y)∥

∥

∥

∥

Lp(X)

≤∫

Y

‖F ( · , y)‖Lp(X) dν(y)

Proof. This is just using duality principle and Fubini (this requires σ-finiteness):

∥

∥

∥

∥

∫

Y

F ( · , y)dν(y)∥

∥

∥

∥

Lp(X)

= sup‖g(x)‖p′=1

∣

∣

∣

∣

∫

X

g(x)

∫

Y

F (x, y)dν(y) dµ(x)

∣

∣

∣

∣

≤∫

Y

∫

X

|g(x)| |F (x, y)|dν(y)dµ(x)

≤∫

Y

(

sup‖g(x)‖p′=1

∫

X

|g(x)| |F (x, y)|dµ(x)

)

dν(y)

=

∫

Y

‖F ( · , y)‖Lp(X) dν(y)

As a consequence, we have the following fact

Proposition 11. Assume 0< p≤ q <∞, and suppose F ≥ 0. Then

∣

∣

∣|F (x, y)|Lp(X)

∣

∣

∣

Lq(Y )≤∣

∣

∣|F (x, y)|Lq(Y )

∣

∣

∣

Lp(X)

6

Proof. First we note that |fα|p= |f |αpα by definition. Then

∣

∣

∣|F (x, y)|Lp(X)

∣

∣

∣

Lq(Y )=∣

∣

∣|F p|L1(X)1/p

∣

∣

∣

Lq(Y )

=∣

∣

∣|F p|L1(X)

∣

∣

∣

Lq/p(Y )

1/p

≤∣

∣

∣|F p|Lq/p(Y )

∣

∣

∣

L1(X)

1/p

=∣

∣

∣|F p|Lq/p(Y )

1/p∣

∣

∣

Lp(X)

=∣

∣

∣|F |Lq(Y )

∣

∣

∣

Lp(X)

where the inequality follows from Generalized Minkowski and the fact that q/p≥ 1.

From this fact, we can derive the Minkowski inequalities:

Theorem 12. For 1≤ p≤∞, we have that

|f + g |p≤ |f |p+ |g |p

and for 0< p≤ 1, we have that if f , g are nonnegative, then

|f + g |p≥ |f |p+ |g |p

Proof. This follows from the previous proposition using an appropriately defined F (x, y). Choose Y = 1,2 with ν counting measure, and F (x, 1) = |f(x)| and F (x, 2) = |g(x)|, and apply the previous propositionwith exponents p and 1.

For 1≤ p≤∞, we have

|f + g |Lp ≤ ||f(x)|+ |g(x)| |Lp =∣

∣

∣|F |L1(Y )

∣

∣

∣

Lp(X)≤∣

∣

∣|F |Lp(X)

∣

∣

∣

L1(Y )= |f |Lp + |g |Lp

and for 0< p≤ 1, and nonnegative f , g, we have the same sequence of steps except that we need nonnega-tivity for the very first equality below:

|f + g |Lp = | |f(x)|+ |g(x)| |Lp =∣

∣

∣|F |L1(Y )

∣

∣

∣

Lp(X)≥∣

∣

∣|F |Lp(X)

∣

∣

∣

L1(Y )= |f |Lp + |g |Lp

Thus in particular, | · |Lp is not a norm when p < 1, and that for 1 ≤ p ≤∞, Lp(X, µ) is a normed linearspace.

Completeness of Lp

To show that Lp is complete for 1 ≤ p ≤ ∞, given a Cauchy sequence fk, pass to a subsequence so that|fk − fk+1|Lp < 2−i. Then noting fm(x) = f1(x) +

∑

k=1m (fk+1(x) − fk(x)) we show that the sum con-

verges a.e. x by showing that the sum is absolutely convergent for a.e. x. Let

g(x)= |f1(x)|+∑

k=1

m

|fk+1(x)− fk(x)|

7

and note that

|g(x)|Lp ≤ |f1|Lp +∑

k=1

m

|fk+1− fk|Lp ≤ |f1|Lp + 1

so that g ∈ Lp. This implies that g(x) <∞ for a.e. x so that the sum is absolutely convergent for a.e. x.Thus limm→∞ fm(x) exists a.e. x, and denote the limit by f(x). We show that fm→ f in Lp:

limm→∞

∫

|fm− f |p=

∫

limm→∞

|fm− f |= 0

by dominated convergence, since |fm− f |p≤ (|fm|+ |f |)p≤ (2 |g |)p≤ 2p|g |p which is integrable.

Examples of Banach Spaces

Example 13. C(X,R) with the supremum norm, where X is compact, Hausdorff. Here completeness fol-lows from the fact that uniform limit of continuous functions is continuous and the fact that if fn isCauchy in C(X,R), then fn(x) is Cauchy in R.

Example 14. Lp(X, µ) where (X,µ) is a σ-finite measure space, and 1≤ p≤∞.

Example 15. Let Ω⊂Rn be open, bounded.

• Cm(Ω)= u: Ω→R, u has continuous partial derivatives up to orderm

• Cm(Ω)= u∈Cm(Ω) , the partial derivatives of uhave continuous extension toΩ

Denoting α = (α1, , αn) and ∂αu = ∂1α1 ∂nαn, |α| = α1 + + αn (multi-index notation), we have semi-

norms

pk(u)=∑

|α|=k

|∂αu|∞

and from these we can construct a norm (not unique) by

|u|Cm(Ω ) =∑

k=0

m

pk(u)

These two spaces equipped with the norm above are Banach spaces.

Example 16. Let (X, d) be a metric space. Then for 0<α≤ 1, let

Cα(X) =

u:X→R, supx1 x2

|u(x1)− u(x2)|d(x1, x2)α

<∞

be the space of Hölder-continuous functions of order α. Then we have a seminorm

[u]α= supx1 x2

|u(x1)−u(x2)|d(x1, x2)α

8

which we use to construct the norm

|u|Cα(X) = |u|∞ + [u]α

This is a Banach Space.

Example 17. Let Ω⊂C, and consider for 1≤ p≤∞,

Ap(Ω) = f : Ω→C, holomorphic, |f |Lp<∞⊂Lp(Ω)

Then Ap(Ω) equipped with the Lp norm is a Banach space.

Since Ap(Ω) is a subset of Lp(Ω), to show completeness it suffices to show that the Lp limit of a sequenceof functions in Ap is also in Ap, e.g. that Ap is closed under the Lp metric.

Note that for p = ∞, completeness follows from the result that if fn is a sequence of analytic functionsand fn→ f uniformly on all compact subsets of Ω, then f is analytic.

For p<∞, first note that by the Cauchy Integral Formula, and for r chosen so that Br(z0)⊂Ω,

f(z0)=1

2π

∫

0

2π

f(z0 + reit)dt

Multiplying by r and integrating in r, we have that on BR(z0)⊂Ω,

1

2R2f(z0) =

1

2π

∫

BR(z0)

f(z) dA

or

f(z0)=1

|BR(z0)|

∫

BR(z0)

f(z)dA

Now note that

|fn(z0)− fm(z0)| ≤ 1

|BR(z0)|

∫

BR(z0)

|fn(z)− fm(z)|dA

and using Hölder’s inequality, we have that

|fn(z0)− fm(z0)| ≤ 1

|BR(z0)|1−1/p′‖fn− fm‖p

If we consider compact subsets K ⊂Ω, then there is a uniform constant for which

|fn(z0)− fm(z0)| ≤C ‖fn− f ‖p, z0∈K

(by compactness). Therefore, ‖fn − fm‖L∞(K) → 0 on compact K, so that we have upgraded the con-

vergece so that fn→ f uniformly on all compact subsets of K, which means that f is analytic (apply the-orem that says that fn is a normal family of functions, etc...). A detail missing here is that focusing onK, fn converges uniformly to some g, but then we show that fn→ g in Lp as well, using

∫

K

|fn− g |p≤‖fn− g‖L∞(K)p

µ(K)→ 0

and by uniqueness of limits g= f on K. This shows that fn→ f uniformly on compact sets.

9

Example 18. Hardy space 1≤ p≤∞. Define

Hp(B1)=

f :B1→Cholomorphic such that sup0<r<1

∫

0

2π

|f(reiθ)|p dθ <∞

with the norm

|f |Hp =

(

sup0<r<1

∫

0

2π

|f(reiθ)|p dθ)1/p

Then Hp with the norm is a Banach space.

Example 19. Let Ω⊂Rn and fix m∈N. Let 1≤ p≤∞ and define

Em=

u∈Cm(Ω) such that |∂αu|Lp(Ω)<∞ for all |α| ≤m

Then define

|u|Em=

(

∑

|α|≤m

|∂αu|Lp(Ω)p

)1/p

Then (Em, | · |Em) is not complete. Its completion will be denoted Wm,p(Ω) (Sobolev Space).

For instance, in the case m = 0 it is easy to find a sequence of continuous functions that converge in theLp norm to a discontinuous function.

Completion of a Normed Linear Space

Given a normed linear space X, a completion of X means a pair (ϕ, Y ) where Y is complete, ϕ:X→ Y is

a linear norm-preserving (and thus injective) map, and ϕ(X) =Y .

Example 20. If X = C([0, 1]) with the L1 norm, then we know that (C([0, 1]), L1) is not complete. The

completion is (f f , L1), identifying X as elements of L1, which is complete, and noting that the sub-space of continuous functions are dense in L1.

Proposition 21. Every normed linear space has a unique completion (Though for practical purposes, it iscrucial to have a concrete description)

Proof. The reason is that we can take Z to be the set of all Cauchy sequences in X , which forms avector space. Then define a seminorm |xkk=1

∞ |Z = limk→∞ |xk|, which is subadditive and positivehomogenous. Quotient out the subspace for which |xk|Z = 0, i.e. Z0 = (xk) ∈ Z: lim xk = 0, then Z/Z0

is a Banach space, and the map ϕ:X→Z/Z0 given by ϕ(x) = (x, x, x, )+Z0 gives the completion.

To show uniqueness, suppose (ϕ1, Y1) and (ϕ2, Y2) are two completions of X . Then we show that thereexists a norm-preserving isomorphism between Y1 and Y2. Note that the images ϕ1(X) and ϕ2(X) aredense in Y1, Y2 respectively, and ϕ1 is invertible as a map from X → ϕ1(X), a linear isomorphism. Thenwe define Φ: ϕ1(X)→ ϕ2(X) by Φ(ϕ1 x) = ϕ2x, noting that |ϕ1x|= |x|= |ϕ2x|, so Φ is a linear, norm-pre-serving isomorphism. We then extend by continuity. Given y ∈ Y1, we choose any sequence ϕ1xn ∈ ϕ1(X)converging to y, and define

Φ(y)= limn→∞

Φ(ϕ1xn)

10

where the limit exists because |Φ(ϕ1xn)−Φ(ϕ1xm)|= |ϕ1xn− ϕ1xm| → 0 so that Φ(ϕ1xn) is Cauchy. This

definition is well defined since if ϕ1xn′ → y then limn→∞ Φ(ϕ1xn− ϕ1xn

′ ) = 0 (the norm goes to 0 with thesame computation). Note |Φ(y)| = limn→∞ |Φ(ϕ1xn)| = limn→∞ |ϕ1xn| = |y | so that Φ is a linear normpreserving map from Y1 → Y2. The same construction can be used to construct a linear norm preservingmap Ψ from Y2→Y1. Then ΨΦy= y essentially by construction...

Separability

A linear metric space (X, d) is called separable if there exists a dense, countable subset, i.e. exists a setE with countably many points such that E=X . There is the following fact:

Proposition 22. (X, d) is separable if and only if X has a countable base, i.e. there exists a countablecollection of open sets B such that for all open sets U, U can be expressed as a union of sets in B.

Proof. If (X, d) is separable, then take balls with rational radii centered around each point in the densesubset. Conversely, given a countable base, take an arbitrary point in each set of the base and denote thecollection by E. Then, for any open ball, there exists an element of the base contained inside, which con-tains a point of E, and thus E is dense.

Properties:

• If (X, d) separable, and Y ⊂X , then (Y , d) is separable.

• If (X, d) separable, and⋃

α∈Λ Uα =X is an open cover, then there exists Λ0 ⊂ Λ, a countable set

of subindices such that⋃

α∈Λ0Uα=X.

To see this, let B be a countable base. Every Uα is a union (at most countable) of elements in B.Let B ′ consist of all the sets B ∈ B for which B ⊂Uα for some α∈Λ. Note that the union of all ele-ments in B ′ is also X. Then for each B ∈ B ′, choose any αB ∈ Λ for which B ⊂ UαB

. Then thedesired subcover is UαB

, B ∈B ′.

Examples of Separable Spaces

• (C[0, 1], | · |∞) is separable (polynomials with rational coefficients)

• (Lp(Ω), | · |Lp) is separable for Ω⊂Rn (step functions, continuous functions, etc)

• (L∞(Ω), | · |L∞) is not separable for Ω with infinitely many elements. (biject to the sequence of 0’sand 1’s)

A Variational Problem

Let X be a Banach space, and K be a convex, closed subset of X , and let y K. We wish to find x0 ∈Kthat minimizes |x − y | for x ∈K, i.e. |x0 − y | ≤ |x − y | for all x ∈K. Does such a point exist? And if so,how many points?

In Rn, the answer is simple. Let d = infx∈K |x − y |. By definition of inf, we can find a sequence xn ∈ K

such that |xn− y | → d. Noting that |xn| ≤ |xn− y |+ |y | ≤ d+ |y | so that the sequence is bounded, and bysequential compactness of bounded sequences in Rn, there exists a subsequence xn′ converging to somex0 ∈ K. Then |x0 − y | = d. Here we used the assumption that K is closed and the Heine-Borel property,that every bounded sequence has a convergent subsequence. Unfortunately this does not work in infinitedimensions.

11

Proposition 23. If X is a normed linear space, and every bounded sequence has a convergent subse-quence, then dimX <∞.

The proof follows quite directly from the following lemma:

Lemma 24. If Y is a proper, closed subspace of X, then for any ε> 0, there exists x∈X such that |x|= 1and infy∈Y |x− y | ≥ 1− ε.

Given this lemma, and the fact that every finite dimensional space is closed, we can construct a sequence

so that |xn|= 1 and infy∈spanx1, ,xn |xn+1 − y | ≥ 1

2, so that d(xj , xk)≥ 1

2. Note at the induction step we

apply the lemma to Y = spanx1, , xn and ε =1

2. Such a sequence cannot have a convergent subse-

quence, but is bounded.

Proof. (of Lemma) First we note that X /Y can be equipped with the norm

‖x+Y ‖= infy∈Y

|x− y |

for x+Y ∈X/Y . For any y1, y2∈Y , we have

infy1+y2∈Y

|x1 + x2− (y1 + y2)| ≤ |x1− y1|+ |x2− y2|

taking inf over y1 and y2 gives subadditivity. Suppose infy∈Y |x − y | = 0. Then there exists a sequence ynsuch that |x− yn|→ 0, and since Y is closed, we have that x∈ Y so that x+ Y = 0 + Y (the zero vector inX/Y ). (Thus, if Y is not closed, ‖ · ‖ is just a seminorm).

Also, if X is complete, so is X/Y . This is because if xn + Y is Cauchy, then ‖xn − xm + Y ‖ =infy∈Y |xn − xm − y | → 0, and can find ym ∈ Y such that |(xn − yn) − (xm − ym)| → 0, so that xn − yn isCauchy, and by completeness xn− yn→ z, and so xn+Y → z+Y .

Now there exists x0 +Y ∈X/Y such that ‖x0 +Y ‖= 1. For ε> 0, there exists y such that

1≤ |x0 + y |< 1+ ε

Then let x=x0 + y

|x0 + y |, so that |x|=1, and furthermore,

infy∈Y

|x− y |= ‖x+Y ‖=‖x‖

|x0 + y | ≥1

1+ ε

as desired.

Remark 25. As an aside, the fact that all finite dimensional subspaces are closed is a consequence of thecompleteness of Lp(1, , n, counting) and the fact that all norms are equivalent in finite dimensionalvector spaces. Let V =Rn and | · |V be a norm for V , and let | · |2 be the standard Euclidean norm.

1. First we show that | · |V is continuous on (Rn, | · |2). Note

|x|V =

∣

∣

∣

∣

∣

∑

k=1

n

xkek

∣

∣

∣

∣

∣

V

≤∑

k=1

n

|xk| |ek|V ≤(

∑

k=1

n

|ek|V)

|x|2 =C |x|2

Then∣

∣|x|V − |y |V∣

∣≤ |x− y |V ≤C |x− y |2 so that | · |V is continuous on (Rn, | · |2).

12

2. Then, let b=min|x|2=1 |x|V > 0, noting that x: |x|2 = 1 is a compact set, and thus | · |V maps thisset to a compact set in R. Since 0 is not in the image (|x|V = 0 only for x = 0, which has 2-norm0), it must be the case that b> 0. Then for any nonzero x, we have that

∣

∣

∣

∣

x

|x|2

∣

∣

∣

∣

V

≥ b

or that |x|V ≥ b|x|2

These two imply that b |x|2 ≤ |x|V ≤ C |x|2 for two constants b, C. This implies that the topologiesinduced by | · |V and | · |2 are the same.

Strict and Uniform Convexity

Back to the problem about minimizing |x − y | over x ∈ K where y K is fixed. K is convex and closedsubset of X . First we show that the problem does not always have a solution.

Example 26. (Minimum not achieved) Let X =C[0, 1] and |f |∞ =maxx∈[0,1] |f(x)| the ∞ norm, andlet

K =

f ∈C[0, 1],

∫

0

1

f(x)dx=1, f(0)= 0, f(1) =1

This is a convex, closed (uniform convergence on compact set allows swapping limits) set. Note that 0 K, and d= inff∈K |f − 0|∞ is not achieved.

First, we show that d= 1.

For any f ∈K, we note that

1 =

∫

0

1

f(x)dx≤ |f |∞

so that d ≥ 1. To show d ≤ 1 we consider specific f ∈ K. Consider the function fε constructed by modi-fying the constant function f(x) = 1 by forcing fε(0) = 0 and compensating for the loss in the integral byraising the maximum value of the function slightly to 1+ ε, i.e.

fε

0 1

1ε

0

This can be chosen so that∫

fε = 1. Then we have that d ≤ ‖fε‖∞ = 1 + ε and since ε is arbitrary, wehave that d≤ 1.

Now we show that d is not achieved. Suppose that f ∈ K and ‖f ‖∞ = 1, so that∫

0

1f(x)dx = 1. Then

∫

0

1(1− f(x))dx= 0, which means that 1− f(x)≡ 0 so f(x)= 1, contradicting f ∈K.

The issue here is compactness. We can take a sequence whose norms tend to 1, but this is a boundedsequence which may not have a convergent subsequence to some element in K (if it did, then the limit hasnorm 1)

13

Next we show that the solution may not even be unique.

Example 27. (Solution not unique) In R2, with |x|∞ = max (|x1| , |x2|) let K be the unit ball (whichis the square [−1, 1]2) and y= (2, 0). Then y K, the minimum distance d= infx∈K |x− y |= 1, but everypoint (1, t), |t| ≤ 1 achieves this distance.

The condition for the uniqueness of the solution is strict convexity.

X is strictly convex if for all x, y ∈X ,

|x|= 1, |y |=1,

∣

∣

∣

∣

x+ y

2

∣

∣

∣

∣

= 1 x= y

Proposition 28. If X is strictly convex, and x, y ∈X, x 0, y 0, and |x + y | = |x| + |y |, then x = λy

for some λ> 0.

Proof. Note we may write the above as

∣

∣

∣

∣

|x||x|+ |y |

x

|x| +|y |

|x|+ |y |y

|y |

∣

∣

∣

∣

= 1

Define f(t) =∣

∣

∣tx

|x|+ (1 − t)

y

|y|

∣

∣

∣for t ∈ [0, 1]. Note that by the strict convexity of X , if

x

|x| y

|y|, then f is

also strictly convex so that f(t)< 1 for t ∈ (0, 1) (f(0) = f(1) = 1). This is because the definition of strictconvexity tells us that two different points on a line cannot have the same norm since if the midpoint ofx, y has the same norm as x and y.

Given the claim, then the above shows that f(

|x|

|x|+ |y|

)

= 1 whereas|x|

|x|+ |y|∈ (0, 1). This implies that

x

|x|=

y

|y|and x=

|x|

|y |y as desired.

Proposition 29. (Uniqueness) Let X be a Banach space, K ⊂X be strictly convex and closed, and y K. Suppose d= dist(y,K)> 0, and x0 x1 satisfies |x0− y |= |x1− y |= d. Then x0 = x1.

Proof. Without loss of generality, we may assume y = 0 by translation invariance. Then we note that|x0|= |x1|= d. This implies that x0 is not a multiple of x1, so that by Proposition 28,

∣

∣

∣

∣

x0 + x1

2

∣

∣

∣

∣

<|x0|+ |x1|

2= d= dist(0,K)

But this is a contradiction since by convexityx0 +x1

2∈ K but its distance from y = 0 is smaller than d.

Thus we conclude that x0 = x1.

For existence, we need to talk about a still stronger notion of convexity. X is called uniformly convex iffor all ǫ> 0, there exists δ > 0 such that for all |x|= |y |= 1,

∣

∣

∣

∣

x+ y

2

∣

∣

∣

∣

> 1− δ |x− y |<ǫ

(i.e. if the midpoint of two unit vectors is close to a unit vector, then the two unit vectors are close)

Remark 30. Let (X, | · |) be strictly convex. If dimX <∞. then it is also uniformly convex.

14

Proof. Suppose X is finite dimensional but not uniformly convex. Then there exists ǫ > 0 such that for

all δ > 0, we can find unit vectors xn, yn such that∣

∣

∣

xn + yn

2

∣

∣

∣> 1 − δ but |xn − yn| > ǫ. Since X is finite

dimensional and we have a bounded sequence, we can find a subsequence xn′, yn′ where xn′, yn′, andxn′ + yn′

2all converge to x∞, y∞,

x∞ + y∞

2. But this implies that

∣

∣

∣

x∞ + y∞

2

∣

∣

∣= 1, and |x∞ − y∞| ≥ ǫ, which

means that X is not strictly convex.

Proposition 31. Let (X, | · |) be uniformly convex. Then it is strictly convex.

Proof. Let |x| = |y | = 1 such that∣

∣

∣

x+ y

2

∣

∣

∣= 1 Let ǫ > 0, and δ be chosen as in the definition of uniform

convexity. Then since∣

∣

∣

x+ y

2

∣

∣

∣= 1> 1 − δ, we have that |x − y |< ǫ. Since ǫ is arbitrary, we have that |x −

y |=0 and thus x= y.

This tell us that if a solution exists to the variational problem in the case that X is uniformly convex,then the solution is unique. Now we can show existence.

Theorem 32. Let X be a uniformly convex Banach space, and let K ⊂X be convex and closed. Then forall y ∈X, there exists a unique x0∈K such that

|y− x0|= infx∈K

|y−x|= dist(y,K)

Proof. First we note that if y ∈K, the theorem is trivial. Let y K. Then let

b= infx∈K

|y− x|> 0

By the definition of b, we can find a sequence xn ∈ K such that |xn − b| → b. We now show that thissequence is Cauchy, in which case we have found the vector of minimal distance to y.

Now we make the following observation: By uniform convexity, if we have two sequences xn, yn such that

|xn| → b, |yn| → b, and∣

∣

∣

xn + yn

2

∣

∣

∣→ b, then |xn− yn| → 0. This follows from looking at the unit vectors

xn

|xn|,

yn

|yn|and

∣

∣

∣

∣

∣

xn

|xn|+

yn

|yn|

2

∣

∣

∣

∣

∣

→ 1

and applying the definition of uniform convexity.

