Elements of Functional Analysis - Texas A&M Universitythomas.schlumprecht/course...Elements of Functional Analysis 7.1 Normed Linear Spaces and Banach spaces In the following K denotes

Chapter 7

Elements of FunctionalAnalysis

7.1 Normed Linear Spaces and Banach spaces

In the following K denotes either the real number field R or the complexnumber field C.

Definition 7.1.1. (Normed Linear Spaces)Let X be a vector space over K. A map

k · k : X ! [0,1), x 7! kxk,

is called a seminorm if it has the following properties:

1. (homogeneity) For every x 2 X and � 2 K it follows that

k� · xk = |�| · kxk.

2. (triangular inequality) For x, y 2 X it follows that

kx+ yk kxk+ kyk.

Note that (1) implies that k0k = 0. we call a seminorm k · k a norm on X if

3. For x 2 X it follows that

kxk = 0 () x = 0.

125

126 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS

In that case we say (X, k · k) is a normed linear space.Note that If (X, k · k) is a normed space it follows by (2) and (3) that

d : X ⇥X ! [0,1), (x, y) 7! kkx� yk

is a metric on X. We call the topology generated by this metric the strongtopology or the norm topology on X and we call

BX = {x 2 X : kxk 1}, the unit ball of X, andSX = {x 2 X : kxk 1}, the unit spheer of X.

If (X, ·k) is a normed linear space which is complete with respect to thenorm topology we call (X, ·) a Banach space.

Let k · k and ||| · ||| be two norms on X. We call k · k and ||| · ||| equivalentif there is a constants b � a > 0 so that

akxk |||x||| bkxk, for all x 2 X.

Note that two equivalent norms generate the same strong topology on X.

Example. Recall that if S is any set we denote

B(S,K) =�f : S ! K : sup

s2S|f(s)| < 1

and if T is a topology on S,

Cb(S,K) =�f : S ! K : continuous and sup

s2S|f(s)| < 1

.

Then by Proposition 6.2.9 B(S,K) and Cb(S,K) are with k · k1 defined by

kfk1 = sups2S

|f(s)| for f 2 B(S,K), respectively f 2 C(S,K)

normed linear spaces which are complete with respect of the above intro-duced metric, and, thus, Banach spaces.

A special case is S = N in that case we denote B(S,K) by

`1 =�(⇠n)

1n=1 : k(⇠n)1n=1k = sup

n2N|⇠n| < 1

.

More generally, sometimes B(S,K) is denoted by

`S1 =

�(⇠s)s2S ⇢ K : k(⇠n)1n=1k = sup

s2S|⇠n| < 1

.

7.1. NORMED LINEAR SPACES AND BANACH SPACES 127

.

Definition 7.1.2. (Absolutely Convergent Series) If (xn) is a sequence in anormed linear space (X, k · k) we call the (formal) series

P1n=1 xn absolutely

convergent if1X

n=1

kxnk < 1.

Theorem 7.1.3. A normed linear space (X, ·) is a Banach space if and onlyif every absolute converging series converges.

Proof. “)” This follows from the fact that ifP1

j=1 xj is absolute convergentthen the sequence of partial sums (sn), i.e., sn =

Pnj=1 xj is a Cauchy

sequence.“(” Assume that every absolute converging series converges and let (xn)be a Cauchy sequence. Then there is a subsequnce (xnk)

1k=1 so that kxn �

xnkk 2�k, for all k 2 N, and all n � nk. Put xn0 = 0 It follows thatP1j=1 xnj � xnj�1 is an absolute converging series, and therefore convergent

to some x 2 X Since xnk =Pk

j=1 xnj�xnj�1 , it follows that x = limk!1 xnkand since for any n 2 N, n � n1 and k 2 N such that nk < nk+1

kx� xnk kx� xnkk+ 2�k

it follows that the whole sequence (xn) converges to x.

The following argument shows how to get from a seminorm to a norm.

Proposition 7.1.4. Assume that on a Vector space X k · k is a seminormbut not necessarily a norm. We consider the following relation on X:

x ⇠ y () kx� yk = 0.

Then ⇠ is an equivalence class on X, the set of equivlaence classes

X/⇠=�x̄ : x 2 X

, where x̄ = {z 2 X : x ⇠ z}, for x 2 X,

is a vector space with

x̄+ ȳ = {x0 + y0 : x0 ⇠ x, y0 ⇠} = x+ y for x, y 2 X,

and� · x̄ = {�x0 : x0 ⇠ x} = �x for x 2 X and � 2 K.

128 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS

On X/ ⇠, the map

k · k : X/ ⇠! [0, infty), x̄ 7! kxk,

is well defined and a norm on X/ ⇠.

Example. Let (⌦,⌃, µ) be a measure space, and let

L1(µ) =nf : ⌦ ! K measruable : kfk1 =

Z|f | dµ < 1

o.

By Proposition 3.3.2 L1(µ) is vector space over K and k · k1 is a seminormon L1(µ), with the property that

kfk1 = 0 () f = 0, µa.e.

So put L1(µ) = L1(µ)/ ⇠, where we let f ⇠ g if f = g, µ-a.e.Then k · k1 is a norm on L1(µ), and from Corollary 3.3.6 it follows that

L1(µ) is complete.Formally speaking, the elements of L1(µ) are not functions but equiva-

lence classes of function. But we still think of them being functions, by tak-ing representants of each equivalence class. But then we mean by “f = g”,actually “f = g a.e”.

We now mention how to construct new linear normed spaces from givenones

Definition 7.1.5. (Complemented sums) Let (X, k · kX) and (Y, k · kY ) benormed linear spaces

ThenX⇥Y = {(x, y) : x2X and y 2 Y } is (using term by term additionand multiplication) and

k · k1 : X ⇥ Y ! [0,1), (x, y) 7! max(kxk, kyk),

is a norm. That space is often denoted by X �1 Y .X �1 Y is a Banach space if X and Y are Banach spaces.More generally if (X, k · kn) are normed linear spaces then

`1((Xn)