Note that

b≤∣

∣

∣

∣

xn+xm2

− y

∣

∣

∣

∣

≤ |xn− y |2

+|xm− y |

2→ b as n,m→∞

Thus∣

∣

∣

∣

xn− y

2+xm− y

2

∣

∣

∣

∣

→ b, |xn− y |→ b, |xm− y |→ b

so that by our observation

|(xn− y)− (xm− y)|→ 0

Therefore the sequence xn is Cauchy, converging to x∞ where

|xn− y |→ |x∞− y |= b= dist(y,K)

15

The uniqueness of the solution follows from Proposition 29 and the fact that uniform convexity impliesstrict convexity.

Remark 33. For a simpler proof using weak convergence with the assumption that X is reflexive, seeExample 71, in the section about lower semicontinuity.

Uniformly Convex Spaces

What sort of spaces are uniformly convex?

Theorem 34. Lp(X) is uniformly convex for 1< p<∞

It is easy to see (in 〈R〉2 say) that in the cases p= 1 and p= ∞ the space is not even strictly convex (theunit circles in the respective norms are composed of straight lines). The theorem follows easily fromClarkson inequality (not proved).

Theorem 35. (Clarkson Inequality) Let u, v ∈Lp(X). Then we have the following inequalities:

1. If 1≤ p< 2, then∣

∣

∣

∣

u+ v

2

∣

∣

∣

∣

Lp

q

+

∣

∣

∣

∣

u− v

2

∣

∣

∣

∣

Lp

q

≤(

|u|Lpp + |v |Lp

p

2

)q−1

2. If 2≤ p<∞, then

∣

∣

∣

∣

u+ v

2

∣

∣

∣

∣

Lp

p

+

∣

∣

∣

∣

u− v

2

∣

∣

∣

∣

Lp

p

≤(

|u|Lpq + |v |Lp

q

2

)p−1

In particular, for p=2, this becomes

|u+ v |L22 + |u− v |L2

2 ≤ 2(|u|L22 + |v |L2

2 )

Proof. For reference, see Adams, Sobolev Spaces .

The proof of uniform convexity from the Clarkson inequality is now easy. If |u|, |v |= 1, then the RHS is 1,

and thus if∣

∣

∣

u+ v

2

∣

∣

∣is near 1, then by the inequality

∣

∣

∣

u− v

2

∣

∣

∣is near 0.

Another example of uniformly convex spaces are Hilbert spaces.

Hilbert Spaces

A Hilbert space is a complete inner product space. There is the Cauchy Schwarz inequality

|〈u, v〉| ≤ |u| |v |

and the norm is |x|= 〈x, x〉√

.

Examples are L2 with the inner product 〈f , g〉=∫

fg.

16

Proposition 36. Every Hilbert space is uniformly convex.

Proof. From properties of inner product, we have the parallelogram equality:

|x+ y |2 + |x− y |2 = 2(|x|2 + |y |2)

If we now write this as∣

∣

∣

∣

x+ y

2

∣

∣

∣

∣

2

+

∣

∣

∣

∣

x− y

2

∣

∣

∣

∣

2

=|x|2 + |y |2

2

then we see that if |x|= |y |= 1, the RHS is 1, and thus if∣

∣

∣

x+ y

2

∣

∣

∣

2→ 1 then

∣

∣

∣

x− y

2

∣

∣

∣

2→ 0 and |x− y |→ 0.

Orthogonal Decomposition

Proposition 37. Let H be a Hilbert space over the real or complex scalar field, and let X ⊂H be a closedsubspace. Then H =X ⊕X⊥ where

X⊥= y ∈H : 〈x, y〉= 0

Furthermore, X⊥ is a closed subspace of H , by the continuity of the inner product.

Proof. First note that X ∩X⊥= 0 since if x∈X ∩X⊥ then 〈x, x〉= 0 so that x=0.

Then, we show that H = X + X⊥. Let z ∈ H . Since every Hilbert Space is uniformly convex, using The-orem 32, there exists a x0 ∈ X such that |x0 − z | ≤ |x − z | for all x ∈X . We show that x ⊥ z − x0 for all

x∈X so that z=x0 + (z − x0) where x0∈X and z − x0∈X⊥.

If we consider

ϕ(t)= |z − (x0 + tx)|2 = |z −x0|2− 2tRe 〈z − x0, x〉+ t2|x|2

then the minimum is achieved when t= 0, and so

0= ϕ′(0)=− 2Re〈z − x0, x〉

Thus Re〈z − x0, x〉= 0 for all x∈X . This implies that 〈z −x0, x〉= 0 if we consider both x, ix∈X .

Properties: Let H be a Hilbert space over real or complex scalar field.

• Let S be any subset of H . S⊥= (span(S))⊥ and S⊥ is closed.

Proof. Note that by linearity, if 〈x, z〉= 〈y, z〉= 0, then 〈αx+ βy, z〉 = 0, and thus if z ∈ S⊥ then

z ∈ (span(S))⊥, so S⊥⊂ (span(S))⊥. Since opposite inclusion is immediate, S⊥=(span(S))⊥.

If yn ∈ span(S) converges to some y in span(S), and 〈yn, z〉= 0, then by the continuity of the innerproduct 〈y, z〉= 0 also. This implies the result.

• Let X ⊂H be a closed subspace. Then (X⊥)⊥=X .

Proof. First, if x∈X , then for all y ∈X⊥, 〈x, y〉= 0. Thus x∈ (X⊥)⊥ and so X ⊂ (X⊥)⊥

17

Next, we use the decomposition H =X ⊕X⊥. If z ∈ (X⊥)⊥ then write z = x+ y where x ∈X and

y ∈X⊥. Then

0= 〈z, y〉= 〈x, y〉+ |y |2 = |y |2

so that y= 0 and z ∈X. Thus (X⊥)⊥⊂X and we have the result.

• Let X ⊂H be any subspace. Then (X⊥)⊥=X .

Proof. Use (X⊥)⊥= (X⊥)⊥=X.

Riesz Representation Theorem

Denote the dual space of a vector space X by X ′ =L(H,K) where K is the scalar field. Note that for anyy ∈H we can define a linear functional in the dual space by ly(x) = 〈x, y〉. This is linear, and by Cauchy-Schwarz, |ly(x)| ≤ |x| |y | so that the operator norm is |ly | = |y |. It turns out that all linear functionalsarise in this manner.

Theorem 38. Let H be a real or complex Hilbert space, and let l be a linear functional in H ′. Then thereexists a unique vector y ∈H such that l(x)= 〈x, y〉.

Proof. Sketch: Consider Nl⊥, which is one dimensional since Nl

⊥ @ H/Nl @ K. Then take the unit vectorin Nl

⊥.

Application: This can be used to provide a proof of the Radon-Nikodym theorem and the LebesgueDecomposition.

Theorem 39. Let µ, ν be two finite measures on X. Then ν can be decomposed as

ν = νac + νsing

where νac ≪ µ and νsing ⊥ µ, i.e. there exists a nonnegative integrable function g such that

νac(A)=

∫

A

g dµ

and a set N ⊂X such that

µ(N) = νsing(X\N)= 0

Also, the decomposition is unique.

Proof. Let H =L2(X, µ+ ν). Then define l(f)=∫

Xf dν, and note that

|l(f)| ≤∫

X

|f |dν ≤∫

X

d(µ+ ν)≤ |f |L2(X,µ+ν) (µ+ ν)(X)√

so that l∈H ′. By the Riesz Representation theorem, there exists h∈H such that

∫

X

fdν =

∫

X

fhd(µ+ ν)

18

The claim is that 0≤h≤ 1 for a.e. µ+ ν. To see this, let E = h≥ 1. Then note that using f = 1E,

∫

E

d(µ+ ν)≥∫

E

dν=

∫

E

hd(µ+ ν)

and so∫

E

(h− 1)d(µ+ ν)≤ 0

Since h> 1 on E, we see that (µ+ ν)(E) =0. Likewise, for F = h< 0, we have

0≤∫

F

dν=

∫

F

hd(µ+ ν)

and since h < 0 on F we have that (µ + ν)(F ) = 0. Now assume without loss of generality that 0 ≤ h ≤ 1everywhere (modifying h on a set of measure zero does not affect the integral).

The intuition here (formal computation) is that “dν = hdµ + hdν so that dν =h

1−hdµ”, which is okay if

h< 1. Thus, let us take the set N = h= 1.

Note that if f = 1N , ν(N) = µ(N)+ ν(N) so that µ(N)= 0. Then we set

νsing(A) = ν(A∩N)

νac(A) = ν(A\N)

Now we show that νac(A) =∫

Agdµ where

g(x) =

h(x)

1− h(x)x∈X\N

0 x∈N

(since µ(N) = 0 the value of g on N does not matter, and h< 1 for x∈X\N). From the definition of νsingand the usual approximation steps (simple to nonnegative measurable functions) we note that

∫

f dνac =

∫

Nc

fdν

so that for any nonnegative meausrable f ≥ 0,

∫

fdνac(A) =

∫

A\N

fdν

=

∫

A\N

fhd(µ+ ν)

=

∫

A

fhdµ+

∫

A\N

fh dνac +

∫

A\N

fhdνsing

=

∫

A

fhdµ+

∫

A

fh dνac

noting that νac(N)= 0 and νsing(A\N) = ν(N ∩A\N)= ν(∅)= 0. Then we have that as in the intuition,

∫

A

f (1−h)dνac =

∫

A

fhdµ

19

Now take f =1

1− h1Nc, which is nonnegative and measurable. Then

νac(A)=

∫

A

dνac =

∫

A

gdµ

as desired, and in particular g is µ-integrable.

For uniqueness, suppose we have two decompositions

ν= νac + νsing = νac′ + νsing

′

where νsing is supported on X\N and νsing′ on X\N ′, νac, νac

′ ≪ µ. If we then consider N0 = N ∪ N ′, wehave that

νsing(A)= νsing(A∩N0) = ν(A∩N0)− νac(A∩N0)= ν(A∩N0)

noting that µ(N0) ≤ µ(N) + µ(N ′) = 0 so that νac(A ∩N0) =∫

A∩N0gdµ ≤

∫

N0gdµ= 0. The same holds

for νsing′ , νac

′ so that νsing = νsing′ and νac = νac

′ .

Fun List of Applications:

• Radon-Nikodym and Lebesgue Decomposition

• Dirichlet Forms

Adjoints

Let A∈L(H1, H2) where H1, H2 are Hilbert spaces.

Fixing y ∈H2, x 〈Ax, y〉H2is a bounded linear functional on H1 since |〈Ax, y〉| ≤ |A| |x|H1

|y |H2. Thus

by Riesz representation there exists some element A∗y ∈H2 such that

〈Ax, y〉= 〈x,A∗y〉

for all x ∈H1. This allows us to define a bounded linear operator A∗ from H2 →H1 which maps y A∗y.(Linearity is straightforward to prove)

Properties:

• |A∗|= |A|Proof:

|A∗|= sup|y|≤1

|A∗y |= sup|x|≤1|y|≤1

|〈x,A∗y〉|= sup|x|≤1|y|≤1

|〈Ax, y〉|= sup|x|≤1

|Ax|= |A|

• A∗∗=A

Proof:

〈Ax, y〉= 〈x,A∗y〉= 〈A∗∗x, y〉

for all x, y, and thus A=A∗∗.

• (AB)∗=B∗A∗

20

Proof:

〈x,ABy〉= 〈A∗x,By〉= 〈B∗A∗x, y〉

for all x, y, thus by definition of adjoint, (AB)∗=B∗A∗.

• RA⊥=NA∗, RA=NA∗

⊥

Proof:

y ∈RA⊥ 〈y,Ax〉=0 for allx∈H1 〈A∗y, x〉= 0 for allx∈H1 A∗y=0 y ∈NA∗

Thus RA⊥=NA∗. Taking the ⊥ of both sides gives the second statement.

Example 40. (Integral Operators) Let K ∈ L2(µ × ν) where (X, µ) and (Y , ν) are measure spaces,

and define A:L2(µ)→L2(ν) by

(Af)(y)=

∫

X

K(x, y)f(x) dµ(x)

K is called the kernel of the integral operator.

Then A is a bounded linear operator, |A| ≤ |K |L2(µ×ν) and A∗: L2(ν) → L2(µ) is also an integral operator

with kernel K∗(y, x)=K(x, y).

(easy computation)

Orthonormal Bases

Let H be a Hilbert space. Then (eα)α∈A is called an orthonormal system if 〈eα, eβ〉= δαβ for α, β ∈Λ.

It is an orthonormal basis if in addition eα, α∈Λ⊥=0, i.e. spaneα=H .

Examples:

• L2(S1), the exponentials ek= e2πikθ is an orthonormal basis.

• Let B1 = z ∈C: |z |< 1, A2(B1)= f holomorphic onB1, ‖f ‖2<∞ with inner product

〈f , g〉=

∫

B1

f g dµ

Then ckzk is an orthonormal base (ck an appropriate normalization constant)

Proposition 41. Every Hilbert space has at least one orthonormal basis.

Proof. Use Hausdorff Maximality Principle (Zorn’s lemma) on the poset of orthonormal systems with theinclusion ordering (ek ≤ fk ek ⊂ fk). Clearly every linearly ordered set has an upper bound

(take the union), and the maximal element B is a basis since otherwise B⊥ 0 so we can add another ele-ment to B preserving orthonormality, and this contradicts the maximality of B.

21

Proposition 42. If H is separable, then every orthonormal set is countable.

Proof. Given an orthonormal system eα, we note that for α β, |eα − eβ | = 2√

, and so taking a suffi-

ciently small ball around each eα gives mutually disjoint balls Bα, α ∈ Λ. Then given a countable densesubset xn, each Bα necessarily contains at least some point in xn. Thus we have a correspondencebetween Bα and a subset of xn, and since xn is countable, there are countably many balls, and thuseα is countable.

Proposition 43. Let H be a separable Hilbert space, and ek and orthonormal basis. Then

1. For all x∈H, x=∑

k 〈x, ek〉ek

2. (Bessel) |x|2 =∑

k|〈x, ek〉|2

Proof. To prove (1), consider x − y =∑

k=1n 〈x, ek〉ek for some y, then taking the norm squared and

taking limits gives

∑

k

|〈x, ek〉|2≤ |x|2

(Bessel’s inequality). This shows that∑

k 〈x, ek〉ek is absolutely summable, and since H is separable, thismeans that it is summable, i.e. converges to some limit z. Note

〈x− z, ej〉= 〈x, ej〉−∑

k

〈x, ek〉〈ek, ej〉= 〈x, ej〉− 〈x, ej〉= 0

Since this is true for all ej, x− z = 0 so x= z. For (2), apply the norm squared to partial sums in (1), andtake the limit.

Dual Spaces

Given a normed linear space X , the dual space X ′ =L(X,K) is the space of linear functionals, and underthe operator norm |l |= sup|x|≤1 |l(x)| the dual space (X ′, | · |) is a Banach space.

Proof. Let ln be a Cauchy sequence in (X ′, | · |). |ln − lm| → 0. Then for a given x ∈X, |ln(x) − lm(x)| ≤|ln− lm| |x|→ 0, so that ln(x) is Cauchy. This allows us to define a linear functional l by limits,

l(x)= limn→∞

ln(x)

which is linear by the linearity of limits. Then,|ln(x)− l(x)|

|x|→ 0 for all x 0, so that |ln− l | → 0, so that ln

converges to l.

Note that X is not required to be complete.

Recall Definitions: Let X be a locally compact, Hausdorff space.

• τ = U open,B= smallest σ-algebra containing τ .

• Borel measure is a measure µ on B such that µ(K)<∞ for K compact.

22

• Regular Borel measure is a Borel measure such that

µ(E)= infU openE⊂U

µ(U)= supK compactK⊂E

µ(K)

i.e. measure can be approximated from above by open sets and below by compact sets.

Examples:

1. In Hilbert Spaces, H ′@ H by Riesz Representation.

2. For 1 ≤ p <∞ and (X, µ) a σ-finite measure space, Lp(µ)′ @ Lp′

(µ), where1

p+

1

p′= 1. The isomor-

phism is given by θ:Lp′→ (Lp)′ defined by

θg (f)=

∫

fg dµ

This mapping is norm-preserving by Hölder’s inequality and duality principle, and is onto by the

Radon-Nikodym theorem (need to show that the resulting function is in Lp′

.

3. Let X be compact and Hausdorff, then C(X,C), the space of continuous functions from X→C isa Banach space under the supremum norm, and

C(X,C)′@ µ: µ regular complexBorelmeasure onX

with norm ‖µ‖ = |µ|(X), where |µ| is the total variation measure of µ. Here the isomorphism isgiven by

θµ(ϕ) =

∫

X

ϕdµ

4. (Generalizes (3)). Let X be locally compact (every point has a compact neighborhood) and T2.Then

C0(X,C)= ϕ:X→Cbounded continuous function, ∀ε> 0, ∃K ⊂X compact, |ϕ| ≤ ε onX\K

is a Banach space under the supremum norm, and as above,

C0(X,C)′@ µ: µ regular complexBorelmeasure onX

Remark 44. We will show (4) using the subspace Cc, which is the space of continuous functions withcompact support. This is a normed linear space under the supremum norm, but it is not a Banach space.The dual space is more complicated here...

Also, if X is locally compact metric space, then every Borel measure is regular. Regularity is importantbecause this implies that Cc(X)⊂L1(X) is dense.

We will make use of the following theorem:

Theorem 45. Let X be a locally compact Hausdorff space, and let l: Cc(X, R) → R be a positive linearfunctional, i.e. l(ϕ)≥ 0 for ϕ≥ 0. Then there exists a unique regular Borel measure µ such that

l(ϕ)=

∫

X

ϕdµ

23

Also, if l:Cc(X,C)→C is a linear functional such that for all ϕ∈Cc(X,R), ϕ≥ 0,

supψ∈Cc(X,C)

|ψ |≤ϕ

|l(ψ)|<∞

then there exists a unique regular Borel measure µ and a measurable function θ: X → C with |θ | = 1 suchthat

l(ϕ)=

∫

X

ϕθdµ

Using this theorem, we see that if l ∈C0(X)′, then l is also a linear functional on Cc satisfying the bound-edness property of the theorem above since given ϕ≥ 0, |ψ | ≤ ϕ with ϕ, ψ ∈Cc, we have that

|l(ψ)| ≤ |l | |ψ |∞≤ |l | |ϕ|∞

and thus there exists θ, µ with |θ |= 1 such that

l(ϕ)=

∫

X

ϕθdµ

and it remains to justify that µ is finite so that θdµ is a complex measure. Since X is locally compact,Hausdorff, µ is regular, and so taking a seqeuence of compact sets Kn that increase to X, we have that

µ(X)= limn

∫

1Kndµ≤ sup

ψ∈Cc(X,C)|ψ |≤ϕ

|l(ψ)|<∞

Existence of nontrivial elements in X′

Proposition 46. Given a normed linear space X with Y ⊂X. If l0∈ Y ′, then there exists an extension toX ′ which preserves norm, i.e. l∈X ′ such that l|Y = l0 and |l |X ′ = |l0|Y ′.

Proof. Since |l0(y)| ≤ |l0| |y | for y ∈ Y , let p(x) = |l0| |x|, subadditive positive homogeneous function. Thensince |l0| ≤ p, we apply Hahn Banach to obtain an extension to l ∈ X ′ with |l | ≤ p and l |Y = l0. Then|l(x)| ≤ |l0| |x| shows that |l| ≤ |l0| and since l= l0 on Y , |l0| ≤ |l |.

Corollary 47. X normed linear space, Y ⊂X with dim Y <∞. Then there exists Z closed such that X =Y ⊕Z. (Useful for Fredholm Theory)

Proof. Let (e1, , en) be a basis for Y . Then we can find linear functionals l1, , ln ∈ X ′ such thatli(ej) = δij, noting that li(ej) = δij defines linear functionals on Y ′ (defined for all of Y by linearity, and iscontinuous since all norms on finite dimensional spaces are equivalent). Then we can extend to X ′ usingthe Proposition above. Let Z =

⋂

j=1n

Nlj, the intersection of the null spaces. Then we can express anyx∈X as

x=∑

j

lj(x)ej∈Y

+

(

x−∑

j

lj(x)ej

)∈Z

so that X =Y +Z. Also, if x∈ Y ∩Z, then x=∑

jcjej, but lk(x)= ck=0 so that x= 0.

24

Z is closed since the null space of a continuous linear functional is closed, and Z is the intersection of nullspaces, and hence is also closed. (null space is inverse image of 0, which is closed)

Duality Principle

Proposition 48. (Duality Principle) For y ∈X,

|y |= sup|l|≤1l∈X ′

|l(y)|

and furthermore there is an l∈X ′, |l| ≤ 1 that achieves the supremum, i.e. |l(y)|= |y |

Proof. Note that |l(y)| ≤ |y | when |l | ≤ 1 so that |y | ≥ RHS. To get the other inequality, we note we candefine a linear functional on spany = Cy by l0(λy) = λ |y |. Then |l0| ≤ 1, and we can extend to l ∈ X ′

such that |l |= |l0| ≤ 1, and |l(y)|= |y |. Thus |y |= |l(y)| ≤ sup|l|≤1 |l(y)|.

Corollary 49. Let X be a normed linear space, Y ⊂X a closed subspace. Then if x0 Y, then there existsl∈X ′ such that l(x0) 0. More specifically, we can find l so that |l | ≤ 1 and |l(x0)|= d(x0, Y ).

Proof. Use X/Y , and π: X → X/Y be the injection π(x) = x + Y ∈ X/Y . Then x0 + Y 0, and by theproposition we have an l ′∈ (X/Y )′ such that |l′| ≤ 1, |l ′(x0 +Y )|= |x0 +Y |, and we can define l∈X ′ by

l(x0) = l ′(x0 +Y )

where |l(x0)|= |l′(x0 +Y )|= |x0 +Y |= d(x0, Y ). Furthermore, since l= l ′ π, we have that

|l(x)|= |l′(π(x))| ≤ |l ′| |π | |x|

so that |l | ≤ 1.

Application of Duality to Density Problems

Let X be a normed linear space, and let S ⊂ X be a subset. We would like to know whether spanS isdense in X . Then we have the following fact:

Proposition 50. Suppose that l ∈ X ′, l|S = 0 = 0, e.g. that the only linear functionals on X that

vanish on S is the zero functional, then spanS=X.

Proof. Suppose note, then Y = span S X. Pick x0 ∈ X\Y (in X but not Y ), then using Corollary 49there exists a linear functional l∈X ′ such that l(x0) 0 but l |S= 0, which is a contradiction.

Application: Runge Approximation Theorem

Recall that given f : B1 → C holomorphic, there exists a sequence of polynomials converging uniformly tof on B1/2 (Taylor).

On the annulus A1,2 = 1 < |z | < 2, given f : A1,2 →C, there may not exist a sequence of polynomials con-verging uniformly to f on A1+ε,2−ε, but there does exist a sequence of rational functions converging uni-

formly to f on A1+ε,2−ε (Laurent).

There is a further generalization to nbd(K) for K compact.

25

Theorem 51. (Runge) Let K compact, f : K → C holomorphic, and if E ⊂ C ∪ ∞ such that everycomponent of C ∪ ∞\K contains at least one element of E, then there exists rational functions Rj withpoles in E such that Rj→ f uniformly on K.

From this, Taylor is with K =B1/2, E = ∞ and Laurent is with K =A1+ε,2−ε and E= 0,∞.

Proof. We need to show that for all l ∈ C(nbd(K))′, l(R) = 0, where R is the space of rational functionswith poles in E, that l=0. Let ϕ be a holomorphic function from nbd(K)→C.

Using a contour Γ ⊂ nbd(K)\K which has winding number 0 on C\nbd(K) and 1 on K, we have that byCauchy’s integral formula,

f(z)=1

2πi

∫

Γ

f(ξ)

ξ − zdξ, z ∈K

and by linearity (and dominated convergence),

l(f(z)) =1

2πi

∫

Γ

f(ξ)l

(

1

ξ − z

)

dξ

Thus it suffices to show that ϕ(ξ) = l(

1

ξ − z

∣

∣

∣

z∈K

)

= 0 for ξ ∈ Γ. Note ϕ is holomorphic on C\K, and for

ξ ∈E, ξ ∞ we note that

ϕ(n)(ξ) = l

(

(− 1)nn!

(ξ − z)n+1

)

= 0, ξ ∈E

But this implies that ϕ≡ 0 on the bounded components of C\K containing ξ, and in particular for ξ ∈ Γ,as desired.

Note if ∞ E, we are done. Otherwise, we need to consider the unbounded component of C\K and show

that ϕ= 0 on this component. Note that1

ξ− z=

1

ξ· 1

1− z/ξ=

1

ξ

∑

j

(

z

ξ

)j

for z ∈K and |ξ | sufficiently large.

Since (z/ξ)j has a pole at ∞, l(

1

ξ − z

)

= 0 for ξ sufficiently large. Note that we can modify the contour Γ

so that the part in the unbounded component of C\K is just |ξ |=R for R large enough.

Dual Variational Problem

Let X be a normed linear space over R and A a subset. Define

SA(l)= supx∈A

l(x)

Properties:

• If A⊂B then SA≤SB

Proof: This property is immediate: supx∈A l(x) ≤ supx∈B l(x) since the supremum on the RHS istaken over a larger set.

• SA+B=SA+SB.

Proof. This is also immediate:

supx∈A,y∈B

l(x+ y)= supx∈A

l(x)+ supy∈B

l(y)

26

since l(x + y) = l(x) + l(y), and taking the supremum over A first, then the supremum over Bsecond gives the equality.

• SA=Sco(A) =Sco(A)

Proof: The first equality holds since l(λa + (1 − λ)b) = λl(a) + (1 − λ)b ≤ max (l(a), l(b)). Thesecond holds by continuity of l.

• If A,B are closed, convex, then SA≥SB A⊃B

Proof: The first property shows ( ) direction. Suppose the other direction does not hold. Thenthere exists x0 ∈B\A and r > 0 such that Br(x0) ∩A= ∅. Then there exists l ∈X ′ such that l 0and l(x)≤ l(y) for x ∈A, y ∈Br(x0) (Hahn Banach separation theorem). Now choose z so l(z) = 1,and then for ε sufficiently small,

l(x)≤ l(x0− εz)= l(x0)− ε

Taking the supremum over x shows that SA(l) ≤ l(x0) − ε < SB(l), which is a contradiction (note,the strict inequality is because this works for any ε sufficiently small).

• Let C be any set. Then x0∈ co(C) l(x0)≤SC(l)

Proof:

x0∈ co(C) l(x0)=Sx0(l)≤Sco(C)(l)=SC(l)

Theorem 52. Let K be a closed, convex set, x0 K. Then

d(x0,K)= infy∈K

|x0− y |= supl∈X ′

|l|≤1

(l(x0)−SK(l))

Proof. Note that for |l | ≤ 1, applying the third property above shows that

l(x0)−SK(l)≤ l(x0)− l(y)≤ |x0− y |

and taking the infimum over y ∈K and the supremum over l∈X ′, |l | ≤ 1 shows that

infy∈K

|x0− y | ≥ supl∈X ′

|l|≤1

(l(x0)−SK(l))

To get the opposite inequality, let d= d(x0, K), and let r < d be arbitrary. Consider an r-neighborhood ofK, i.e. K + Br (Br = Br(0)). By the hyperplane separation theorem (Theorem 3) on K + Br and x0, wecan find a functional l where l(x) ≤ l(x0) for x ∈ K + Br, and without loss of generality we may choose|l| ≤ 1 by scaling. This means that

SK(l)+SBr(l)=SK+Br

(l)≤ l(x0)

Then because SBr(l)= sup|x|≤r l(x)= r |l |= r, and rearranging, we have that

r=SBr(l)≤ l(x0)−SK(l)≤ sup

|l|≤1

(l(x0)−SK(l))

27

and since r is arbitrarily close to d(x0,K), we have the result.

This is in Lax, Functional Analysis , Theorem 8.18

Remark 53. if K is a closed subspace, then from Corollary 49 we can find l ∈ X ′ with |l | ≤ 1, wherel(y) =0 for y ∈K and l(x0) = d(x0,K), and thus SK(l) =0 and

infy∈K

|x0− y |= d(x0,K) = l(x0)−SK(l)≤ supl∈X ′

|l|≤1

(l(x0)−SK(l))

But this completely ignores the role of SK.

Remark 54. This is the picture in R2. The term sup (l(x0) − SK(l)) probes “directions”. Given a direc-tion vector v of unit length, we study the lines formed by v · x = t. We increase t until we hit the lastpoint of K. Then we subtract this from v · x0. This gives us a rough estimate (lower bound) of how far x0

is from K. We then choose the direction v that maximizes this quantity. See picture.

v

v · x= 1 v ·x= t

0x0

v ·x= v ·x0

In the picture above, l(x0)− SK(l) is v · x0− t. If we rotate v counterclockwise, we will get the correct dis-

tance. Using a nonconvex K in R2 we can find an example where the theorem is false.

Weak Convergence

Recall that for infinite dimensional normed linear spaces, closed and bounded sets are not necessarilysequentially compact (Heine-Borel property not satisfied) in the topology induced by the norm. However,we still hope for some type of convergence using a different topology.

Definition 55. X normed linear space, xj ∈X. We say xj x weakly if

l(xj)→ l(x) for all l∈X ′

(Sometimes will write “xj→x weakly” also...)

We will discuss the topology associated with this notion of convergence later.

Remark 56. If xj → x weakly and xj → y weakly, then x = y (i.e. the topology of weak convergence isHausdorff). The reason is because then l(x− y) = 0 for all l, and by duality principle (Proposition 48) x−y= 0.

28

Example 57. For 1 ≤ p <∞, eikx→ 0 weakly in Lp(0, 1) by the Riemann-Lebesgue lemma, which saysthat

∫

f(x)eikx→ 0

for f ∈Lq, 1≤ q ≤∞. Note (Lp)′@ Lq for 1≤ p<∞. The dual of L∞ is not L1.

Remark 58. Note that if xn→x in norm, i.e. |xn− x|→ 0, then xn→x weakly. This follows from

|l(xn)− l(x)| ≤ |l | |xn− x|→ 0

Even with this weaker notion of convergence, it not always true that bounded sequences have convergentsubsequences. The condition for this is that the space be reflexive.

Reflexive Spaces

First, we note a natural map from X → X ′′; that is, given x ∈ X, there is a natural mapping κ: X → X ′′

which maps x to a linear functional κx: X′ → K given by κx(l) = l(x). Note that ‖κx‖ = sup|l|≤1 |l(x)| =

|x|, so that κ is an isometric embedding of X→X ′′.

A Banach space X is said to be reflexive if κ(X)=X ′′ (i.e. that κ is onto).

Facts About Reflexive Spaces:

1. For 1< p<∞, Lp(X, µ) is reflexive.

This follows since (Lp)′′@ (Lp′

)′@ Lp.2. L1(Ω),Ω⊂Rn is not reflexive, and neither is C[0, 1].

These use the following fact:

Proposition 59. If X ′ is separable, then so is X (in their respective norm topologies)

Proof. Let li, i = 1, 2, be a countable dense subset of X ′ with li 0. Then for all i, there

exists xi ∈ X such that |xi| = 1 and |l(xi)| ≥ 1

2|li|. The claim is that spanxi =X , which we show

using Proposition 50. If not, then there exists l ∈ X ′ nonzero such that l(xi) = 0 for all i. By den-sity of li, given ε> 0 there exists li such that |li− l | ≤ ε. Then

1

2|li| ≤ |li(xi)|= |li(xi)− l(xi)| ≤ ε

and thus

|l | ≤ |l− li|+ |li| ≤ ε+ 2ε= 3ε

so that l=0, a contradiction.

Now given the Proposition, the fact that L1(Ω) is not reflexive follows easily, since the dual ofL1(Ω) is isometrically ismomorphic to L∞(Ω), which is not separable (Ω infinite). Now if L1 were

reflexive, then the double dual would also be L1, and since L1 is separable, the previous propositionimplies L∞ is separable, a contradiction.

The fact that C[0, 1] is not reflexive follows from the fact that C[0, 1] ⊂ L1, so that the dual (C[0,1])′ ⊃ (L1)′ = L∞, and since L∞ is not separable, neither is the dual of C[0, 1], and thus C[0, 1] isnot reflexive, by the same reasoning as for L1.

29

3. Closed subspaces of reflexive spaces are reflexive.

Let X be reflexive, and Y ⊂ X is a closed subspace. We only need to show that κ: y κy mapsonto Y ′′. Let ϕ ∈ Y ′′. Then given l ∈X ′, we consider ϕ(l|Y ) = l(x) for some x. Note that for any l

such that l∣

∣

Y= 0, we have that l(x) = ϕ

(

l |Y)

= 0. Since Y is closed, this implies that x ∈ Y (apply

Proposition 50 to Y ⊂Y +Kx)

Now for any l0∈Y ′, we can extend arbitrarily to some l∈X ′ so that l |Y = l0. Then

ϕ(l0) = ϕ(l |Y )= l(x)= l0(x)

where the last equality follows since x∈Y . Thus κ(x)= ϕ for x∈Y , and so κ is onto.

4. Hilbert spaces are reflexive.

In Hilbert spaces, by Riesz representation we have that H @ H ′ @ H ′′ (isometric isomorphism fol-lows from duality principle (Proposition 48))

5. Uniformly convex spaces are reflexive. This is proved in the section about Double Polar Theorem(Corollary 117).

Theorem 60. Let X be a Banach space. Then X is reflexive if and only if every bounded sequence has aweakly convergent subsequence.

Proof. First suppose that X is reflexive. Let xi be a bounded sequence in X. Let Y = spanxi. This isa separable subspace of X , and since Y is a closed subspace, Y is also reflexive. Then Y ′′ = κ(Y ) is sepa-rable and thus Y ′ is separable. This means that there exists lj , j = 1, 2, which is dense in Y ′. Now foreach j, lj(xi), i = 1, 2, is a bounded sequence and therefore has a convergent subsequence. By usingdiagonal procedure, we can find a single convergent subsequence xi′ for which lj(xi′) converges for all j.

Define ϕ:Y ′→K by ϕ(lj)= limi′ lj(xi′), and we can extend to Y ′ by continuity (lj is dense in Y ′).

Then ϕ ∈ Y ′′, and thus for some x ∈X, κ(x) = ϕ, i.e. ϕ(l) = l(x) for all l ∈ Y ′. Then since l(xi′)→ ϕ(l) =l(x) for all l, xi′→ x weakly.

The converse is true, but relies on more machinery.

Application to Galerkin’s Method

Galerkin’s method is about finding solutions to PDEs by approximation in finite dimensional subspaces.

Let Ai(x) be an n× n matrix for each x ∈Rn and 1 ≤ i≤ n, and B(x) be an n× n matrix for each x. Foru:Rn→R, define L as the operator

Lu=∑

i=1

n

Ai ∂iu+Bu

Let Ai, B be smooth and using the standard basis in Rn, e1, , en, suppose Ai(x+ ej) =Ai(x) and B(x+ej)=B(x) (i.e. periodic).

Given f :Rn→R periodic, we wish to solve Lu= f for some periodic u.

Define

H = u∈Lloc2 (Rn):u(x+ ej) =u(x)∀j

which is a Hilbert space with the inner product 〈u, v〉=∫

Quv dx with Q the unit cube in Rn.

30

Proposition 61. If AiT = Ai and there exists λ > 0 such that B + BT −∑

i∂iAi ≥ λI, then for every f ∈

H, there exists u∈H such that

〈u, L∗v〉= 〈f , v〉

for every v ∈C1∩H. (this condition says that u is a weak solution to Lu= f).

Notational Warning: For convenience we will start writing Ai∂i in place of∑

iAi∂i. It is taken to be

understood that we are summing over i.

Proof. A simple computation using the definition 〈u, L∗v〉= 〈Lu, v〉 and integration by parts shows that

L∗v=(−Ai∂i− ∂iAi+BT)v

Since Lu=Ai∂iu+Bu, we have that

L∗u=−Lu+ (B+BT − ∂iAi)u

Then

〈Lu, u〉 = −〈u, Lu〉+⟨

u, (B+BT − ∂iAi)u⟩

〈Lu, u〉 =1

2

⟨

u, (B+BT − ∂iAi)u⟩

≥ 1

2〈u, λu〉

=λ

2|u|L2(Q)

2

where the inequality follows from assumption. Thus the assumption tells us that L is a positive (semi)def-inite operator.

Now we cosnider a subspace X ⊂H ∩C1, dimX <∞. Let πX:H→X be the orthogonal projection to X.Instead of solving Lu= f directly, we solve πXLu=πXf , or πX(Lu− f)= 0.

First we note that the mapping u πXLu (u∈X) is injective, since if πXLu= 0, then Lu∈X⊥ and

0 = 〈Lu, u〉≥ λ

2|u|L2(Q)

2

for u ∈ X, and thus u = 0. Since X is finite dimensional, this implies that u πxLu is bijective. Thismeans that there exists a unique uX ∈X such that πXLuX =πXf .

For this uX, we estimate |uX |L2(Q). Note that

λ

2|uX |L2(Q)

2 ≤〈LuX , uX 〉= 〈πXLuX , uX 〉= 〈πXf , uX〉= 〈f , uX 〉≤ |f |L2(Q)|uX |L2(Q)

so that |uX |L2(Q)≤ 2

λ|f |L2(Q), and estimate independent of X. Here we have used the fact that orthogonal

projections whose ranges are closed are self adjoint, so that 〈πXLuX , uX 〉= 〈LuX , πXuX 〉= 〈LuX , uX〉.Then we take X larger and larger, closer to C1∩H and then take a weak limit.

Let (ϕi)i=1∞ be an orthonormal basis for H , with ϕi smooth (for instance, take trigonometric polynomials),

and let Xm = spanϕ1, , ϕm. Then πmLum = πmf with |um|L2(Q) ≤ 2

λ|f |L2(Q). um is then a bounded

sequence of functions in H , and therefore there exists a subseqeuence um′ that converges to some u

weakly, by the reflexivity of H and Theorem 60. Now we show that the weak limit u is the desired solu-tion.

31

Note for v ∈Xm0 for m0≤m,

〈um, L∗v〉= 〈Lum, v〉= 〈πmLum, v〉= 〈πmf , v〉= 〈f , v〉

and taking the limit as m→ ∞, this shows that 〈u, L∗v〉 = 〈f , v〉 for v ∈ Xm0. Thus this is true for v ∈⋃

k=1∞

Xk. Now for any v ∈C1∩H , we can find vk ∈⋃

kXk with vk→ v and Dvk→Dv uniformly (Weier-

strass type theorem) Then

〈u, L∗v〉= limk

〈u, L∗vk〉= limk

〈f , vk〉= 〈f , v〉

as desired. It turns out that u∈C1 as well... (not proved)

Uniform Boundedness Principle

First we start with a topological theorem. Let (X, d) be a complete metric space.

Theorem 62. (Baire Category Theorem) The intersection of a collection of dense open subsets isdense in X.

The reason for the name is because of an immediate corollary and the definition of “first category” sets.First, we say that a set E is nowhere dense if

(

E)c

is dense in X . We say that E is of first category if it

is the countable union of nowhere dense sets. All other sets are of second category.

Corollary 63. Let (X, d) is a complete metric space. Then X is of second category.

Consequently, if Fk is a sequence of closed sets with empty interior, then the union also has empty inte-rior.

And if Fk is a sequence of closed sets whose union is X, then at least one Fk has nonempty interior.

Proof. Take complements in the Baire Category Theorem. Note a closed set is nowhere dense if and onlyif it does not contain an open ball.

Proposition 64. (Uniform Boundedness Principle) Let X, Y be Banach spaces, and let Aα for α∈Λbe a family of bounded linear operators from X→ Y for which

supα∈Λ

|Aα(x)|<∞

for all x. Then supα∈Λ ‖Aα‖= supα∈Λ,|x|≤1 |Aα(x)|<∞ (i.e. the bound is uniform)

Proof. Take Fk= x ∈X: |Aα x| ≤ k for all α. Then⋃

kFk=X by assumption. Since Fk is closed, one

of them has nonempty interior, so there exists Br(0)⊂Fk. (each Fk is symmetric and convex)

Now for any |x|< 1, we note that rx∈Br(0), and thus

|Aα x|= 1

r|Aα(rx)| ≤ k

r

for all α, and taking the supremum over |x|< 1, we conclude that ‖Aα‖≤ k

r, as desired.

Applications:

32

Proposition 65. If fi→ f strongly in Lp and gi→ g weakly in Lp′

, then fi gi→ f g weakly in L1.

Proof. We will show that l(fi gi)→ l(f g) for l∈ (L1)′=L∞. Given ϕ∈L∞, we have that

∣

∣

∣

∣

∫

fi giϕ−∫

f gϕ

∣

∣

∣

∣

≤∫

|fi− f | |gi| |ϕ|+∣

∣

∣

∣

∫

f giϕ−∫

f ϕ g

∣

∣

∣

∣

≤ |fi− f |p |gi|p′ |ϕ|∞+

∣

∣

∣

∣

∫

fϕ gi−∫

f ϕ g

∣

∣

∣

∣

≤ C |fi− f |p+

∣

∣

∣

∣

∫

f ϕ gi−∫

f ϕ g

∣

∣

∣

∣

→ 0

where we note that |gi|p′ is uniformly bounded by the uniform boundedness principle.

Remark 66. There exists fi∈L2[0, 1] such that fi→ f weakly in L2, fi2→ g weakly in L1 but g f2.

Try fi = sign(sin(2πx)) (square wave oscillating between 1 and −1). Then fi → 0 weakly, justifying firstfor continuous functions and then approximating arbitrary L2 functions by continuous functions. Also,fi

2→ 1 strongly (and hence weakly), but g f2.

Proposition 67. Suppose xj→x weakly. Then supj |xj |<∞, i.e. the sequence is bounded.

Proof. Recall the mapping κ: X → X ′′ given by x κx where κx(l) = l(x), which is an isometric map-ping, so that |x|X = |κx|X ′′. Then κxj ∈X ′′ and given l ∈X ′,

supj

|κxj(l)|= supj

|l(xj)|<∞

since l(xj) converges. Thus the conditions in Proposition 64 are satisfied, and

supj

|xj |X = supj

|κxj |X ′′<∞

as desired.

Lower Semicontinuity under Weak Convergence

Remark 68. If K is closed with respect to the norm, and xj ∈ K, xj → x weakly. It is not necessarily

true that x ∈K. A quick example is for instance K = ej , j ∈N ⊂ l2. Then ej → 0 weakly since for any(xk)k=1

∞ ∈ l2,

limj→∞

〈ej , (xk)k=1∞ 〉= lim

j→∞xj= 0

but 0 K. This says that the weak-closure of a set is not necessarily the same as the norm-closure.

However, if we add the condition that the set is convex, then the property does hold. That is, the weak-closure of a convex set is the same as the norm-closure.

Proposition 69. Let K is convex and norm-closed, and suppose that xj ∈ K and xj → x weakly. Thenx∈K.

33

Proof. Suppose that x K. Then there exists r > 0 such that Br(x) ∩ K = ∅. Then there exists l ∈ X ′

such that l 0 and Re l |K ≤ Re l |Br(x)

. Then using y such that Re l(y) = 1 (pick arbitrarily and scale

appropriately), we have that

Re l|K ≤Re l(x− εy)=Re l(x)− ε

for ε < r. However, this implies that Re l(xj) ≤Re l(x)− ε, and taking limits, Re l(x) ≤Re l(x) − ε, a con-tradiction.

Corollary 70. If f :X→R convex, continuous, and xj→x weakly, then

f(x)≤ liminfj

f(xj)

Proof. Suppose that c = liminfj→∞ f(xj). Consider the set K = x: f(x) ≤ c + ε. This is a (norm)-closed, convex set by assumption. By definition of liminf, we can find a subsequence xj ′ such thatf(xj ′)< c+ ε. Since xj ′ → x weakly and K is convex and closed, Proposition 69 shows that x∈K so that

f(x)≤ c+ ε. Since ε is arbitrary, we have proved the result.

Example 71. Let X be reflexive, K ⊂ X be closed, convex, and nonempty. Suppose y K. Then thereexists x0∈K such that |y− x0|= dist(y,K).

Proof. Recall dist(y, K) = infx∈K |y − x|. By definition of dist, there exists xn such that |y − xn| →dist(y, K). Then since |xn| ≤ |y − xn| + |y |, we note that xn is a bounded sequence in a reflexive space.Thus by Theorem 60 there exists a subsequene xj ′ that converges weakly to some x0. Since K is closedand convex, x0 ∈ K. Then by Corollary 70, and the fact that the mapping x |y − x| is continous andconvex,

|y−x0| ≤ liminfj

|y− xj ′|= dist(y,K)

and since dist(y,K)≤ |y− x0| by definition of dist, so that |y− x0|= dist(y,K).

Upgrading to Norm Convergence

Recall that a sequence that converges in norm also converges weakly, and that the converse is not true.However, with an additional condition, it is true.

Proposition 72. If xj→ x weakly, and |xj |→ |x|, and X is uniformly convex, then xj→x in norm.

In particular, this is true for reflexive spaces such as Lp for 1< p<∞.

Proof. If x=0, then there is nothing to prove.

Suppose that x 0. Then consider uj=xj

|xj |and u=

x

|x|. Note uj→ u weakly since

l(uj) =l(xj)

|xj |→ l(x)

|x| = l(u)

Furthermore,uj +uk

2→ u weakly as j , k→∞. By lower semicontinuity of the norm,

1= |u| ≤ liminfj,k

∣

∣

∣

∣

uj+ uk2

∣

∣

∣

∣

≤ limsupj,k

∣

∣

∣

∣

uj+ uk2

∣

∣

∣

∣

≤ 1

34

This implies that∣

∣

∣

uj +uk

2

∣

∣

∣→ 1 and by uniform convexity, |uj − uk| → 0. Therefore uj is Cauchy in norm

and uj→u in norm (recall the weak limit is unique, so uj must converge to u). This means that

|xj− x|= ||xj | uj − |x|u| ≤ |xj | |uj − u|+ ||xj | − |x|| |u| → 0

Remark 73. To see what can happen in L1[0, 1] say, which is not uniformly convex (not even strictlyconvex), consider ϕn= sign(sin(2πnx)) (square waves). Then since ϕn→ 0 weakly, 1 + ϕn→ 1 weakly. But|1 + ϕn|1 = 1 since ϕn is 1,− 1 each on a set of measure 1/2, thus 1 + ϕn is 0, 1 each on a set of mea-sure 1/2. Thus 1 + ϕn→ 1 weakly, |1 + ϕn|1 → |1|1, but |1 + ϕn− 1|1 = |ϕn|1→ 0, so there is no norm con-vergence.

(condition for L1... possibly involving the additional assumption that fn→ f in measure for all finite sets.Recall, norm convergence in L1 implies that fn → f in measure on finite subsets. The converse is true ifwe add weak convergence.)

Weak* Convergence

Let X be a Banach space, and lj ∈X ′. We say that lj converges to l in the weak* sense if lj(x)→ l(x) forall x∈X .

Theorem 74. If X is separable, then every bounded sequence lj ∈ X ′, |lj | ≤ c has a subsequence lj ′ thatconverges in the weak* sense.

Example 75. Let X be a compact metric space, and (X, µ) with µ a Borel measure. Then any functionin L1(X, µ) can be considered a functional in C(X)′ by considering f lf = fdµ. In other words, given a

continuous function ϕ ∈C(X), lf(ϕ) =∫

ϕdµ. Then, given a sequence fj ∈ L1 that is bounded |fj |L1 ≤ c,there exists some measure ν on X such that fj dµ→ ν in the weak* sense (or a subsequence does), i.e.

∫

X

ϕ fj dµ→∫

X

ϕ dν

for all ϕ∈C(X).

Remark 76. Proposition 64 (Uniform Boundedness Principle) implies that if lj → l in the weak* sense,then supj |lj |<∞. In the hypotheses of the proposition, Y =K, and for x∈X , supj lj(x)<∞.

Example 77. Quadratures for Integral Evaluation . Suppose we want to find a suitable rule for approxi-

mating∫

0

1f(x)dx for continuous f using a family of weighted sums

∑

k=1n

f(xj(n)

)wj(n)

. Note that a

necessary condition for

limn→∞

∑

k=1

n

f(xj(n))wj

(n) =

∫

0

1

f(x)dx

for all f is the uniform boundedness of the functionals ln(f) =∑

k=1n

f(xj(n))wj

(n), since weak*-ly conver-

gent sequences are bounded. Defining l(f) =∫

0

1f(x)dx, we are saying that if ln→ l in the weak* sense,

then ln is a bounded sequence.

Remark 78. Note that on X ′ we can also talk about weak convergence. That is, if ln ∈ X ′, l ∈ X ′, thenln→ l weakly if for all ϕ∈X ′′, ϕ(ln)→ ϕ(l).

35

Because κ: X → X ′′ is an isometric embedding, if ln → l weakly, then ln → l in the weak* sense. To seethis, for any x∈X, we have that for κx (defined by κx(l) = l(x)), κx(ln)→κx(l), and thus ln(x)→ l(x).

The converse is not true in general, but it holds when X is reflexive. This is because κ(X)=X ′′, so that ifln → l in the weak* sense, then given ϕ ∈ X ′′, we can find x ∈ X so that ϕ(l) = l(x) for all l, and thusϕ(ln)→ ϕ(l) for all ϕ, and ln→ l weakly.

Weak and Weak* Convergence in L1

Example 79. Consider C[0, 1]′, which is the space of regular Borel measures on [0, 1], where

µ(f)=

∫

f dµ

and |µ|C[0,1]′ = (µ+ + µ−)([0, 1])= |µ|([0, 1]) (|µ| is the total variation measure).

Note that L1[0, 1] can be isometrically embedded into C[0, 1]′ by identifying f with fdm where m is theLebesgue measure. Then consider fn=n1[0,1/n]∈L1[0, 1]. Then for every ϕ∈C[0, 1], we have that

∫

fnϕ→ ϕ(0)=

∫

ϕdδ0

where δ0 is the dirac delta measure δ0(A) =

1 0∈A0 0 A . Thus fn dm→ δ0 in the weak* sense. However, fn

does not converge to any f ∈L1 in the (L1)-weak sense (recall (L1)′@ L∞).

Proof. Suppose towards a contradiction that fn does converge weakly to some function f ∈ L1. Using1[0,t]∈L∞, weak convergence of fn shows that

1 = limn→∞

∫

0

t

fn=

∫

0

t

f

Then letting F (t) =∫

0

tf , we have that F (t) = 1, and because f is integrable, by the Fundamental The-

orem of Calculus F ′= f . But F ′= 0 (since F is constant), and thus f = 0, which contradicts∫

0

tf =1.

Moreover, fn dm does not converge to δ0 in the (C[0, 1]′)-weak sense either!

Proof. Denote M =C[0, 1]′, the space of Borel measures. Note that M ′⊂L∞ since M ⊃L1. Also, for anymeasurable set B, we can define a linear functional lB on M ′ by lB(µ) = µ(B). Note lB is linear andbounded since |lB(µ)|= |µ(B)| ≤ |µ|([0, 1])≤ 1 for |µ| ≤ 1. Thus, using (0, t], we have that

l(0,t](fn dm) =

∫

0

t

fn dm→ 1

But

l(0,t](δ0)= 0

Thus fn dm does not converge to δ0 weakly. (Note that if fn dm converges weakly to some ν, then it con-verges to ν in the weak* sense, and since the weak* limit is unique ν = δ0. Thus, it cannot converge tosome other ν δ0 weakly)

36

This situation also provides an example of a bounded sequence in L1 with no weakly-convergent subse-quence.

When L1 functions have a weak limit in L1

A natural followup question is then, when does a sequence fi ∈ L1 have a weakly-convergent subsequencein L1? We already know that boundedness is not enough to have a weakly-convergent subsequences. Itturns out it is enough to add the property that the functions are uniformly integrable.

Proposition 80. Let (X, µ) be a finite measure space where L1(X) is separable. Let fj ∈ L1(X) be abounded sequence, i.e. sup |fj |L1<∞, and furthermore, suppose that fj satisfy

• (Uniform Integrability) For any ε> 0, there exists δ > 0 such that if µ(A)<δ, then

∫

A

|fj |dµ≤ ε

for all j.

Then there exists a subsequence fj ′ converging weakly to some f ∈L1.

Proof. The idea is to show that for measurable A,∫

Afj ′ dµ → ν(A) for some measure ν and subse-

quence fj ′ and then that ν≪ µ.

Step 1. First we find a countable collection of sets Ai for which we can approximate an arbitrary measur-able set A; in other words, given ε > 0, we can find Ai such that µ(A∆Ai)< ε. To do this, consider the setS = 1A:Ameasurable ⊂ L1(X). This is a closed subset, since given a sequence 1Aj

for which 1Aj→ f in

L1,

µ(|f |>ε∩ |f − 1|>ε)≤ µ(|f − 1Aj|>ε)≤ 1

ε

∫

|f − 1Aj|→ 0

so that µ(f 0, 1) = 0 , and f = 1A for some A. Thus S is closed, and since L1 is assumed to be sepa-rable, S is also separable, and there exists a countable collection 1Ai

, i ∈ N ⊂ S which is dense in S.This means given an arbitrary measurable A, and ε> 0, there exists Ai such that

µ(Ai∆A)=

∫

|1Ai− 1A|dµ<ε

Step 2. The next step is to extract a subsequence of fj for which∫

Aifj ′ dµ converges for all Ai. This is

accomplished via a diagonalization process. Since∫

A1fj dµ is a bounded sequence,

supj

∣

∣

∣

∣

∫

A1

fj dµ

∣

∣

∣

∣

≤ supj

|fj |L1<∞

there is a subsequence f1,j for which∫

A1f1,j dµ converges. Likewise, there is a subsequence f2,j of f1,j

for which∫

A2f2,j dµ converges. The key is that since f2,j is a subsequence of f1,j, we also have that

∫

A1f2,j dµ converges. Continuing onwards, we get nested subsequences f1,j ⊃ f2,j ⊃ and if we

take the sequence (f1,1, f2,2, f3,3, f4,4, ) this gives the desired sequence for which we can define

ν(Ai)= limj→∞

∫

Ai

fj,j dµ

37

Step 3. The third step is to show that∫

Afj,j dµ converges to some ν(A) using density of Ai and the

uniform integrability of fj,j. To show this, given ε > 0, let δ be as in the uniform integrability property inthe hypotheses of the theorem. Choose Ai such that µ(A∆Ai) < δ, and let N > 0 such that for j , k > N ,∣

∣

∣

∫

Aifj dµ−

∫

Aifk dµ

∣

∣

∣<ε (note

∫

Aifj dµ converges, and thus is Cauchy). Then for j , k >N ,

∣

∣

∣

∣

∫

A

fj dµ−∫

A

fk dµ

∣

∣

∣

∣

≤∣

∣

∣

∣

∫

A

fj dµ−∫

Ai

fj dµ

∣

∣

∣

∣

+

∣

∣

∣

∣

∫

Ai

fj dµ−∫

Ai

fk dµ

∣

∣

∣

∣

+

∣

∣

∣

∣

∫

Ai

fk dµ−∫

A

fk dµ

∣

∣

∣

∣

≤∫

A∆Ai

|fj |dµ+

∣

∣

∣

∣

∫

Ai

fj dµ−∫

Ai

fk dµ

∣

∣

∣

∣

+

∫

A∆Ai

|fk|dµ

≤ 3ε

Thus∫

Afj dµ is a Cauchy sequence, and we set the value of ν(A) to be the limit.

Step 4. The fourth step is to show that ν is a (signed) measure with ν ≪ µ. Finite additivity followsfrom linearity of limits and integrals. For countable additivity, note that ν(X) < ∞, and given a nestedsequence of sets Aj ⊃ Aj+1 converging to the empty set ∅, we see that |ν(Aj)| → 0 by uniform integra-

bility of fj. Also, if µ(A) = 0, then ν(A) = limj

∫

Afj dµ = 0, so that ν ≪ µ. Thus by Radon-Nikodym,

there exists f ∈L1 for which ν = f dµ.

Step 5. Finally, we have found f ∈L1 for which

∫

A

fj dµ→∫

A

f dµ

for all measurable sets A. By linearity we have that∫

Afj ϕ dµ→

∫

Af ϕ dµ for simple functions ϕ, and

by density we conclude that this holds for all ϕ∈L∞, and this shows that fj→ f weakly, as desired.

The converse is also true, that the weak convergence of fj in L1 implies the uniform integrability of fj.This follows as a corollary to the following theorem:

Theorem 81. (Vitali-Hahn-Saks) Assume (X, µ) is a finite measure space, and νi is a sequence ofsigned measures on X with νi ≪ µ and νi converges setwise to some set function ν, i.e. νi(A) → ν(A) forall measurable A. Then:

1. For all ε > 0, there exists δ > 0 such that if µ(A) < δ, then |νi|(A) < ε for all i. (this is absolutecontinuity, except uniform in i)

2. ν is also a signed measure

Proof. Let Y = A⊂X,Ameasurable under the metric d(A,B) = |1A− 1B |L1 = µ(A∆B). Then (Y , d) isa complete metric space (we proved that Y is a closed subspace of L1 in Proposition 80). With this

metric, note that νi:Y →C is continuous. This is because νi≪ µ so that νi= fi dµ for some fi∈L1, fi≥ 0.Then

|νi(A)− νi(B)|= νi(A∆B) =

∫

A∆B

fi dµ

and by absolute continuity of the integral, if d(A, B) = µ(A∆B) is small, we can make the RHS small aswell. In particular, if B = ∅, in the context of property (1) above, given ε we can find a δi that works forindividual i, and also for finitely many i (by taking the minimal δi). It remains to study what happens asi→∞.

38

Now let ε> 0. Define

Fn= A∈Y : |νk(A)− νn(A)| ≤ ε, k ≥n

Fn is a nested sequence of closed subspaces increasing to Y . By the corollary to Baire’s Category The-orem (Corollary 63), one of the Fn has non-empty interior. So we can find Br(A0) ⊂ Fn for some n, A0, r.This means that for any set B ∈Br(A0), |νk(B)− νn(B)| ≤ ε.

Then note that for any set A with µ(A) < r, we can write A = (A ∪ A0)\(A0\A) where A ∪ A0 and A0\Aare both in Br(A0):

d(A∪A0, A0)= µ((A∪A0)∆A0)= µ(A\A0)≤ µ(A)<r

d(A0\A,A0)= µ((A0\A)∆A0)= µ(A∩A0)≤ µ(A)<r

Thus for k ≥n, we have that

|νk(A)− νn(A)| = |νk(A∪A0)− νk(A0\A)+ νn(A∪A0)− νn(A0\A)|≤ |νk(A∪A0)− νn(A∪A0)|+ |νk(A0\A)− νn(A0\A)|≤ 2ε

To finish the result, by the absolute continuity of νn, we can find a δn for which µ(A) < δn νn(A)< ε.Then for k≥n, this same δn shows that for µ(A)<δn,

|νk(A)| ≤ |νk(A)− νn(A)|+ |νn(A)| ≤ 3ε

and so we have found a δn in property (1) that works for all k ≥ n. Combining this with the observationthat we can already find δ1, , δn−1 that works for 1, , n− 1, choosing δ=min (δ1, , δn) proves property(1).

Showing that ν is a measure is straightforward, and is proved in the same manner as in step 4 in theproof of Proposition 80.

Corollary 82. Let (X, µ) be a finite measure space. Let fi be a sequence of L1 functions converging tof ∈ L1 weakly. Then fi is uniformly integrable, i.e. for all ε > 0 there exists δ > 0 such that for all A withµ(A)<δ,

∫

A|fi|dµ<ε.

Proof. Note if fi→ f weakly, then using ϕ= 1A gives∫

Afi dµ→

∫

Afdµ. Since fi dµ is a sequence of

signed measures converging to a set function f dµ (which is already a measure), property (1) of Theorem81 gives the result. Note |fi dµ|(A) =

∫

A|fi| dµ.

Upgrading Weak Convergence to Norm Convergence

Recall Proposition 72, about when weak convergence implies convergence in norm. For L1, we have thefollowing result:

Theorem 83. Let (X, µ) be a finite measure space. Suppose fn, f ∈L1(X, µ) and fn→ f weakly. Then ifin addition, fn→ f in measure, then fn→ f in norm.

Proof. The idea for this particular proof is to use the following real analysis lemma (there is probably abetter way):

39

Lemma 84. Let (X, µ) be a finite measure space. Suppose fn, f ∈L1(X) and fn→ f a.e. Then

|fn− f |L1→ 0 |fn|L1→|f |L1

Proof. ( ) This direction is obvious by triangle inequality∣

∣|fn|L1− |f |L1

∣

∣≤ |fn− f |L1

( ) This direction is by generalized Dominated Convergence Theorem. Note that we just need to jus-tify the limit swap in

limn

∫

|fn− f |=∫

limn

|fn− f |= 0

Note that |fn − f | ≤ |fn| + |f | and∫

|fn| + |f | →∫

2|f |, i.e. |fn − f | is dominated by an integrable

sequence gn for which∫

gn→∫

limn gn. In addition, fn→ f a.e., and we have satisfied all conditions inthe generalized Dominated Convergence Theorem (e.g. see Royden). Thus the limit swap is valid, and wehave proved the result.

Note that although we need fn→ f a.e. to use the Lemma, if fn→ f in measure, there is a subsequencefn′ for which fn′ → f a.e. Thus, the goal is to show that for this subsequence, |fn′|L1 → |f |L1. For the restof the prove, we will use fn to denote the subsequence that converges a.e.

Let ε > 0, and let En= |fn− f |>ε. Then convergence in measure means µ(En)→ 0. Also, since fn→ f

weakly, the corollary to Vitali-Hahn-Saks (Corollary 82) implies that fn is uniformly integrable. That is,we can find δ so that µ(A)<δ⇒

∫

A|fn|<ε.

Weak convergence (lower semicontinuity of norm) also implies that |f |L1 ≤ liminfn |fn|L1. Now for n suffi-ciently large, µ(En)<δ, and in this case

∫

|fn| =

∫

En

|fn|+∫

Enc

|fn|

≤ ε+

∫

Enc

|f |+ ε

≤∫

|f |+ 2ε

where in the second line, we have used uniform integrability and the fact that |fn| ≤ |f |+ |fn− f | ≤ |f |+ε on En

c . Now taking the limsup, we have that

limsupn

|fn|L1≤ |f |L1 + 2ε

and since ε is arbitrary, we have that

|f |L1≤ liminfn

|fn|L1≤ limsupn

|fn|L1≤ |f |L1

or in other words, |fn|L1→|f |L1. Then by the Lemma, this implies that |fn− f |L1→ 0, as desired.

What we have shown is that given the hypotheses of this Theorem, we can find a subsequence for whichfn′ → f in norm. A technical point is that this implies that the original sequence fn→ f in norm. This isdue to the fact that if every subsequence of fn has a further subseqeuence that converges to f in norm,then fn converges to f in norm.

Remark 85. The proof above can be extended to σ-finite measure spaces, but changing fn→ f in mea-sure to fn|A→ f |A in measure (with respect to µ|A) for all finite measure subsets A ⊂X. Then it can be

proved in the usual manner by splitting X into a countable union of (disjoint) finite measure sets andapplying the above results to each individual piece.

40

Corollary 86. In l1, weak convergence is the same as norm convergence.

Proof. Recall l1 = L1(N, ν) with ν being the counting measure, so that convergence in measure (on finitesubsets) is the same as weak convergence. Then apply the previous theorem.

Sketch of Alternate Proof: (do not need all the machinery above). It suffices to show that if xn → 0weakly, then |xn|l1 → 0. Assume towards a contradiction that |xn|l1→ 0. Then since xn converges weakly,|xn|l1 is uniformly bounded, and thus there exists a subsequence for which |xn|l1 ≈ C > 0 for all n.Another consequence of weak convergence is that xn→ 0 pointwise (use functionals ϕ= ek). The idea is tofind a subsequence xkn

for which the essential supports are disjoint (where the l1 norm is concentrated).Here is the construction:

Let xk1 be arbitrary, and since xk1 ∈ l1 there exists some set A1 = [0, n1] outside of which xk1 has very

small l1 norm.

Now since xn→ 0 pointwise, we can find xk2 which almost vanishes in A1 (a finite set). Then there existssome set A2 = [n1 + 1, n2] outside of which xk2 has small l1 norm.

Continuing in this manner, we have extracted a subsequence xknfor which xkn

is essentially supported onAn, where An are disjoint. Now let ϕ(i)i=0

∞ be the sequence defined by ϕ(i) = sgn(xkn) when i ∈ An

(matches the sign of ϕknwhere ϕkn

is essentially supported). Then∑

ixkn

(i)ϕ(i) ≈ |xkn|l1 ≈C > 0 for all

n, which contradicts weak convergence (which says that∑

ixkn

(i)ϕ(i)→ 0 as n→∞).

Summary of Results: Let (X, µ) be a finite measure space.

• For 1< p<∞, bounded sequences in Lp have weakly convergent subsequences.

• For p= 1,

Let X be compact, Hausdorff, and µ be a regular Borel measure. Then if we consider L1 asa subspace of C(X, C)′ with the (norm-preserving) identification f fdµ, then a bounded

sequence in L1 has a subsequence that converges in the weak* sense to some measure ν ∈C(X,C)′.

A bounded sequence in L1 has a weakly convergent subsequence if and only if the sequenceis uniformly integrable.

Suppose a sequence in L1 converges weakly. Then it also converges in norm if and only if itconverges in measure.

Open Mapping Theorem

Theorem 87. Let X, Y be Banach spaces, A ∈L(X, Y ) such that AX = Y, then A is an open map (mapsopen sets to open sets)

Proof. To show A is an open map, we only need to show that A(B1) ⊃ Bδ for some δ > 0. The reason isthat for arbitrary open sets U , given y ∈A(U), there exists some x such that Ax= y. Since U is open, wecan find x+Br⊂U . But then we can find y+Bδ ′⊂A(Br(x))⊂A(U) so that A(U) is open.

41

Now note Y =⋃

k=1∞

A(Bk), and by the corollary to Baire Category Theorem (Corollary 63), one of the

A(Bk) has nonempty interior, so that Br(y)⊂A(Bk) for some r, k > 0, y ∈A(Bk)

• First show we can find Br(0) ⊂ A(B1) for some different r. (Since A(Bk) is convex, symmetric andcontains 0, we should expect this to be true). Note that

Br/2(0)⊂Br/2(y)−Br/2(y)⊂Br(y)⊂A(Bk)

and since A(Bk)= kA(B1), we note that Br/2k(0)⊂A(B1).

• Now we show that Br(0)⊂A(B2) given above. Let |x|< r. We want to find |z |< 2 such that A z =x.

Since Br(0)⊂A(B1), we can find |z0|< 1 with |x−Az0|< r

2.

Then x−Az0∈Br/2, and since Br/2(0)⊂A(B1/2), we can find |z1|< 1

2with

|x−Az0−Az1|< r

4

Continuing in this manner, we have a sequence |zk|< 1

2kwith

∣

∣

∣

∣

∣

x−∑

j=0

k

Azk

∣

∣

∣

∣

∣

<r

2k

Since∑

k|zk|<∞,

∑

kzk is Cauchy and converges to some z, and then

|x−Az |= limk

∣

∣

∣

∣

∣

x−∑

j=0

k

Azk

∣

∣

∣

∣

∣

= 0

with |z |<∑k|zk|= 2 and Az= x.

We conclude that Br/2(0)⊂A(B1) as desired.

This has many important consequences.

Example 88. If X, Y Banach, A∈L(X, Y ) is bijective, then A−1 is also continuous, and there exists c >0 for which

c−1|x| ≤ |Ax| ≤ c|x|

Since A is an open map, the preimage of A−1 of any open set is open, and thus A−1 is continuous.

Theorem 89. (Closed Graph Theorem) Let X, Y be Banach spaces, and let A: X → Y be a linearmap. Let the graph of A, GA be defined by

GA= (x,Ax), x∈X⊂X ×Y

where X ×Y is also a Banach space. Then if GA is closed, then A is bounded.

42

Remark 90. Note that if A is bounded, then GA is closed, since (xj , A xj) → (x, y) implies that xj→ x

and Axj→ y. But since A is bounded, and hence continuous, Axj→Ax so Ax= y.

Proof. Consider the projection operator π: GA→ X defined by π(x, A x) = x. π is continuous and sinceGA is closed π is a bijection between two Banach spaces GA and X . Thus π is an open map, and we canfind c such that

c−1|(x,Ax)|X×Y ≤ |π(x,Ax)|= |x|

and

c−1(|x|+ |Ax|)≤ |x|

so that

|Ax| ≤ (c+1)|x|

and thus A is bounded.

Example 91. Let H be a Hilbert space over R, and let A: H → H be a linear operator such that thereexists B:H→H linear with 〈Ax, y〉= 〈x,B y〉, then A is bounded.

To prove, we show that GA is closed. Let (xk, A xk) → (x, y). We then want to show that y = A x. Notethat xk→x and Axk→ y, and then for any z ∈H , we have

〈Axk, z〉= 〈xk, B z〉

we take limits to obtain

〈y, z〉= 〈x,B z〉= 〈Ax, z〉

for all z. Then 〈y − A x, z〉 = 0 and taking z = y − A x shows that y = A x. This implies that A isbounded.

Example 92. Let X be a Banach space. Let X =X1 ⊕X2 with X1, X2 closed. Take π1:X→X1 and π2:X→X2. Then π1, π2 are continuous. That is, if we decompose x= x1 + x2 with xi∈Xi, then

|x1| ≤ c |x| and |x2| ≤ c |x|

for some constant c.

Note if we define a new norm on X by ‖x‖4 |π1 x|+ |π2 x|, then π1 and π2 are continuous with respect to‖ · ‖, since ‖π1x‖+ ‖π2x‖= |x1|+ |x2|= ‖x‖, so individually ‖πix‖≤ ‖x‖. Then it suffices to show that theidentity map from (X, ‖ · ‖)→ (X, | · |) is a homeomorphism.

Note that (X, ‖ · ‖) is complete. Suppose that x(j) is Cauchy in (X, ‖ · ‖) so that ‖x(j) − x(k)‖→ 0. Then

individually, |xi(j) − xi(k)| → 0 for i ∈ 1, 2. Then by completeness of X1, X2 as closed subspaces of (X, | ·

|), there exists xi for which |xi(j) − xi| → 0. But then ‖x(j) − (x1 + x2)‖→ 0, so that x(j) converges to x1 +x2 in ‖ · ‖.Then the identity map between two complete spaces is a homeomorphism by the open mapping theorem,and thus there exists c for which c−1‖x‖≤ |x|. This means that

|xi| ≤ |x1|+ |x2| ≤ c |x|

43

for i∈ 1, 2, as desired.

Example 93. Consider l∞⊃ c0 = xk: xk→ 0 as k→∞. The claim is that c0 has no closed complement.Thus it is not necessarily true that a closed subspace has a closed complement.

Suppose towards a contradiction that l∞ = c0 ⊕ Z for some Z closed. Then by the previous example, theprojection π: l∞→ c0 must be continuous.

First we claim that for ϕi∈ (l∞)′, if ϕi→ 0 weak*-ly, then

limi→∞

∑

k=1

∞

|ϕi(ek)|= 0

where ek(i)=

1 i= k

0 i k (standard basis).

Given the claim for now, we define ψi: c0 → C by ψi(x) = xi, and note that ψi→ 0 weak*-ly, since givenany x ∈ c0, ψi(x) = xi→ 0 by definition of c0. Note |ψi(x)| = |xi| ≤ |x|l∞ so that ψi are bounded, and thusin (c0)

′. Then if we define ϕi = ψi π, we have ϕi: l∞ → C and ϕi→ 0 weak*-ly as well. Also, as ϕi is a

composition of two bounded operators, ϕi∈ (l∞)′.

However,∑

k=1∞

ϕi(ek) =1 since ϕi(ek)= ψi(ek)= δik, which contradicts the claim.

To prove the claim, we need to prove that ai = (ϕi(ek))k=1∞ is a sequence in l1 which converges to 0

strongly. We will use the result that in l1, weak convergence is the same as norm convergence (Corollary86), and show that ai→ 0 weakly in l1. First of all, each ai∈ l1 by applying ϕi to bk= sgn ai(k)∈ l∞

ϕi(bk)=∑

k=1

∞

sgn ai(k)ϕi(ek)=∑

k=1

∞

|ai(k)|

and ϕi(bk)<∞ since bk ∈ l∞. The infinite sum above is valid by continuity of ϕi. Now given an arbitraryx(k)∈ l∞ we show that

∑

k=1∞

ai(k)x(k)→ 0 as i→∞. Note

ϕi(x)=∑

k=1

∞

ϕi(ek)x(k)=∑

k=1

∞

ai(k)x(k)

and since ϕi→ 0 weak*-ly, ϕi(x) → 0 as i→ ∞, and we have proved the claim, since ai(k) converges to 0weakly in l1, and thus in l1 norm as well.

Note that continuity of π was needed to show that ϕi∈ (l∞)′

Interpolation Theory

Besides Open Mapping Theorem, there are other methods to verify that some linear maps are bounded.In Harmonic Analysis for instance, a basic tool is interpolation. Suppose A: Lp→ Lq. In many cases, itis easy to show that A is bounded for specific choices for p and q. Given these, interpolation tells us thatA is bounded for choices of p and q “in-between” the choices that we can prove easily. Here we prove onesuch interpolation theorem. It is not the most useful, but can be applied to various operators.

Theorem 94. (Riesz-Throin) Let p0 ≥ 1 and p1, q0, q1 ≤∞, 0 < θ < 1. Suppose that T is a linear oper-ator from Lp0(X) +Lp1(X)→Lq0(X)+Lq1(X), and that

|T u|Lq0≤A0 |u|Lp0 and |T u|Lq1≤A1 |u|Lp1

44

i.e. T is bounded from Lp0→Lq0 and from Lp1→Lq1. Then we have that

|Tu|Lqθ ≤A01−θA1

θ |u|Lpθ

where

1

pθ=

1− θ

p0+θ

p1and

1

qθ=

1− θ

q0+θ

q1

i.e. that T is bounded from Lpθ→Lqθ.

Before the proof, we consider some applications of this interpolation theorem.

Example 95. (Convolution) Consider the operator

(f ∗ g)(x)=

∫

Rn

f(x− y)g(y) dy=

∫

Rn

f(y)g(x− y) dy

Then if 1≤ p, q, r ≤∞ with1

p+

1

q=

1

r+ 1, then

|f ∗ g |Lr ≤ |f |Lp + |g |Lq

Note there are two easy cases. For instance, if r=∞, then1

p+

1

q= 1 and

|f ∗ g |∞≤∫

|f(x− y)| |g(y)| dy≤‖f ‖p‖g‖q

by Hölder’s inequality. Also, if r=1, then p= q= 1 and

|f ∗ g |1≤∫ ∫

|f(x− y)| |g(y)| dy dx≤∫

‖f ‖1 |g(y)| dy≤‖f ‖1 ‖g‖1

by Fubini. Then, using interpolation,

1. Using the operator Tf(g) = f ∗ g, with f ∈ L1, we know that |Tf g |1 ≤ |f |1 |g |1 and that |Tf g |∞ ≤|f |1|g |∞, we can show that |Tf g |p ≤ |f |1 |g |p. In the theorem, take p0 = q0 = 1 and p1 = q1 = ∞,then for θ,

1

pθ=

1− θ

p0+θ

p1= 1− θ

so that pθ=1

1− θ. The same holds for qθ= pθ. Riesz-Thorin tells us that

|Tf g |pθ≤ |f |11−θ|f |1θ |g |pθ

= |f |1 |g |pθ

Thus we have |f ∗ g |p≤ |f |1 |g |p.

2. For the full range of p, q, we apply interpolation again. Using Tgf = g ∗ f this time, note that|Tg f |q ≤ |g |q |f |1 from above and that |Tg f |∞≤ |g |q |f |q ′. Then we apply the theorem with p0 = 1,q0 = q and p1 = q ′, q1 =∞. Then for θ,

1

pθ=1− θ+

θ

q= 1− θ

q ′

45

and

1

qθ=

1− θ

q ′

Riesz-Thorin tells that

|Tg f |qθ≤ |g |q1−θ |g |qθ |f |pθ

= |g |q |f |pθ

so that |g ∗ f |qθ≤ |g |q |f |pθ

where

1 +1

qθ= 1+

1− θ

q ′=

1

q+

1

pθ

as desired.

Now we turn to the proof of Riesz-Thorin.

Proof. First, note Lpθ ⊂ Lp0 + Lp1 (cutoff the function u at some level λ. Then u1|u|<λ ∈ Lp1 and

u1|u|>λ ∈ Lp0) Given u ∈ Lpθ, we want to estimate |T u|qθ. Note by duality, and density of simple func-

tions,

|Tu|qθ= sup

|v|qθ′≤1

v simple

∫

Y

(Tu) v dy

It also suffices to bound |T u| for simple functions u, because otherwise we can take simple un converging

to u in Lpθ, and un = un′ + un

′′ with un′ converging to some u′ in Lp0 and un

′′ converging to some u′′ in Lp1.This implies by boundedness on Lp0 and Lp1 that Tun

′ → Tu′ in Lq0 and Tun′′ → Tu′′ in Lq1. Then we can

find a subsequence un′ of un for which T un→ T u almost everywhere, and then we apply Fatou’s lemma sothat

∫

(T u) v dy ≤ liminfn→∞

∫

(T un) v dy

so that the bound for |T u| for nonsimple functions can be obtained with bounds for simple functions.Now define

F (z) =

∫

Y

T

(

|u|pθ/p(z) u

|u|

)

|v |qθ′/q(z)′ v

|v | dy

where |v |L

qθ′ ≤ 1, and

1

p(z)=

1− z

p0+

z

p1and

1

q(z)=

1− z

q0+

z

q1,

1

q(z)′= 1 − 1

q(z). Note that noting that z = 0,

|u|pθ/p0 ∈ Lp0 and |v |qθ′/q0

′ ∈ Lq0, and likewise for p1 and q1. For u, v simple, F is holomorphic, since forinstance if v=

∑

kvk1Bk

, then

|v |qθ′/q(z)′ =

∑

k

|vk|qθ′/q(z)′

1Bk(y)

|vk|vk

where z only appears in the exponent terms. Also, for 0≤Re z ≤ 1, |F (z)| ≤C bounded since the sums arefinite. Then since |F (0 + i t)| ≤A0 (apply T :Lp0−Lq0) and |F (1 + i t)| ≤A1 (apply T :Lp1−Lq1), we applythe Hadamard’s three lines theorem on the region z ∈C: 0≤Re z ≤ 1 to get that

|F (θ)| ≤A01−θA1

θ

46

but F (θ) =∫

YT (u) v dy so we are done after taking the supremum over v with norm 1.

Locally Convex Spaces

There are two motivations for considering locally convex spaces.

• Earlier we introduced weak convergence, and a natural question to ask is, “Under which topology isconvergence the same as weak convergence?” The result will not be a normed space.

• We know that C(K, C) for K compact is normed with the L∞ norm. What about C(Ω, C) forΩ ⊂ Rn open? We already have notions of convergence, for instance, uniform on every compactsubset. The corresponding topology will not be a normed space here either.

Weak Topology

Let X be a Banach space. Recall weak convergence, where xn→ x weakly if l(xn)→ l(x) for all l ∈X ′. Wenow investigate which topology corresponds to this notion of convergence.

In general, let fα:X→ Yα α ∈ Λ be a family of functions mapping from a set X to a topological space Yα.Consider

S= fα−1(Vα):Vα⊂Yα is open in Yα

preimages of open sets. Then let τS be the smallest topology containing S. We call τS the topology gener-ated by fα.

Remark 96. A few notes:

• τS is the weakest topology on X for which all fα are continuous maps from (X, τS) → Yα. This is

by definition of τS. fα is continuous means fα−1(V ) is open for V open in Yα, and we already set

up S so that fα−1(V )∈S.

• If xj , x∈X, then xj→x in τS if and only if fα(xj)→ fα(x) for all α.

Proof. Note that if xj → x in τS, then fα(xj) → fα(x) for all α since fα are continuous from (X,τS) → Yα. Conversely, suppose that fα(xj) → fα(x) for all α. We want to show that for any open

neighborhood U(x) (containing x) in (X, τS), that xj ∈ U for all sufficiently large j. Since τS isgenerated by S, there exist αi for which

x∈ fα1

−1(Vα1)∩ ∩ fαn

−1(Vαn)⊂U(x)

This implies that fαi(x) ∈ Vα1. But then since fαi

(xj) → fαi(x) for α1, , αn, we see that for suffi-

ciently large j, fαi(xj)∈ Vα1 as well. This implies that

xj ∈ fα1

−1(Vα1)∩ ∩ fαn

−1(Vαn)⊂U(x)

for all sufficiently large j. Thus xj→ x in (X, τS).

This tells us how to generate the topology consistent with the weak topology.

47

Weak and Weak* Toplogies:

For the Banach space X , and its dual X ′= l:X→C bounded, define

σ(X,X ′)4 topology on X generated byX ′

(above, X is the same, and X ′ is the family of functions from X → C). By the remarks above, conver-gence in this topology is the same as weak convergence.

Likewise, for X ′, if we consider the family ϕx:X ′→C defined by ϕx(l)= l(x), we denote

σ(X ′, X)4 topology on X ′ generated by (ϕx)x∈X

And lj → l in this topology if and only if lj(x) = ϕx(lj) → ϕx(l) = l(x) for all x ∈ X . This is simply the

weak* topology.

Remark 97. The weak topology does not have a countable neighborhood base (i.e. is not A1) if dimX =∞. Recall for (X, τ ) topological space, having a countable neighborhood base means that for every x∈X,we can find a countable collection of open neighborhoods Vj(x) around x for which for which every neigh-borhood U(x) contains some Vj(x).

The reason this is significant is that if (X, τ ) has a countable neighborhood base, then a mapping f :X→Y is continuous if and only if for every xj→x in (X, τ) we have that f(xj)→ f(x).

Note: If f is continuous, then f(xj) → f(x) for every sequence xj → x already. Take any neighborhood around f(x), the

preimage is open, containing x, and thus xj is inside the preimage for j sufficiently large, and thus f(xj) is in the neighbor-

hood for all j sufficiently large.

The issue is the converse. If we know f(xj) → f(x) for all xj → x, it is not necessarily true that f is continuous. But if (X,

τ) has a countable neighborhood base, it is true. It is a consequence of the fact that for such spaces, x is in the closure of A

if and only if there exists a sequence in A converging to x.... and etc. See texts on topology / wikipedia.

This issue is resolved in general by the introduction of “nets”, which generalize sequences.

Proof. (of Remark) Let X/R be a real Banach space, let dimX = ∞, and consider the weak topology(X, σ(X, X ′)). Suppose (Vi)i=1

∞ is a base for 0, then for all j, by definition of the weak topology thereexists li,1 , , li,mi

∈ X ′ for which⋂

j=1mi li,j

−1(−1, 1) ⊂ Vi (by scaling we can use the preimage of the same

set (−1, 1)). Then, given any l∈X ′, since Vi is a countable base for 0, for some Vi we have that

⋂

j=1

mi

li,j−1(−1, 1)⊂Vi⊂ l−1(−1, 1)

This implies that for all λ> 0,

⋂

j=1

mi

li,j−1(−λ, λ)⊂ l−1(−λ, λ)

and hence taking λ→ 0,

⋂

j=1

mi

N(li,j)⊂N(l)

There are now two claims to be made.

1. The null space containment above implies that l is a linear combination of li,j.

2. If X is a Banach space and dimX = ∞, X cannot be expressed as the span of a countable basis S(i.e. not every x can be written as a finite linear combination of elements of S).

48

Given the first claim, every l is in the span of li,1, , li,m for some i, and thus

X ′= spanli,j: 1≤ i≤∞, 1≤ j ,≤mi

But given the second claim, this is a contradiction.

Proof of First Claim: Suppose l1, , ln and l are in X ′ and⋂

i=1n

N(li)⊂N(l). Then construct a map-

ping A:X→Rn by A(x) = (l1(x), , ln(x)). Then we wish to show that there exists a linear functional ϕ:Rn→R for which ϕ A= l, in which case

l(x) = ϕ(l1(x), , ln(x)) =∑

i=1

n

ϕ(ei)li(x)

so that l is a linear combination of the li. Note that A x = 0 implies that l(x) = 0, so N(A) ⊂ N(l). Thismeans that it is well defined to define ϕ(y) = l(A−1(y)). Note that A−1(y) =x+N (A) for some x. Then

l(A−1(y))= l(x+N(A)) = l(x)

so that this is well defined. Clearly it is linear, so we are done.

Proof of Second Claim: We prove the contrapositive. Suppose that there exists a countable basis ei forwhich every x ∈X can be written as a finite linear combination of ei. Note that Kn= spane1, , en is asequence of closed subspaces of X (since finite dimensional) and whose union is X. By Baire’s CategoryTheorem (Corollary 63), one of the Kn has nonempty interior, Br ⊂Kn. But this implies that X ⊂Kn bylinearity, and since Kn is finite dimensional, X is finite dimensional.

Thus, the weak topology (X, σ(X,X ′)), without a countable neighborhood base does not necessarily havean easy characterization of continuous functions from X→Y in terms of sequences.

Nets

First, a directed set is a set I and a relation ≤ on I such that:

• α≤α

• (Transitive) α≤ β, β ≤ γ α≤ γ

• For all α, β ∈ I, there exists γ ∈ I such that α≤ γ and β ≤ γ.

Example 98. (N,≤ ) is a directed set.

Example 99. For (X, τ ), consider x0∈X, then if we take

I = U(x0): U neighborhood of x0

then (I ,⊃ ) is a directed set, i.e. U ≤V if U ⊃V . In fact, this is what we will be using to construct “nets.”

Example 100. If (I ,≤I ) and (J ,≤J ) are directed sets, then so is (I ×J ,≤ ) where

(α1, α2)≤ (β1, β2) α1≤ β1 and α2≤ β2

49

A net in X is a mapping from a directed set (I ,≤ )→X , where we will be using the notation xα∈X withα∈ I. Note that a special case is sequences, where we use xn∈X with the directed set (N,≤ ).

Net convergence: In (X, τ ), let (Λ, ≤ ) be a directed set, and consider the net (xα)α∈Λ ∈ X . Then wesay that xα→x iff for all U(x), there exists α0 such that xα∈U(x) for α≥α0.

Proposition 101. Let f : (X, τX) → (Y , τY ) be a mapping, and let x0 ∈X. Then f is continuous at x0 ifand only if f(xα)→ f(x0) for all convergent nets xα→x0.

Proof. First suppose f is continuous at x0. Then let V = V (f(x0)) some neighborhood of f(x0). Thenf−1(V ) contains some neighborhood U(x0) of x0. Now suppose xα → x0. This means that there is someα0 for which xα ∈ U(x0) for all α ≥ α0. But since U(x0) ⊂ f−1(V ) this implies that f(xα) ∈ V for all α ≥α0 and therefore f(xα)→ f(x0).

Conversely, suppose that f is not continuous at x0. Then there exists V = V (f(x0)) for which given anyneighborhood U(x0), there exists xU ∈U(x0) with f(xU) V . Note that using the directed set (U(x0),⊃), xU is a net with xU → x0, since given any U ′(x0), xU ∈ U ′ for all U ⊂ U ′ (U ≥ U ′). However, given anyU ′(x0), f(xU ′) V , so f(xU)→f(x0).

Two facts written in notes, not sure if relevant:

1. X is Hausdorff if and only if the (net) limit is only unique, i.e. if xα→ x and xα→x′ then x=x′.

2. (There exists?) fα:X→Yα for α∈Λ for which

xβ→x fα(xβ)→ fα(x) for all α

Subnets. Given (xα)α∈I a net, then (yβ)β∈I is a subnet of xα if there exists ϕ: J→ I for which

1. yβ= xϕ(β)

2. For all α0∈ I, there exists β0∈J such that β ≥ β0 ϕ(β)≥α0.

Properties:

1. In (X, τ ),

A= x∈X, exists xα∈A s.t. xα→x

2. If xα, α∈ I has x as a limit point, then there exists a subnet (yβ)β∈J for which yβ→x.

3. X is compact if and only if every net in X has a convergent subnet.

Reference: See Reed-Simon, Chapter 2.

In summary, the main idea is that in a topological space, we can describe continuity of functions withpreimages of open sets, but also as the preservation of net convergence.

Topological Vector Spaces

A topological vector space is a vector space X with a topology (X, τ) such that addition and scalarmultiplication are continuous maps. In other words,

1. Addition, X ×X→X with (x, y) x+ y is continuous

50

2. Scalar multiplication, K×X→X with (λ, x) λx is continuous.

Banach spaces (X, | · |) with topology generated by the norm is a topological vector space (addition andscalar multiplication are continuous with respect to the norm).

Example 102. Now that we have characterized continuity with net convergence, we can show easily that(X, σ(X, X ′)) is a topological vector space. For instance, if we take a net on X ×X, with (xα, yα) → (x,y) weakly, which means that xα → x and yα → y (product topology), then by the definition of weaktopology, for all l ∈X ′, l(xα)→ l(x) and l(yα)→ l(y). But by linearity, this implies that l(xα+ yα)→ l(x+y), and thus xα+ yα→ x+ y weakly, and this implies that addition is continuous with respect to the weaktopology, since net convergence is preserved. The same computation holds for scalar multiplication.

Proposition 103. Let X be a Hausdorff topological vector space, and let dimX = n <∞. Then X is iso-morphic to Rn

Proof. Let u1, , un be a basis for X . Define ϕ:Rn→X by (t1, , tn) ∑

i=1n

tiui. Note ϕ is continuous

because addition and scalar multiplication are continuous. We want to show that ϕ is a homeomorphism(preserves topology). Let ξ ∈ Rn, and let S = |ξ | = 1. Then S is compact, and ϕ(X) ⊂ X is also com-pact, therefore closed. Since 0 ϕ(S), we can find V = V (0) such that V ∩ ϕ(X) = ∅, i.e. V ⊂ X\ϕ(S).Now for every 0≤ t≤ 1, t V ⊂V , and the claim is that this implies ϕ(B1)⊃V .

This is because given x ∈ V , x= ϕ(ξ) for some ξ, and since for 0≤ t ≤ 1, ϕ(t ξ) = t x ϕ(S) (since t x ∈ Vand V does not intersect with ϕ(S)), we have that t ξ S for any 0≤ t≤ 1 so that |ξ |< 1, so x∈ ϕ(B1).

Locally Convex Spaces

A locally convex space is a topological vector space (X, τ) where every neighborhood U(0) contains anopen convex neighborhood V (0) of 0.

For instance, Banach spaces are locally convex since Br= |x|<r are convex, and the weak topology (X,σ(X,X ′)) is also a locally convex space since every neighborhood contains a set of the form

l1−1(−ε1, ε1)∩ lk−1(−εk, εk)

which is convex and open.

Example 104. Another example is furnished by a family of seminorms. Let X be a real vector space,and let pα for α ∈Λ be a family of seminorms (norms, but without the property that |x|= 0⇒ x= 0). Wedefine the topology

τ =

U : for all y ∈U , there exists αi, ri, 1≤ i≤n s.t.⋂

i=1

n

Bri

pαi(y)⊂U

where Brp(y) = x: p(x− y)<r (seminorm ball)

(As a special case, the weak topology is generated by the family of seminorms pl(x)= |l(x)| for l ∈X ′)

Then (X, τ ) is locally convex. Note that Remark 96 and the same proof as in Example 102 show that (X,τ) is a topological vector space. By definition of τ , every open neighborhood contains a finite intersectionof seminorm balls, which are convex. Also, Remark 96 shows that all pα are continuous with respect to(X, τ).

Proposition 105. Every locally convex space is generated by a family of seminorms.

51

Proof. Let X be a locally convex space. Let V ⊂X be a convex neighborhood of 0 with − V = V (sym-metric). Recall the Minkowski functional (gauge) from Remark 4:

pV (x) = infλ>0x∈λV

λ

This is a seminorm: Let a= pV (x), b= pV (y). Then x∈ (a+ ε)V and y ∈ (b+ ε)V for all ε. Then

x+ y ∈ (a+ b+ 2ε)V

for all ε, and thus pV (x+ y)≤ a+ b+ 2ε for all ε. Thus pV is subadditive, and hence a seminorm (since itis nonnegative and pV (0) = 0). Now we claim that (X, τ ) = (X, (pV )V convex, symmetric nbd), i.e. that theMinkowski functionals generate the topology of X .

(⊂ ) Let U =U(0) be an open neighborhood in (X, τ ). Then there exists a convex set V with V ⊂U , andfurthermore we can make V symmetric by replacing V with V ′=(−V )∩V . Then

B1pV ′(0)⊂V ⊂U

Note x ∈B1pV ′ means that x ∈αV ⊂ V some α< 1. Thus U is in the topology generated by the seminorms

pV .

(⊃ ) Let W be a neighborhood of 0 in (X, (pV )). Then

⋂

i=1

n

pvi(x)<εi⊂W

where the LHS is a finite intersection of open sets in (X, τ ), and hence is open in (X, τ ).

Example 106. Let Ω ⊂Rn be open, and let X = C(Ω,R). Recall that X is not normable. However, wecan generate the topology of uniform convergence on compact sets by a family of seminorms. Let K ⊂ Ω,and define

pK(ϕ)= supx∈K

|ϕ(x)|

Then (pK)K⊂Ω generate a topology on C(Ω), where

ϕα .C(Ω)ϕ ϕα|K ϕ|K uniformly

In fact, C(Ω) has a countable neighborhood base, and can be made into a metric space (Take a countable

sequence Kn which increases to Ω, with Kj ⊂ int(Kj+1), and use d(ϕ, ψ) =∑

n

pKn(ϕ− ψ)

1+ pKn(ϕ− ψ)2−n. It is an

exercise to show that the two topologies are equivalent, i.e. (X, (pK)j)= (X, d))

Frechet Spaces

A Frechet Space is a locally convex Hausdorff space which has a countable neighborhood base, andevery Cauchy sequence has a limit. In fact, under these conditions, such a space is metrizable under atranslation invariant metric, and under this metric the space is a complete metric space. (See Reed-Simon,Ch 3-4).

For instance, (C(Ω,R), d) in the previous example is a Frechet Space.

52

Example 107. Consider the Schwarz class,

S(Rn) =

ϕ∈C∞(Rn) s.t. supx∈Rn

|α|≤m

(1+ |x|)m|∂αϕ(x)|<∞

For instance, ϕ(x)= e−|x|2 is in S(Rn)

Taking pm(ϕ) = supx∈Rn

|α|≤m

(1 + |x|)m|∂αϕ(x)| which are seminorms, (S(Rn), (pm)m=0∞ ) is a Frechet Space.

Fourier Transform and Tempered Distributions.

For ϕ∈S define

ϕ(ξ)=

∫

Rn

ϕ(x) e−2πix·ξ d ξ

Recall F : ϕ ϕ is an isomorphism from S → S. The space of continuous linear functionals on S is calledthe space of tempered distributions S ′. By duality we can find F ′ so that the diagram commutes:

S ′ .F ′

S ′ S .F S

where the map S→S ′ is the usual mapping taking ϕ∈S to the functional (ψ ∫

ϕψ)∈S ′.

i.e. that [F ′(l)](ϕ) = l(Fϕ) for l ∈S ′. So we have a “Fourier transform” F ′ on the space of tempered distri-butions which agrees with the usual Fourier transform on L1, L2 and on measures µ(Rn). For instance,

L1(Rn) S ′

∧ FC0(R

n) S ′

commutes. The same situation occurs with L2 and the space of measures.

We can define “derivatives” for distributions in a similar manner. Recall

∂jϕ = i ξj ϕ(ξ)

call the natural map J : S → S ′ where Jϕ(ψ) =∫

ϕ ψ. Note that J is injective, and integration by partsgives

〈J(∂kϕ), ψ〉= (−1)k〈J(ϕ), ∂kψ〉

(here 〈 · , · 〉 denotes functional evaluation, not inner product). Then for T ∈S ′,

〈∂kT , ψ〉=(−1)k〈T , ∂kψ〉

and in particular, for T = J(ϕ) we have ∂kJ(ϕ)= J(∂kϕ).

Further Reference: Rudin, Functional Analysis , Part II.

Example 108. Distribution Spaces . Let Ω⊂Rn, and let

D(Ω) =Cc∞(Ω,C)

53

Let K ⊂Ω be compact, and define

DK(Ω)=

ϕ∈D(Ω): ϕ|Ω\K= 0

pm(ϕ)= supx∈K|α|≤m

|∂αϕ(x)|

and so (DK(Ω), (pm)m=0∞ ) is a Frechet Space. However, the construction of the topology for D(Ω) is

strange. Let

S=

co

(

⋃

K⊂Ω compact

UK(0)

)

, UK(0) is some neighborhood of 0 in DK

(convex hull)

Define

τ = U ⊂D(Ω), for all ϕ∈U , existsV ∈S s.t. ϕ+V ⊂U

Then

• (D(Ω), τ) is a locally convex space

• S is a neighborhood base of 0

• DK(Ω)iK D(Ω), and τ is the largest locally convex topology on D(Ω) such that all iK are contin-uous.

• If T : D(Ω) → Y , where Y is locally convex, is a linear map, then T is continuous if and only ifT |DK

is continuous for all K.

• D(Ω) does not have a countable neighborhood base. (D(Ω), τ ) is called the “inductive topology”.

Why is it so complicated? It turns out that

(D(Ω), (pm,K)m∈N,K⊂Ω)⊂ (C∞(Ω), (pm,K)m∈N,K⊂Ω)

where pm,K(ϕ)= sup|α|≤m,x∈K |∂αϕ(x)| is not complete.

(D ′ is the space of distributions).

Separation Theorem

Returning to general locally convex spaces, recall that in the earlier treatment of Hahn-Banach we showedthe existence of linear functions bounded above by a postiive homogeneous and subadditive function p(x).Using this we have shown the existence of nontrivial continuous linear functionals on normed spaces andalso separation of convex sets with continuous linear functionals. Now in the weaker setting of locallyconvex spaces, we can again show this with an appropriate choice for p(x). The key to show continuity ofthe linear functional in a locally convex space is to show continuity at 0, that if x→ 0 (Note! We areusing net convergence here), then l(x)→ 0 as well.

Proposition 109. If X is a locally convex space and X0 is a subspace with l0 ∈ X0′ , then there exists l ∈

X ′ such that l |X0= l0.

54

Proof. Note there exists U open and convex such that 0∈U and

U ∩X0⊂|l0|< 1

Then

|l0(x)| ≤ pU(x)

for x∈X0. This is because

x∈ λU ∣

∣

∣l0

(

x

λ

)∣

∣

∣≤ 1 |l0(x)| ≤λ |l0(x)| ≤ pU(x)

Then by Hahn-Banach there exists l:X→R for which l(x)≤ pU(x) and l |X0= l0. Then

|l(x)| ≤max pU(x), p−U(x)

and therefore if x→ 0 then l(x)→ 0, so that l∈X ′.

Example 110. Let X be locally convex Hausdorff space, and x0 ∈ X nonzero. Then there exists l ∈ X ′

such that l(x0) = 1 by extension of the finite dimensional (and hence continuous) map tx0 t fromKX0→K. By linearity we also have l(t x0)= t.

Proposition 111. Let X be a locally convex space over R, and A,B ⊂X be convex subsets with intA 0and A ∩B = ∅. Then there exists l ∈X ′ such that l 0 and l(x)≤ l(y) for all x ∈A, y ∈B. (For complexvector spaces, obtain Re l→ l).

Proof. This is exactly Theorem 3, using the previous proposition for the extension instead.

Double Polar Theorem

Let 〈 · , · 〉 be a bilinear function X × Y → R, not neceesarily an inner product. Then we say we have adual pair of spaces X,Y . We say that (X,Y ) is separate (or nondegenerate) if

〈x, y〉= 0 for all y ∈Y x= 0

and the same for the other coordinate.

Weak Toplogies. Given the bilinear function 〈 · , · 〉, for all y we can define a linear functional ly on X byly(x) = 〈x, y〉. Then (ly)y∈Y generates a locally convex space (X, σ(X, Y )) = (X, (ly)y∈Y ). The same is

true for (Y , σ(Y ,X)).

Note: Given a net xα, note that

xα→x in σ(X,Y ) 〈xα, y〉→ 〈x, y〉 for all y ∈Y

Therefore if (X, Y ) is separate, then the spaces (X, σ(X, Y )) and (Y , σ(X, Y )) are Hausdorff (nets have aunique limit).

Example 112. As special cases, we already know (X,X ′) as a separate dual pair with bilinear function

〈x, l〉= l(x)

55

then (X, σ(X, X ′)) is the weak topology on X and (X ′, σ(X ′, X)) is the weak* topology on X ′. Also,(X ′, X ′′) is a separate dual pair with (X ′, σ(X ′, X ′′)) the weak topology on X ′.

Dual Topologies.

Proposition 113. Let (X,Y ) be a separate dual pair. Then (X,σ(X,Y ))′=Y

Proof. Note if y ∈ Y , then ly(x)= 〈x, y〉 defines a linear functional in (X, σ(X,Y ))′, so we have that Y ⊂(X,σ(X,Y ))′. Now suppose l∈ (X,σ(X,Y ))′. Then we know that

|l |< 1⊃⋂

j=1

m

|lyj|<εj

since the topology is generated by (ly)y∈Y . The same proof as in Remark 97 shows that

N(l)⊃⋂

j=1

m

N(lyj)

and that

l= c1ly1 + + cmlym= lc1y1++cmym

i.e. that l can be identified with c1y1 + + cmym∈ Y .

Definition of Polar. Let (X, Y ) be a dual pair with 〈 · , · 〉:X × Y →R. Let A⊂X and B ⊂ Y . Then wedefine

A = y ∈Y : 〈x, y〉 ≤ 1 for all x∈AB = x∈X: 〈x, y〉≤ 1 for all y ∈B

called the polar of A and the polar of B. There is nothing special about which side the is, since wecan swap X,Y (and the bilinear function) to reverse them. This is just a notational convenience.

Note, we will use intA to denote the interior of A, and A to denote the polar of A

Example 114. For X Banach, and X, X ′ the dual pair with 〈x, l〉 = l(x) as before, we note that

denoting B1X by the unit ball in X, B1

X = x∈X: |x| ≤ 1, we have that

(B1X) = B1

X ′

( D1X ′

) = B1X

by duality principle (Proposition 48), since |x|= supl∈X ′,|l|≤1 |l(x)|, and |l|= supx∈X,|x|≤1 |l(x)|.Also, if Y ⊂X a subspace, then

Y = l ∈X ′: l(y)= 0 for all y ∈ Y

because if l(t y)≤ 1 for all t, then l(y) =0.

Observations:

• A⊂Y is convex, and 0∈A.

56

• A is closed in σ(Y , X). This follows since A is the intersection of closed sets y: 〈x, y〉 ≤ 1(inverse image of a closed set for a continuous linear functional)

Theorem 115. (Double Polar Theorem) Given (X,Y ) a separate dual pair with 〈 · , · 〉. For A⊂X,

( A)= co(A∪0)

where co denotes convex hull and the closure is with respect to σ(X,Y ).

Proof. ( ⊃ ) First note that A ⊂ ( A), since if a ∈ A, then for all A ∈ A by definition 〈a, y〉 ≤ 1 andhence a ∈ ( A). Also, 0 ∈ ( A) trivially. Furthermore, ( A) is convex and closed in σ(Y , X) and there-fore taking the convex hull and then the closure shows that

co(A∪0)⊂ ( A)

(⊂ ) Towards a contradiction, suppose we have x0∈ ( A) and x0 co(A ∪ 0). Then there exists U open

in σ(X, Y ) such that x0 +U ∩ co(A∪ 0) = ∅. Since we are in a locally convex space, U can be chosen tobe convex. This implies that there exists l∈ (X,σ(X,Y ))′ nonzero such that

l(z)≤ l(x0 +w)

where z ∈ co(A∪ 0) and w ∈U by the separation theorem. Since l 0 there exists some w ∈U for whichl(w)> 0 (pick arbitrary w ∈ U and scale appropriately, may need U to be balanced as well, which is fine).Then

l(z)≤ l(x0)− l(w)< l(x0)

for all z ∈ co(A∪0), and in particular, using z= 0 shows that l(x0)> 0.

Thus there exists c > 0 for which l(z) ≤ c < l(x0) and by scaling, c−1l(z) ≤ 1 < c−1l(x0). Since (X, σ(X,

Y ))′=Y (identification), we have that c−1l= ly for some y ∈ Y , so that

〈z, y〉 ≤ 1< 〈x0, y〉

again for all z ∈ co(A∪ 0), which implies on one hand that y ∈A, but on the other hand that x ( A),which contradicts our assumption on x.

Remark 116. Note that the theorem also works for ( B ) by symmetry (swapping X,Y for instance)

Corollary 117. If X is a uniformly convex Banach space, then X is reflexive.

Proof. Recall that uniformly convex means that for all ε > 0 there exists δ > 0 such that if |x| = |y | = 1

and∣

∣

∣

x+ y

2

∣

∣

∣> 1− δ then |x− y |<ε.

Let κ: X → X ′′ be the natural identification map given by κx(l) = l(x). We wish to show that κ is onto.

Let B1X = x ∈X : |x| ≤ 1, and likewise for B1

X ′

and B1X ′′

. Consider the dual pair (X ′, X ′′) with 〈l, ϕ〉 =

ϕ(l). Note κ(B1X)⊂X ′′. Now

( κ(B1X))= l ∈X ′:κx(l)≤ 1 for all |x| ≤ 1= l ∈X ′: |l(x)| ≤ 1 for all |x| ≤ 1=B1

X ′

57

and that by the Double Polar Theorem (Theorem 115),

κ(B1X)= ( ( κ(B1

X)))=(B1X ′

)=B1X ′′

where the closure above is in σ(X ′′, X ′). Now if ϕ∈X ′′, with |ϕ|X ′′ 1 1, we need to find x such that κx=

ϕ. Since ϕ ∈B1X ′′

= κ(B1X), we can find xα ∈B1

X (net) with κxα→ ϕ in σ(X ′′, X ′) (the weak* sense). Wewant to show that xα is a Cauchy net, from which we conclude that xα → x for some x (Exercise, butshould follow from simply extracting a Cauchy sequence from the Cauchy net).

Note liminfα |κxα| ≥ |ϕ| = 1 by lower semicontinuity under weak* convergence, and thus |xα| → 1 since

xα∈B1X. Replace xα by

xα

|xα|, and since |xα|→ 1 we still have κ

xα

|xα|→ ϕ in the weak* sense. Thus now we

have xα unit vectors, and note

κ

(

xα+xβ2

)

→ ϕ

in the weak* sense as well. Then

1≤ liminfα,β

∣

∣

∣

∣

xα+ xβ2

∣

∣

∣

∣

≤ limsupα,β

∣

∣

∣

∣

xα+xβ2

∣

∣

∣

∣

≤ 1

so that∣

∣

∣

xα + xβ

2

∣

∣

∣→ 1. By uniform convexity, this implies that |xα − xβ | → 1, so that xα is a Cauchy net.

Therefore xα→x for some x and κx= ϕ.

Krein Milman Theorem

Theorem 118. Let X be a locally convex Hausdorff space, and K ⊂X a convex, compact subset. Then

1. The extremal set of K is nonempty.

2. K = co(ext(K)) (i.e. K is the closure of the convex hull of the extremal points of K)

Before we prove the theorem, we make a few remarks and applications.

Remark 119. If K ⊂Rn, then there is a simple result by Carathéodory which states that every compactsubset K in Rn has extreme points, and every point of K can be written as a convex combination of n+1 extremal points.

The proof is by induction on the dimension n. For n= 1 the statement is trivial, as K is just an interval.Suppose the result is true for all subspaces of dimension less than n. There are two cases for K.

• If K has empty interior in Rn, then K is contained in a lower dimensional affine subspace, and(translating to a linear subspace) by induction we are done.

• Otherwise, first we claim that every boundary point of K can be represented as a convex combina-tion of n extreme points of K. Given x∈ ∂K, we can find a linear functional l for which l(x)≤ l(y)for all y ∈ K. Consider Kx = y ∈ K: l(y) = l(x), which is a hyperplane intersected with theboundary of K. Kx is an extreme subset of K, as all other points z satisfy l(z)< l(x).

Since this is lower dimensional, by induction every point in Kx can be written as a convex combi-nation of n extreme points of Kx. Note that the extreme points of Kx are also extreme points ofK. Thus any x∈ ∂K can be written as a convex combination of n extreme points of K.

58

To finish the proof, now consider an arbitrary point y ∈K. If y is an extreme point already, thereis nothing to prove. Otherwise, pick an arbitrary extreme point x1 of K, and form the line z: z =t x1 + (1 − t)y, t ∈ R. This line necessarily intersects ∂K at some x2. By above, x2 is a convexcombination of n extreme points. Since y is contained in the line segment connecting x1 and x2,then y is a convex combination of x1 and x2, and thus y is a convex combination of x1 and n otherextreme points, and we are done.

Where to get Compactness

Proposition 120. (Banach-Alaoglu) If X is Banach, then B1X ′

, the unit ball in the dual is compact inthe weak* topology σ(X ′, X).

Proof. Consider the map ϕ: B1X ′ → ∏

|x|≤1x∈X

[−1, 1] defined by ϕ(l) = (l(x))x∈X. The infinite product is

compact by Tychonoff’s Theorem (the topology is the topology of pointwise convergence). Then

1. ϕ is a homeomorphism.

lα σ(X ′,X)

l lα(x)→ l(x) for all x∈X lα(x)→ l(x) for all |x| ≤ 1, x∈X ϕ(lα)→ ϕ(l) pointwise

2. ϕ(B1X ′

) is closed.

Suppose that ϕ(lα) → f , so that lα(x) → f(x) for all |x| ≤ 1. Then for x 0, let f(x) = f(

x

|x|

)

|x|(extension). Thus

lα(x)= |x| lα(

x

|x|

)

→|x| f(

x

|x|

)

= f(x)

so that lα(x)→ f(x) for all x. By linearity of limits, f is linear, and also |f(x)| ≤ 1 for |x| ≤ 1 since

f is in the product of [−1, 1]. Thus f ∈B1X ′

, and f = ϕ(f).

This shows that ϕ(B1X ′

) is a closed subset of a compact space, and hence is compact, and since ϕ is a

homeomorphism, B1X ′

is also compact.

Application of Krein Milman

Theorem 121. L1(0, 1) is not the dual of any space Y.

Proof. Suppose towards a contradiction that L1 @ Y ′. Then consider B1L1

, by the previous Proposition

120 B1L1

is compact in the weak* σ(L1, Y ) topology. However, we note that ext(B1L1

) = ∅. Note given any

f ∈B1L1

, we can split f = f1[0,b] + f1[b,1] where b satisfies

∫

0

b

|f |=∫

b

1

|f |= |f |L1

2

(since∫

0

x |f |dx is continuous in x, we can find such a b). Then we have that

f =1

2

(

2f1[0,b]

)

+1

2

(

2f1[b,1]

)

59

where |2f1[0,b]|L1 = |2f1[b,1]|L1 = |f |L1, i.e. both parts are in B1L1

and so f is not extremal.

Now by Krein-Milman, (Theorem 118), we have that B1L1

being compact in σ(L1, Y ) is the closure of theconvex hull of its extreme points, but since the set of extreme points is empty, this is a contradiction.Thus L1 cannot be the dual of any space.

(A technical detail is that we need to identify L1 homeomorphically to Y ′, but this is not an issue sincewe assumed L1@ Y ′)

Proof of Krein Milman

Proof. (of Theorem 118) Let E = L: L is a closed, extremal subset of K. Noting that K ∈ E , weknow that E is nonempty. Now we have an ordering on E defined by

L1≤L2 L1⊃L2

Then every linearly ordered subset of E has an upper bound, since the intersection is an upper bound,which is in E since the intersection of nonempty compact sets is also nonempty and compact. Also, theintersection of extremal subsets is also extremal. Thus by Zorn’s Lemma, there exists a maximal elementK0 in E . The claim is that K0 = x, i.e. consists of a single point. This implies that x ∈ ext(K) so thatext(K) ∅.

Suppose that K0 consists of multiple points, and let x1, x2 ∈ K0 with x1 x2. Then there exists a linearfunctional l ∈X ′ for which l(x1)< l(x2). Then let K1 = x ∈K0: l(x) = maxK0 l which is also an extremalsubset of K0 and hence an extremal subset of K. Since l(x1)< l(x2), x1 K1 and hence K1>K0, contra-dicting the maximality of K0.

Now we show that K = co(ext(K)). Suppose not, then let x0 ∈ K\co(ext(K)). Then there exists l ∈ X ′

such that l(x) ≤ c < l(x0) for all x ∈ co(ext(K)). Then define K0 = x ∈ K: l(x) = maxK l. This is anextremal subset of K, which is nonempty by continuity of l. Now since K0 is nonempty and convex, takey ∈ ext(K0) (exists by previous part). Then y is also an extreme point of K (extreme point of an extremesubset is still an extreme point of the original set). However, this is a contradiction since y ext(K) byhyperplane separation: points in ext(K) satisfy l(y)≤ c< l(x0)≤maxK l, and hence cannot be in K0.

Spectrum of Bounded Linear Operators

Recall from Linear Algebra that for a matrix A,

λ is an eigenvalue of A λI −A not invertible det(λI −A) =0 N(λI −A) 0

In particular, A has exactly n eigenvalues, and we have

• In general, have the Jordan Form

• If A∗=A (self-adjoint), then A=U

λ1 λn

U∗ for U unitary and λj real.

• If A∗A=AA∗ (normal), then A=U

λ1 λn

U∗ for U unitary and λj complex.

60

We hope to have similar results for operators between Banach spaces under some conditions.

Let X/C be a Banach space, and let A∈L(X) (bounded operator from X→X). We define

σ(A) 4 λ∈C:λI −A is not invertible= λ∈C:N(λI −A) 0

⋃

x∈C:R(λI −A) X

σ(A) is the spectrum, and it consists of two parts, the first part λ, λI − A 0 are the eigenvalues orpoint spectrum, which we denote σp(A), and the second part will remain unnamed.

The complement, ρ(A)=C\σ(A) is called the resolvent, and

ρ(A) = λ∈C:λI −A is invertible= λ∈C: N(λI −A)= 0, R(λI −A)=X

Example 122. Let X = C([0, 1],C) with the supremum norm |f |∞ = max0≤x≤1 |f(x)|, and consider themultiplication operator A(f)= x f . This is a bounded linear operator. Consider the equation

(λI −A)f =(λ− x)f

First we note that if (λI − A)f = 0, then f(x) = 0 for x λ, and by continuity f ≡ 0. Thus N(λI −A) =0 for all λ, and thus A has no point spectrum (eigenvalues).

Note that if λ [0, 1], then it is easy to solve (λ− x)f = g by f =g

λ− x(which is continuous now that λ

[0, 1]). Then for λ [0, 1], λI −A is one-to-one and onto, and hence bijective. Thus C\[0, 1]⊂ ρ(A)

On the other hand, if λ ∈ [0, 1], note that (λI −A)X ⊂ f ∈C[0, 1], f(λ) = 0 which is a strict subspace ofX, and thus [0, 1]⊂σ(A).

This implies that σ(A)= [0, 1] and ρ(A)=C\[0, 1].

Example 123. Let H =L2[0, 1], and study the multiplication operator again Af = xf .

If (λI −A)f = 0, then (λ− x)f(x)= 0 for all x. Then (λ− x)f(x)= 0 except on a set of measure zero, andsince λ − x is only zero at x = λ, this implies that f(x) must be zero except on a set of measure zero, sof = 0 (in L2).

From the same reasoning as above, we have that if λ [0, 1] then (λI −A)g

λ− x= g so that C\[0, 1]⊂ ρ(A).

Now suppose λ ∈ [0, 1]. Then we show that λI −A is not onto. More specifically, if we choose g = 1 (con-stant function), then there does not exist f ∈L2[0, 1] such that

(λI −A)f = 1

Otherwise, we must have f =1

λ− xwhich is not in L2. Thus σ(A)= [0, 1] and ρ(A)=C\[0, 1] and A has no

point spectrum since the null space is trivial for all λ.

Holomorphic Functions from C→ X

Before we turn to facts about σ(A), we develop a tool that will be useful in many of the proofs to follow.

We will be using results in complex analysis. Recall that a function f :C→C is holomorphic at z0 if

f ′(z0)= limz→z0

f(z)− f(z0)

z − z0

61

exists (the limit is in C). Likewise, we can use the same definition for functions from C→X . A functionf :C→X is (strongly) holomorphic at z0 if

f ′(z0)= limz→z0

f(z)− f(z0)

z − z0

where the limit is in X.

A function f :C→X is weakly holomorphic at z0 if for every linear functional l ∈X ′, l f is holomorphicas a function from C→C. We can also see this as the same definition as above but replacing the limit inX with the weak limit. With this view, it is clear that strongly holomorphic functions are weakly holo-morphic.

For Banach spaces, it turns out that if f is weakly holomorphic on some open domain Ω, then it is alsostrongly holomorphic on Ω. (Also true for Frechet spaces)

Proposition 124. If f is weakly holomorphic in an open domain Ω, then it is strongly holomorphic in Ω.

Proof. See Lax, Section 11.4

Many results in complex analysis will carry through.

• (Cauchy’s Theorem) The Cauchy-Goursat Theorem says that if f is holomorphic in some domainΩ, then

∫

γ

f(z)dz=0

for γ a closed curve in Ω. There is a proof that uses solely the differentiability of f and firstproving the Theorem for rectangles and repeatedly cutting the rectangle into fourths.

The Cauchy Integral Formula

f(z)=1

2πi

∫

γ

f(ξ)

ξ − zdξ

can be deduced from this when applying the Cauchy-Goursat Theorem tof(ξ)− f(z)

ξ− z.

Thus, Liouville’s Theorem, Residue Theorem, Power Series, Laurent Series all hold. Below we pre-sent different approaches for the proofs.

• (Liouville’s Theorem) Suppose that f(z): C→X is entire (holomorphic on C) and bounded. Thenthe same is true for l f , for any l, and thus f must also be constant.

• (Power Series) A useful tool is that if we can express f(z) has a power series∑

k=0∞

ck zk where

ck ∈ X , then f is holomorphic within the radius of convergence (recall in a Banach space X , anyabsolutely summable series is summable).

• (Laurent Series) We can use power series to transfer to Laurent series which are valid in an annulus

r1 < |z | < r2, f(z) =∑

k=−∞∞

ck zk with ck ∈ X . Furthermore, we can still talk about residues,

Resz=0f(z)= c1, and

1

2πi

∫

γ

f(z) d z=Resz=0f(z)

62

This is justified by term by term integration, and using the usual complex analysis identity(Residue Theorem)

1

2πi

∫

γ

ckzk dz=

ck2πi

∫

γ

zk dz=

ck k=− 10 else

Facts about σ(A)

Now we turn to basic facts about the spectrum σ(A).

Proposition 125. Let A∈L(X). Then

1. σ(A) is compact

2. σ(A) ∅

3. The spectral radius, rA4 supz∈σ(A) |z |=maxz∈σ(A) |z | (max since σ(A) is compact) satisfies

rA= limn→∞

|An|1/n

We will be using the following fact:

Proposition 126. If |A|< 1, then I −A is invertible, and

(I −A)−1 =∑

j=0

∞

Aj

The same formula holds if |A|m< 1 for some m as well.

Proof. Since |Aj | ≤ |A|j and |A| < 1, the RHS series is absolutely summable, and hence converges tosome element. Note that a computation involving telescoping series shows that

(I −A)∑

j=0

n−1

Aj=

(

∑

j=0

n−1

Aj

)

(I −A)= I −An

and as m→∞, we have that I −An→ I. (note |An| ≤ |A|n→ 0). Thus

∑

j=0

∞

Aj=(I −A)−1

Now to show that the same formula holds when |A|m< 1 for some m, we simply show that the RHS seriesis absolutely summable, and then the same computation carries through (replacing n by multiples of m).

Write θ= |Am|1/m. Then for any j note we can write j= km+ l so that

|Aj | = |Akm+l|≤ |Am|k |Al|= θmk+l · |A

l|θl

≤ Cθj

63

where C =max0≤l≤m|Al|

θl.

Now we return to the facts about the spectrum σ(A).

Proof. (of Proposition 125)

(1) Compactness of σ(A): First we show that σ(A) is compact, and since it is a subset of C, it sufficesto show that σ(A) is closed and bounded. Note that for |z | > |A|, we have that z I − A = z(I − z−1A) is

invertible, so z ∈ ρ(A) and (taking complements) σ(A)⊂B|A|. Thus σ(A) is bounded.

Now we show that ρ(A) is open, which shows σ(A) is closed. Let z0∈ ρ(A). Then

z I −A= (z0I −A)[

I +(z − z0)(z0I −A)−1]

where for |z − z0| sufficiently small (smaller than |(z0I −A)−1|), I + (z − z0)(z0I −A)−1 is invertible, andhence z I − A is invertible for all |z − z0| sufficiently small. Thus we can find Bε(z0) ⊂ ρ(A) and henceρ(A) is open

(2) σ(A) nonempty: To show that σ(A) is nonempty, we will use the fact that z I − A is holomorphicon ρ(A) (as a function from C→X , see previous section). This is because for any z0∈ ρ(A), by the repre-sentation above we can expand (z I −A)−1 in a power series about z0 for |z − z0|<ε, ε sufficiently small:

(z I −A)−1 =

[

∑

j=0

∞

(−1)j(z0I −A)−j(z − z0)j

]

(z0I −A)−1

Since we can express (z I − A)−1 as a power series around any z0 ∈ ρ(A), (z I − A)−1 is holomorphic on

ρ(A). Now, suppose towards a contradiction that σ(A)= ∅, or ρ(A)=C, so that (z I −A)−1 is entire.

The claim is that (z I −A)−1 is also bounded. First note that for any linear operator B ∈L(X), if

inf|x|=1

|Bx|= ρ

then

|B−1|X =1

ρsup|x|=ρ

|B−1x| ≤ 1

ρ

In particular, for |z |> 2|A|X, we have that inf|x|=1 |(z I −A)x| ≥ |z | − |A|X so that

|(z I −A)−1|X ≤ 1

|z | − |A|X≤ 1

|z |

For |z | ≤ 2|A|X, we note that z |(z I − A)−1|X is continuous, and hence is bounded on the compact set|z | ≤ 2 |A|X. Thus (z I −A)−1 is bounded. By Liouville’s Theorem (generalized), (z I −A)−1 reduces to a

constant, and since by the above estimate |(z I −A)−1|X→ 0 as z→∞, (z I −A)−1 must be identically 0,which is a contradiction (0 is not invertible). Thus it must be the case that ρ(A) C, so σ(A) isnonempty.

(3) Spectral Radius: We will show this in three steps.

Step 1. First we show that the limit exists, and in particular

limn→∞

|An|1/n= infm≥0

|Am|1/m

64

Now fixing m, let n= km+ l for 0≤ l≤m− 1. Then

|An| ≤ |Akm+l|≤ |Am|k|Al|= |Am|n/m · |Al|

|Am|l/m≤ C |Am|n/m

where C = max0≤l≤m−1|Al|

|Am|l/m. Taking n-th roots gives |An|1/n ≤ C1/n|Am|1/m, and thus taking limsup

in n and inf over m we have

limsupn→∞

|An|1/n≤ infm≥0

|Am|1/m≤ liminfn→∞

|An|1/n

and the right inequality follows from liminfn→∞ |An|1/n= supn infm>n |Am|1/m. Furthermore, since liminfis bounded above by limsup, we have that

liminfn→∞

|An|1/n= limsupn→∞

|An|1/n= infm≥0

|Am|1/m

Step 2. Now we show that rA ≤ limn→∞ |An|1/n. It suffices to show that if |z |> limn→∞ |An|1/n then z ∈ρ(A). If |z | > limn→∞ |An|1/n then for some m we have that |z | > |Am|1/m for some m. We can rewrite

this as∣

∣

∣

(

A

z

)m∣∣

∣

1/m< 1. Now by Proposition 126, this implies I − A

zis invertible, in which case z I − A is

invertible so that z ∈ ρ(A)

Step 3. Finally we show that rA≥ limn→∞ |An|1/n. For |z |> |A| we have the Laurent series expansion

(

I − A

z

)−1

=∑

k=0

∞Ak

zk

or

(z I −A)−1

=∑

k=0

∞Ak

zk+1

and if we integrate over the curve |z |=R⊂ ρ(A)

1

2πi

∫

|z|=R

zk(z I −A)−1 dz=Ak

(see previous section) Then we have the bound

|Ak | ≤ 1

2π

∫

0

2π

C(R)Rk+1 dθ=C(R)Rk+1

with |(z I −A)−1| ≤∑k=0∞ 1

|z |·∣

∣

∣

A

z

∣

∣

∣

k

=C(R). Taking k-th roots and taking the limit as k→∞ shows that

limk→∞

|Ak|1/k≤R

and since |z |=R⊂ ρ(A) for all R>rA, letting R→ rA we conclude that

limk→∞

|Ak|1/k≤ rA

65

Example 127. Note that if A is nilpotent, i.e. Am= 0, then rA= 0. This is easily seen in the case whereX =Rm and

A=

0 1 10

where all the eigenvalues are 0.

Functional Calculus

As in finite dimensional linear algebra, we can define A2 =A ·A and if p(z) =∑

k=0n

ck zk then

p(A) =∑

k=0

n

ckAk

Furthermore, if f is entire with power series f(z) =∑

k=0∞

ck zk, we can define

f(A) =∑

k=0

∞

ckAk

For instance, the exponential of an operator is

eA=∑

k=0

∞Ak

k!

We can even go further and define f(A) for f holomorphic from Ω → C where Ω contains σ(A) via theintegral formula

f(A) 4 1

2πi

∫

γ

f(ξ)(ξI −A)−1 dξ

where n(γ, z) = 1 for all z ∈ σ(A) and n(γ, z) = 0 for all z Ω. (This integral can be defined as a limit ofRiemann sums). This corresponds to the Residue Formula in complex analysis, and will allow us to definefor instance log(A) in a region containing σ(A) that is simply connected and does not contain 0.

Note in the case that f is entire, then we can use a power series to see that this definition matches withthe previous. Choose γ to be |z |=R with R> |A|, so that

1

2πi

∫

γ

f(z)(z I −A)−1 =1

2πi

∫

γ

f(z)

(

∑

k=0

∞

z−k−1Ak

)

dz

=∑

k=0

∞

Ak(

1

2πi

∫

γ

f(z)

zk+1dz

)

=∑

k=0

∞

ckAk

= f(A)

Now in finite dimensions, we know that if A is n × n matrix, and λ1, , λn are its eigenvalues, then for fentire, we know that f(λ1), , f(λn) are the eigenvalues of f(A). This follows from computing f(A) ofthe Jordan form of A. The diagonal terms of f(A) are precisely f(λk).

66

A natural question is to ask in the general setting, whether σ(f(A))= f(σ(A)).

Proposition 128. Let A ∈ L(X), and let f , g be holomorphic functions from Ω → C with Ω containingσ(A). Then

1. f(A) g(A)= (f g)(A)

2. σ(f(A))= f(σ(A))

3. g(f(A))= (g f)(A), where g is holomorphic in a region containing σ(f(A)) instead.

Proof. First we show that (1) (2). Assume that f(A) g(A) = (f g)(A). First we show that f(σ(A)) ⊂σ(f(A)) Let z0∈ σ(A). We want to show that f(z0)∈ σ(f(A)). Define g(z) =

f(z)− f(z0)

z− z0which is holomor-

phic for the same region as f (the singularity at z= z0 is removable). Now

f(z)− f(z0)= (z − z0) g(z)

and applying (1) we have

f(A)− f(z0) I = (A− z0 I)g(A) = g(A) (A− z0 I)

and since A − z0I is not invertible, f(A) − f(z0)I is not invertible either: If N(A − z0 I) 0, thenN(g(A)(A − z0 I)) 0, and if R(A − z0I) X, then R((A − z0I)g(A)) X. This implies that z0 ∈σ(f(A)) so that (σ(A))⊂ σ(f(A)).

To show that σ(f(A))⊂ f(σ(A)), suppose that w f(σ(A)). We will show that w I − f(A) is invertible so

that w σ(f(A)). If we choose g(z) =1

w− f(z), then g is holomorphic in a neighborhood of σ(A). Then

since (w− f(z))g(z) =1, we apply (1) to get

(w I − f(A))g(A)= g(A)(w I − f(A)) = I

so that w I − f(A) is invertible (left and right inverse).

For a proof of (1) and (3), see Lax, Theorem 5 in Section 17.2.

Fredholm Theory

Note: This is a rearrangement of material covered in class.

In finite dimensions, if A:X→X is a linear map, then we have the following two properties:

1. A is injective if and only if A is surjective.

2. We have that

Ax= b 〈Ax, y〉= 〈b, y〉 for all y ∈X⟨

x,ATy⟩

= 〈b, y〉 for all y ∈X

so that if x is a solution to Ax= b, then a necessary condition is that 〈b, y〉 = 0 for all y such thatATy= 0, i.e. that b∈ (NAT)⊥. This turns out to be a sufficient condition also. In other words,

RA= (NAT)⊥

67

This is almost immediate when examining the matrices, particularly the fact the operation Ax canbe viewed as taking the inner product of the rows of A with x:

RA= col(A)= row(AT )= (NAT)⊥

(A technical note: to use complex inner product we use the adjoint A∗ instead of the transpose)

However, these properties may not hold in infinite dimensions:

Example 129. Let l2 = (an)n=1∞ :

∑

n=1∞ |an|2<∞, and define A: l2→ l2 by

A(a1, a2, )= (0, a1, a2, )

(the right shift operator). Note that A is injective, but not surjective. This is an example of a property infinite dimensions that does not hold in infinite dimensions (the equivalence of injectivity and surjectivity).

The second property still holds: The range looks like (0, a1, a2, ). AT is precisely the left shift, since

〈A(a1, a2, ), (b1, b2, )〉 = 〈(0, a1, a2, ), (b1, b2, )〉= 〈(a1, a2, ), (b2, b3, )〉=⟨

(a1, a2, ), AT(b1, b2, )⟩

Then the null space of AT consists of elements of the form (r, 0, 0, ), and the orthogonal complement isprecisely the range of A, with elements of the form (0, a1, a2, ).

Example 130. Consider X = L2([0, 1]) =

f :∫

0

1 |f(x)|2 dx <∞

, and the multiplication operator M

defined by

Mf(x) =xf(x)

Note that M is a self-adjoint operator:

∫

Mf(x) g(x) dx=

∫

xf(x) g(x) dx=

∫

f(x)Mg(x) dx

Note that NM = 0, so that (NM)⊥=X, but RM X = (NM)⊥ since 1∈RM.

Later, however, we will show that it is the case that RM = ker(NM)⊥.

A natural question to consider, then, is for which operators A does the properties (1) and (2) aboveremain true? Fredholm operators enjoy the same properties, and we will see them shortly.

Index and Pseudoinverse

Note: For what follows, X,Y are generic vector spaces.

Let K:X→ Y be a linear operator, with X, Y vector spaces over R. If ran(K) is finite dimensional, thenwe say that K is of finite rank. Note the following properties:

• If X A Y B Z, then B A is of finite rank if either B or A is of finite rank.

• If K1,K2:X→Y are of finite rank, then K1 +K2 is of finite rank.

68

Pseudoinverse. If A:X→ Y is a linear map, and there exists B:Y →X such that

AB = I +K

BA = I +L

where K, L are of finite rank, then we say that A has a pseudoinverse, and B is called the pseudoinverseof A. Note that if the dimensions are finite, every map is pseudoinvertible.

Why is this concept useful? Consider the same question, when does Ax = b have a solution? Wealready know the story for finite dimensions. There are finitely many relations for b in this case (b is inthe column space of the matrix A). However, the same is true for the case that A is pseudo-invertible. Wewill show that if A is pseudo-invertible, then Y = RA ⊕ Z where Z is finite dimensional. Then we onlyneed to check that b has no component in Z.

Another Form: Given A,B linear maps from X to Y , we say that A∼B if A−B is of finite rank. Then

• ∼ is an equivalence relation.

• If A1∼B1, A2∼B2, then A1 +A2∼B1 +B2

• If A1∼B1, A2∼B2, and XA1

B1

Y A2

B2

Z, then A2A1∼B2B1.

Fact: A: X → Y is pseudoinvertible if and only if there exists a linear map B from Y to X such thatAB∼ I and BA∼ I. Note that invertibility is the property above with ∼ replaced by = .

Fact: If A:X→Y linear such that B1A∼AB2∼ I, then B1∼B2 and B1 is a pseudoinverse of A.

Proof. B1 = B1I ∼ B1AB2 ∼ IB2 = B2, so B1 ∼ B2. And since AB1 ∼ AB2 ∼ I , B1 is a pseudoinverse ofA.

Proposition 131. Let A: X → Y linear. Then A is pseudoinvertible if and only if dim NA < ∞ anddim (Y /RA)<∞.

Y /RA is called the “corange” of A and its dimension is the “codimension” of A.

Before we prove the result, let’s look at a few examples:

Example 132. Consider T : l2 → l2 as before, the shift operator T (a1, a2, ) = (0, a1, a2, ). Then T ispseudoinvertible, since if we define the reverse shift T ′(a1, a2, )= (a2, ), we see that T ′ T = I, and

T (T ′(a1, a2, ))= (0, a2, )

so that TT ′= I −K where K(a1, a2, )= (a1, 0, 0, ) has finite rank. Thus TT ′∼T ′T ∼ I.

Example 133. Let X = C([0, 1]) (continuous functions on [0, 1]). Then if we define a linear map T map-ping X→X by

(Tf)(x) = f(x)+

∫

0

1

exyf(y)dy

The equation Tf = g comes up in practice, and we will discuss this later in the Fredholm theory.

69

We will show that T is pseudoinvertible later when we discuss compact operators. It turns out that the

integral operator f ∫

0

1exyf(y)dy is a compact operator (Hilbert-Schmidt kernels, Example 149), and

that an operator in the form I +K with K compact is pseudoinvertible.

Before the proof of the Proposition 131, we introduce a general technique:

Block Decomposition: We use block matrices to organize computation. If we decompose X =X1⊕ ⊕Xn and Y = Y1 ⊕ ⊕ Ym, then letting Pj be the projection of X to Xj and Πi be the projection of Y toYi. We can define Aij:Xj→ Yi by Aij = Πi A Pj (project to Xj, then map A, then project to Yi), andthen A =

∑

ijAij. Furthermore, we can express this action in a matrix. Decomposing any x as x =

x1 + +xn, we have

Π1AxΠmAx

=

A11 A1n Am1 Amn

x1xn

This is a purely algebraic identity, and thus it works for infinite dimensional spaces as well (decomposingthem into a direct sum of finitely many subspaces).

Observation: If X1⊂X is a subspace, then there exists a subspace X2⊂X such that X =X1⊕X2.

These two points will be used in the proof.

Proof. (of Proposition 131) First suppose that A is pseudoinvertible. This means there exists anoperator B from Y →X such that BA= I +K and AB= I +L where K,L are finite rank. Note that

NA⊂NBA=NI+K

Note that if x∈NI+K then x+Kx=0, thus x=K(−x)∈RK, and so NI+K⊂RK and

dimNA≤ dimNI+K ≤ dimRK<∞

As for Y /RA, we make a few observations:

• RA⊃RAB=RI+L

• Y /RA@ (Y /RAB)/(RA/RAB)

This is because we can write RA=RAB⊕W , where W @ RA/RAB, and so

y+RA y+RAB⊕W (y+RAB)+RA/RAB

• dimY /RI+L<∞This is because if we consider the injection map Π from Y to Y /RI+L, defined by

Π(y) = y+RI+L

then we note that given x+RI+L∈Y /RI+L we have that −Lx∈RL and

Π(−Lx) =−Lx+RI+L=−Lx+ (I +L)x+RI+L= x+RI+L

so that

Π(RL)=Y /RI+L

70

and since dimRL<∞, we have that dim (Y /RI+L)<∞ as well.

Finally, we have that

dim (Y /RA)= dim(Y /RAB)−dim(RA /RAB)≤ dim(Y /RAB)= dim(Y /RI+L)<∞

This shows that if A is pseudoinvertible, then dim(NA) and dim(Y /RA) are finite.

Conversely, suppose that dim(NA) and dim(Y /RA) are finite. Decompose X =X1⊕NA and Y =RA⊕ Y1.Then we can write A in block matrix form

A=R

A

Y1

(

A1 0

0 0

)

X1 NA

Then we claim that A1:X1→RA given by A1 =A|X1is invertible. Injectivity follows from the fact that for

any x1 ∈ X1 such that Ax1 = 0, we see that x1 ∈ X1 ∩ NA = 0. Surjectivity follows from the fact thatgiven y ∈RA and x with Ax= y, we have that x=x1 + x2 with x1∈X1 and x2∈NA. In this case

y=Ax=Ax1 =A1x1

Then, we simply define

B=X

1

NA

(

A1−1 00 0

)

RA Y1

in which case

AB=R

A

Y1

(

I 00 0

)

RA Y1

= I +

(

0 00 − I |Y1

)

where the last term above has finite rank, since dim(Y1) = dim Y − dim RA <∞. We can do the same forBA, and this shows that AB∼BA∼ I, so A is pseudoinvertible.

Definition 134. If A:X→ Y is pseudoinvertible, then we define its index to be

indA=dimNA− dimY /RA

Example 135. If A:X→Y where both spaces are finite dimensional, then

indA=dimNA− dimY + dimRA=dimX − dimY

and in particular, if A:X→X with dimX <∞ then indA= 0.

Example 136. For the shift operator, T : l2→ l2, dimNT =0 and dim l2/RT = 1, so the index is −1.

Proposition 137.

1. If A,B are pseudoinvertible, then so is BA and indBA= indB+ indA.

2. If A is pseudoinvertible and K is finite rank, then ind(A+K)= indA

71

Proof.

(1) exact sequence 0→NA→NBA→NB→Y /RA→Z/RBA→Z/RB→ 0

(2) first prove for I +K using decomposition X =X1⊕X2, with dimX1<∞, X2⊂NK and RK⊂X1.

More precisely, X = (RK⊕X4)⊕ (NK ∩X3).

Fredholm Operators

Let X, Y now be Banach spaces. If A∈L(X, Y ) such that dimNA<∞, RA is closed and dim Y /RA<∞,then we say that A is a Fredholm operator. We define the index of such an operator to be

ind(A)= dimNA− dim Y /RA

as with pseudoinvertible operators. Note that we have Banach space structure here, and Fredholm opera-tors have the additional condition that RA is closed, though actually this is not an additional restriction.It turns out that

Proposition 138. Let A∈L(X,Y ). Then

A is Fredholm dimNA<∞, dimY /RA<∞, RA closed dimNA<∞, dimY /RA<∞, A is pseudoinvertible

We have already proved the last equivalence, and the first equivalence is a definition. The middle followsfrom open mapping theorem. We will show that if dim Y /RA <∞, then RA is closed. The reason is thatfirst, without loss of generality NA = 0, otherwise we replace A with A: X/NA→ Y with A(x + NA) =Ax. Then we have Y =RA⊕Y1 with dimY1<∞. Then we can define a map ϕ:X ⊕Y1→Y by

ϕ(x, ξ) =Ax+ ξ

and since Y = RA ⊕ Y1, this is a bijection. ϕ is one-to-one since A is one-to-one, and given any y ∈ Y , wecan decompose y = A x + ξ for some x, ξ. By open mapping theorem, ϕ is a homeomorphism, and sinceRA= ϕ(X), ϕ(X) is closed and therefore RA is closed.

In addition to the pseudoinvertible property that A,B Fredholm implies BA is Fredholm with

ind(BA)= indB+ indA

we also have the following special property. Denote Fred(X, Y ) ⊂ L(X, Y ) the subspace of operators thatare Fredholm.

Proposition 139. Fred(X,Y )⊂L(X,Y ) is an open subset, and specifically,

ind(A+E)= indA

for all |E |<ε with ε sufficiently small.

Proof. Let A ∈ Fred(X, Y ). We will use block decomposition again. Let X = X1 ⊕ NA, with X1 closedand NA finite dimensional (recall finite dimensional subspaces have a closed complement). Also, Y =RA⊕Y1, with Y1 finite dimensional, RA closed since A is Fredholm.

72

Now

X1

NA

A=

(

A1 00 0

)

RA

Y1

where A1 = A|X1. Note that A1: X1 → RA is a bijection and thus by open mapping A1 is a homeomor-

phism. Now we write

X1

NA

A+E=

(

A1 +E11 E12

E21 E22

)

RA

Y1

for |E |< ε, Eij are the appropriate restrictions of E, i.e. E11 = πRAE |X1

. For ε sufficiently small, we note

that A1 +E11 is invertible, more specifically, when |E11|< 1

|A1−1|

. This is because we can write

A1 +E11 =A1(I +A1−1E11)

and use the Neumann series I − B =∑

iBi when |B | < 1. Thus, we now multiply A + E by invertible

operators and use the property that the composition of psueodinvertible operators is also pseudoinvertible.Computation gives

RA

Y1

(

I 0

−E21(A1 +E11)−1 I

)

RA

Y1

X1

NA

(

A1 +E11 E12

E21 E22

)

RA

Y1

X1

NA

(

I − (A1 +E11)−1E12

0 I

)

X1

NA

which reduces to

X1

NA

C =

(

A1 +E11 00 B

)

RA

Y1

where B =E22 −E21(A1 +E11)−1E12. What’s important here is that A1 +E11 is invertible and B is finite

dimensional, and thus this operator C is pseudoinvertible. This implies that A+E is also pseudoinvertibleafter reversing the procedure. Furthermore, we can compute the index. Since the index of invertible oper-ators is 0, we note that ind(A+E) = ind(C) = ind(A1 +E11) + ind(B) (this equality follows by inspection;for instance, the null space is the null space of A1 + E11 plus the null space of B). Since B is a finitedimensional operator, by Example 135 ind(B)= dimNA− dim Y1, and since A1 +E11 is invertible its indexis 0. Thus ind(A+E)= dimNA− dimY1 = ind(A).

Example 140. If t ∈ [0, 1], then if A(t): [0, 1]→ Fred(X, Y ) is continuous, then indA(s) = indA(t) for alls, t∈ [0, 1]. This is referred to as “homotopy invariance of index”.

Compact Operators

Recall in finite dimensions, for every operator (matrix) A, the spectrum consists entirely of eigenvalues(recall that in finite dimensions injectivity implies surjectivity). Compact operators in infinite dimensionsenjoy a similar property.

Let X,Y be Banach spaces. If A∈L(X,Y ) (bounded linear operator) such that

AB1 = Ax: |x|< 1

73

is precompact in Y , then we say that A is a compact operator.

Here precompact means that the closure is compact in the context of a complete metric space. Forinstance, in Rn all bounded sets are precompact.

Proposition 141. Let X be a complete metric space, and let A⊂X. Then

A precompact every sequence xj ∈A has a Cauchy subsequence for all ε> 0, A=⋃

i=1

m

Ai, diam(Ai)<ε for all ε> 0, exists a finite ε-net

We have the following connection to pseudoinvertible operators.

Theorem 142. Let X/C and K ∈L(X) be compact. Then

1. dim(NI−K)<∞.

2. RI−K is closed

3. dimX/RI−K = dimNI−K<∞ i.e. if NI−K = 0 then X =RI−K.

For proving Theorem 142, we will be making use of the following facts:

Proposition 143.

1. If K1,K2 compact, then K1 +K2 is also compact.

2. If K is compact and A is bounded, then KA is compact (also AK is compact)

3. If Ki is a sequence of compact operators, and Ki→K (in operator norm), then K is compact.

4. If K is compact, K ′ is also compact.

Proof. (1) We can find Ai with diam(Ai) < ε, K1B1 =⋃

i=1m

Ai and likewise K2B2 =⋃

j=1n

Cj withdiam(Cj)<ε. Then

(K1 +K2)B1⊂K1B1 +K2B2⊂⋃

i,j

Ai+Bj

where diam(Ai+Bj)< 2ε (i.e. a finite 2ε-net)

(2) The proof uses the same idea. Let |A| = M . We have K B1 =⋃

i=1m

Ci with diam(Ci) < ε. Then note

KA B1 ⊂ KBM =⋃

i=1m

MCi where diam(M Ci) ≤ Mε. We also have A K B1 ⊂ ⋃

i=1m

A Ci, and

diam(ACi)≤Mε also (|Ax−Ay | ≤M |x− y |).(3) The idea here is to take Ki with |Ki−K |<ε, and use KiB1 =

⋃

j=1m

Aj with diam(Aj)<ε, with

KB1⊂KiB1 +(K −Ki)B1⊂⋃

j

Aj+ (K −Ki)B1

where diam(Aj+ (K −Ki)B1)< 2ε, so we have a finite 2ε net for KB1.

74

(4) Recall that K ′:Y ′→X ′ is defined by K ′l(x)= l(Kx). Examine

K ′B1Y ′

= K ′l, |l|< 1= l K, |l |< 1

Let F = K B1, which is a compact subset of Y . We will show the precompactness of K ′B1Y ′

. Note givenl1, l2∈Y ′, we have

|l1 K − l2 K |X ′ = sup|x|≤1

|l1Kx− l2Kx|

= supy∈F

|l1(y)− l2(y)|

The result will follow from Arzela Ascoli, using C(F ,C) and the compactness of F . Note that

l |F , |l |< 1

⊂C(F ,C)

we have l |F uniformly Lipchitz (with constant 1) on a compact set F , and thus this family of functions is

equicontinuous. Given a bounded sequence li K ∈ K ′B1Y ′

, note that li|F is an equicontinuous sequence,

and by Arzela Ascoli there exists a subsequence li′|F which is Cauchy uniformly, i.e.

supy∈F

|li′(y)− lj ′(y)|→ 0

which by above implies that |li′ K − lj ′ K |X ′→ 0, and hence K ′B1Y ′

is precompact and K ′ is compact.

Proof. (of Theorem 142) Let A= I −K.

(1) Here we show that dim NA < ∞. Suppose that dim NA = ∞. Then there exists xj ∈ NA such that

|xj | = 1 and |xj − xk| ≥ 1

2for 1 ≤ k < j. Since K is compact, there exists a subsequence K xn′ which is

Cauchy. Then on one hand, we have that |Kxn′ −Kxm′| → 0 but on the other, |xn′ − xm′| ≥ 1

2, and this is

a contradiction since K is bounded.

(2) Now we show that RA is closed. Recall that this is implied by dim Y /RA <∞, but we will show thisdirectly (without directly appealing to open mapping). Write X =NA⊕X1 where X1 is closed. The claimis that |A x| ≥ c|x| for x ∈ X1. Then this implies that RA = A X1 is closed, since under this condition ifAxj is Cauchy, then xj is Cauchy, and by continuity the limit of xj maps to the limit of Axj.

To prove the inequality, suppose it is false, then there exists a sequence xj ∈X1 where |A xj | < 1

j|xj |. By

scaling, we can take |xj | = 1, and then A xj → 0, which means that xj − K xj → 0. Passing to a subse-quence, by compactness of K we can find Kxj ′ → y, and then xj ′ → y. This implies that Ky→ y and y ∈NA. But since X1 is closed, y ∈X1 also. Thus y= 0. However, we have chosen xj ′ so that |y |= 1, and thisis a contradiction.

(3) Next we show dim X/RA <∞, and that dim X/RA = dim NA. Note that A′ = I − K ′ is compact byProposition 143 (4). By above, dimNA′<∞. We will use the isometries (X/RA)′@ (RA)⊥=NA′,. Then

dim (X/RA)′=dimNA′<∞

and since (X/RA)′ is a finite dimensional Banach space, dimX/RA= dim (X/RA)′<∞.

75

The identification is, given l: (RA)⊥, define l ′(x + RA) = l(x) (well defined since l annihilates RA), and wecan go backwards from l ′∈ (X/RA)′ as well. This is a norm preserving bijection also, since

|l |X ′ = sup|x|≤1

|l(x)|

= sup|x|≤1y∈RA

|l(x+ y)|

= sup|x+RA|≤1

l ′(x+RA)

= |l′|(X/RA)′

recalling that |x+RA|= infy∈RA|x+ y |.

Now consider the map A(t): [0, 1] → Fred(X, Y ) given by A(t) = I − t K. Note that A(0) = I and A(1) =I −K. By homotopy invariance, we have that ind(I −K)= ind(I)= 0, and this implies that

dimNA= dimX/RA

and as a consequence, I −K is injective if and only if I −K is surjective.

Proposition 144. For K compact, we also have

NI−K⊂ ⊂N(I−K)m =N(I−K)m+1 =i.e. N(I−K)m =N(I−K)m+1 for some m.

Remark 145. This property is easy to see in finite dimensions, for instance the matrix

I −K =

0 10

00

where NI−K⊂N(I−K)2 =N(I−K)3 =Proof. Suppose that this property does not hold, and let A = I − K. This means that there exists xnwith |xn|= 1, (I −K)nxn=0 and d(xn, N(I−K)n−1)≥ 1

2. (each xn∈N(I−K)n\N(I−K)n−1).

Note that Kxn = xn + yn where yn ∈ N(I−K)n−1. Then, by compactness of K, we can find a subsequence

Kxn′ such that |Kxn′−Kxm′|→ 0. On the other hand, we have that

|Kxn′−Kxm′| = |xn′ + yn′− xm′− ym′|= |xn′− (yn′ +xm′ + ym′)|≥ 1

2

since yn′ + xm′ + ym′∈N(I−K)n′−1 (if n′>m′), and this is a contradiction.

Alternate Characterization of Fredholm Operators.

Proposition 146. Let A:L(X, Y ). If there exist B, C ∈L(Y , X) such that AB = I +K and CA= I + L

with K,L compact, then A is a Fredholm operator.

76

Remark 147. This is weaker than the condition for pseudoinvertibility. Recall that for pseudoinvert-ibility we need a single operator B ∈ L(Y , X) so that A B = I + K, B A = I + L with K, L finite rank .Thus, note that if A is a Fredholm operator, then this condition certainly holds (since A Fredholm A

pseudoinvertible).

Proof. Note that NA ⊂ NCA = NI+L which is finite dimensional. Thus dim NA <∞. Also we have thatRA ⊃ RAB = RI+K, and this implies that dim Y /RA ≤ dim Y /RI+K < ∞. Thus by the equivalences inProposition 138, A is Fredholm.

Corollary 148. If A:X→ Y is Fredholm and K:X→Y is compact, then A+K is Fredholm with

ind(A+K) = ind(A)

Proof. Since A is Fredholm and thus pseudoinvertible, we can find B: Y →X such that AB= I +K1 andBA= I +K2 where K1,K2 are finite rank. Then we have that

(A+K)B = I +K1 +KB

B(A+K) = I +K2 +BK

and since K1 +KB,K2 +BK are compact (Proposition 143), A+K is Fredholm. Furthermore, by homo-topy invariance, we can take A(t) = A + t K which is continuous operator from [0, 1] → Fred(X, Y ), andthus

ind(A)= indA(0)= indA(1) = ind(A+K)

Example 149. Hilbert-Schmidt Kernels. Let (X, µ) and (Y , ν) be measure spaces, let K(x, y)∈L2(X,Y )in the product space L2(X,Y ) =L2((X ×Y ), µ× ν), and define TK:L2(X)→L2(Y ) by

TKf (y) =

∫

X

K(x, y)f(x)dx

Then TK is a compact operator.

Proof. First we show that TK is bounded. This is simply by Hölder:

|TKf(y)|2 ≤( ∫

X

|K(x, y)|2 dx)

|f |L2(X)2

|TKf(y)|L2(Y )2 =

∫

Y

|TKf(y)|2dy ≤ |K(x, y)|L2(X×Y )2 |f |L2(X)

2

and thus the operator norm is |TK |L(L2(X),L2(Y ))≤ |K |L2(X,Y ).

To show that TK is compact, we will find a sequence of finite rank operators TKn→ TK convergent in

operator norm, and since finite rank operators are compact, by Proposition 143 TK will be compact.

Using an orthonormal basis for L2(X × Y ), we can write K(x, y) =∑

i=1∞

pi(x)ψi(y) where pi∈L2(X) andψi(y) ∈ L2(Y ) (this is accomplished by fixing x and expanding K in terms of a basis for L2(Y ), and thenexpanding each coefficient in terms of a basis for L2(X)). Now if we truncate the series

Kn(x, y)=∑

i=1

n

pi(x)ψi(y)

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and define the operator TKnby

TKnf(y) =

∫

X

Kn(x, y)f(x)dx=∑

i=1

n ( ∫

pi(x)f(x)dx

)

ψi(y)

which is a finite linear combination and thus TKnis finite rank (and thus compact).

Note that Kn → K in L2(X × Y ), and by the same inequality used to show boundedness of TK, thisimplies that TKn

→TK in operator norm, and therefore TK is also compact.

Spectrum of Compact Operators

Theorem 150. Let X/C be a Banach space, K ∈L(X) compact. Then

1. σ(K) is countable.

2. λ∈σ(K)\0 λ∈σp(K) (i.e. an eigenvalue), dimNλI−K<∞ and

NλI−K⊂N(λI−K)2⊂ ⊂N(λI−K)m =N(λI−K)m+1 =3. σ(K)\0 has no nonzero limit point (the spectrum is isolated)

Proof. First note that (3) implies (1). First we prove (2). Let λ ∈ σ(K)\0 This means that λI −K isnot invertible. Note that for Fredholm operators, injectivity is equivalent to surjectivity, so thereforeNλI−K 0 and λ ∈ σp(K). Note that dim NλI−K <∞ since λI − K is Fredholm, and we proved that

the generalized eigenspace⋃

m=1∞

N(λI−K)m is finite dimensional in Proposition 144.

Next we prove (3). Suppose there exists a nonzero limit point in σ(K)\0. Then we have a sequence ofeigenvalues λi ∈ σp(K)\0, λi λj for i j with λi→ λ 0. Let Xi =NλiI−K. Note that Xi is invariantunder K, i.e. KXi⊂Xi,, since λix=Kx λiKx=K2x.

Also, span⋃

iXi

=⊕

iXi since eigenvectors with different eigenvalues are linearly independent. Now

let Yn=⋃

i=1n

Xi so that Y1 ( Y2 ( Y2 ( , and choose yk such that |yk|= 1 and d(yk, Yk−2)≥ 1

2. Since K

is compact, there is a Cauchy subsequence Kyn′. Note that yn′ = xn′ + zn′ with xn′ ∈Xn′ and zn′ ∈ Yn′−1,and that

Kyn′ = Kxn′ +Kzn′

= λn′xn′ +Kzn′

where Kzn′∈ Yn′−1. Thus, on one hand by Cauchy property |Kyn′−Kym′|→ 0, but on the other hand,

|Kyn′−Kym′|= |λn′|∣

∣

∣

∣

xn′ +1

λn′

Kzn′

∣

∣

∣

∣

≥ 1

2|λn′| 1

2|λ| 0

which is a contradiction.

Spectrum of Compact Operators in Hilbert Spaces

Now we let H be a complex Hilbert space. Recall that the adjoint A∗ of A∈L(H) is defined by

〈A∗x, y〉= 〈x,A y〉

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for all x, y ∈H .

Recall in linear algebra:

1. If A∗=A (i.e. self-adjoint), then A=U

λ1 λn

U∗ with U unitary and λi real.

2. If A∗A=AA∗ (i.e. normal), then A=U

λ1 λn

U∗ with U unitary and λi complex.

We will be proving the analogous result for compact self-adjoint and normal operators:

Theorem 151. (Spectral Theorem for Compact Self-Adjoint Operators) If H is a separableHilbert space, K ∈ L(H) compact and self-adjoint, then there exists an orthonormal basis (uj)j=1

∞ suchthat Kuj=λjuj, λj ∈R and λj→ 0.

Theorem 152. (Spectral Theorem for Compact Normal Operators) Let K ∈ L(H) be a normalcompact operator, then there exists an orthonormal basis (uj)j=1

∞ such that Kuj = λj uj, λj ∈ C andλj→ 0.

Note the following characterization of self-adjoint operators on complex Hilbert spaces:

Proposition 153. If A∈L(H), then

A∗=A 〈Ax, y〉= 〈x,Ay〉 for all x, y ∈H 〈Ax, x〉 ∈R for all x∈H

Proof. The first equivalence is by definition. Given that 〈Ax, y〉 = 〈x, Ay〉 for all x, y ∈H , then we havethat 〈Ax, x〉= 〈x,Ax〉 and thus 〈Ax, x〉 ∈R. Given that 〈Ax, x〉 ∈R for all x∈H , note

〈A (x− iy), x− iy〉= 〈Ax, x〉+ i〈Ax, y〉− i〈Ay, x〉+ 〈Ay, y〉 ∈R

This implies that 〈Ax, y〉− 〈Ay, x〉=0.

Definition: If A ∈L(H), A∗ =A and 〈Ax, x〉 ≥ 0 for all x∈H , then we use the notation A≥ 0 (A is non-negative definite).

Proposition 154. If A≥ 0, then |A|= sup|x|=1 〈Ax, x〉.

Proof. Cauchy Schwarz shows sup|x|=1 〈Ax, x〉 ≤ |A|. For the other direction, let M = sup|x|=1 〈Ax , x〉.Note that 〈Ax, y〉 + ε〈x, y〉 is an inner product on H (the ε is necessarily to make a strictly positivequadratic form 〈Ax, x, 〉+ ε〈x, x〉> 0. Then Cauchy Schwarz gives

|〈Ax, y〉+ ε〈x, y〉| ≤ 〈Ax, x〉+ ε|x|2√

〈Ay, y〉+ ε|y |2√

Letting ε→ 0, we have

|〈Ax, y〉| ≤ 〈Ax, x〉√

· 〈Ay, y〉√

≤ M |x| |y |

for all x, y. Taking supremum over x, y shows that |A| ≤M (recall that |x|= supy |〈x, y〉|).

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Corollary 155. For any A∈L(H), |A∗A|= |A|2.

Proof.

|A∗A| = sup|x|=1

〈A∗Ax, x〉

= sup|x|=1

〈Ax,Ax〉

= sup|x|=1

|A2x|

= |A2|

As a direct consequence, we have the following:

Corollary 156. If A is normal, then rA= |A|

Proof. We have that

|A2|2 = |(A2)∗A2|= |(A∗A)2|= |A∗A|2 = |A|4

where we have applied the previous two results. Since A∗A≥ 0, by Proposition 154,

|(A∗A)2|= sup|x|=1

⟨

(A∗A)2x, x⟩

= sup|x|=1

|A∗Ax|2 = |A∗A|2

Thus we have that |A2| = |A|2, and repeating the argument, we have that |A2n| = |A|2n

. Thus, using theformula for spectral radius (Theorem 125) we have that

rA= limn→∞

|An|1/n= limn→∞

|A2n|1/2n

= |A|

Remark 157. Combining the results above, if K is a compact, self-adjoint, non-negative definite oper-ator A≥ 0, then

λ1 = rK = |K |= sup|x|≤1

〈Kx, x〉

(just like in finite dimensions). This also gives a hint for how to find eigenvalues for K that is not neces-sarily non-negative.

The key idea is that this quantity sup|x|=1 〈Kx, x〉 (no loss in changing |x| ≤ 1 with |x| = 1 by linearity)

yields both eigenvalues and eigenvectors. It is directly related to the Rayleigh quotient RK(x)=〈Kx, x〉

〈x, x〉.

Proposition 158. Let K be a compact, self-adjoint operator. Then

λ= sup|x|=1

〈Kx, x〉

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is an eigenvalue, the supremum is achieved by some vector z, and Kz = λz, i.e. z is an eigenvector of Kwith eigenvalue λ.

Proof. (Following Lax, Ch 28) The idea is that by definition of sup, there is a sequence xk with |xk| = 1and 〈Axk, xk〉 → λ. Since xk is a bounded sequence in a Hilbert space, we can replace xk with a conver-gent subsequence so that xk→ z weakly.

Now we claim that the compactness of A implies that Axk → Az in norm. Note that Axk → Az weaklysince l A is a bounded linear functional so that xk→ z weakly implies that l(Axk)→ l(Az) for all l ∈X ′.We already know that by compactness we can replace xk with a further subsequence so that Axk con-verges in norm to some y. But this implies Axk → y weakly as well, and by uniqueness of limits of weakconvergence, y=Az and Axk→Az in norm.

Now we have that 〈Axk, xk〉→ 〈Az, z〉 since

|〈Axk, xk〉− 〈Az, z〉| ≤ |〈Axk−Az, xk〉|+ |〈Az, xk〉− 〈Az, z〉|≤ |Axk−Az | |xk|+ |〈Az, xk〉− 〈Az, z〉|→ 0

noting that |Axk − Az | → 0 by norm convergence and y 〈Az, y〉 is a linear functional so the secondterm vanishes by weak convergence of xk→ z.

Thus, 〈Az, z〉 = λ1. Note that by lower semicontinuity of the norm, |z | ≤ 1, but in fact z is also a unit

vector, since otherwise⟨

Az

|z |,z

|z |

⟩

=λ1

|z |2>λ1 which contradicts λ1 being the supremum of 〈Ax, x〉.

The rest is just like in linear algebra: to show that Az = λ1z, note that z maximizes the Rayleigh quotient

RA(x) =〈Ax, x〉

|x|2(over all x 0). Then taking an arbitrary vector w and f(t) = RA(z + tw) maps R →H

and achieves its maximum at t= 0, thus its derivative vanishes at t= 0. The derivative at t=0 is

limh→0

f(h)− f(0)

h= lim

h→0

1

h

[

〈A(z+ hw), z+hw〉|z+hw |2 − 〈Az, z〉

|z |2]

= limh→0

1

h

[

h〈Az, w〉+h〈Aw, z〉+ h2〈Aw,w〉|z+ hw |2 + 〈Az, z〉

(

1

|z+ hw |2 −1

|z |2)]

=〈Az,w〉+ 〈Aw, z〉

|z |2 +λ1 limh→0

〈z, z〉 − 〈z+ hw, z+ hw〉|z |2|z+ hw |2

=〈Az,w〉+ 〈Aw, z〉

|z |2 −λ1〈w, z〉+ 〈z, w〉

|z |4

Setting this to zero gives

Re 〈Az −λ1 z, w〉= 0

for all w. This implies Az = λ1z (set w = Az − λ1z), and thus z is an eigenvector of A with eigenvalueλ1.

Remark 159. The procedure above works for inf|x|=1 〈Kx, x〉 as well, which will give the negative eigen-values and eigenvectors.

Having shown how to get an eigenvector z, we can prove the main theorem by iterating the procedure onK restricted to the complement of spanz. A technical point is that for an operator K not necessarilynonnegative, we need to consider both sup|x|≤1 〈Kx, x〉 and inf|x|≤1 〈Kx, x〉

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Proof. (of Theorem 151) Essentially, the previous proposition does all the work. Before we do any-thing, first take an orthonormal basis of the null space ker K (recall this is finite dimensional), and set itaside. If K = 0 then we are already done. For the rest of the proof we will then gather the other eigenvec-tors corresponding to nonzero eigenvalues, and do this in such a way that |λn| decreases to 0.

If K 0, then consider sup|x|=1 〈Kx, x〉 or inf|x|=1 〈Kx, x〉, whichever has larger magnitude, and denote

this quantity by λ1. As described above, we can find a unit eigenvector u1 with Ku1 = λ1u1. Note thatsince K is self-adjoint, the orthogonal complement of the eigenvector u1 is invariant under K, since givenv ∈u1

⊥,

〈Kv, u1〉= 〈v,Ku1〉=λ1〈v, u1〉=0

so that Kv ∈ u1⊥ as well. We can then repeat the procedure with K |

u1⊥(the operator restricted to the

orthogonal complement of u1). We will then find a second eigenvalue λ2 with |λ2| ≤ |λ1|, and a corre-sponding eigenvector u2. Continuing this procedure, we generate a sequence of eigenvectors un corre-sponding to eigenvalue λn and λn → 0 (noting there is no cluster point other than 0). This grabs allnonzero eigenvectors since at each stage we always grab an eigenvector with the eigenvalue with largestmagnitude.

Note that our collection of eigenvectors (uk)k=1∞ (now including the vectors in ker K) is complete, since

otherwise the orthogonal complement contains another eigenvector with nonzero eigenvalue, which is acontradiction since such an eigenvector would have been found in our procedure.

Remark 160. (Min-Max Description of Eigenvalues) Let A be compact and self-adjoint, and listthe positive eigenvalues in decreasing order λ1 ≥ λ2 ≥ . Then as in finite dimensional linear algebra,there is a min-max principle for the n-th largest eigenvalue:

λk = infV ⊂H

dim V =k−1

supx∈V ⊥

|x|=1

〈Ax, x〉

= supV ⊂H

dim V =k

infx∈V|x|=1

〈Ax, x〉

The proof is similar to the same result in finite dimensions.

We sketch the method of proof for compact normal operators:

Proof. (Sketch of Theorem 152) The proof is similar to the finite dimensional spectral theory. Weprove a result about commuting compact, self-adjoint, operators being simultaneously diagonalizable, orhaving the same eigenvectors. Then decompose a normal operator K into a self-adjoint operator A and ananti-self-adjoint operator B:

K =

(

K +K∗

2

)A

+

(

K −K∗

2

)B

Since K is compact, K∗ is also compact, and therefore A, B are compact. Since K is normal, A, B com-mute, and we can apply the result for commuting compact self-adjoint operators to A and iB, and thiswill give the eigenvectors for K, since if Av=λv and iBv= µv, then

Kv= (λ− iµ)v

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Remark 161. Note that with the spectral theorem for a compact normal operator K, we can identify Hwith l2 on which K becomes a multiplication operator:

H xKx H

ϕl lϕ

l2 xk λk xk

l2

and the identification is ϕ(x) = (〈x, uk〉)k=1∞ .

Also, we can define for a bounded function f the operator

f(A)x=∑

k=1

∞

f(λk) 〈x, uk〉uk

which is also bounded:

‖f(A)x‖H = ‖f ‖∞‖x‖H

What’s missing: Spectral theorem for bounded operators in Hilbert spaces.

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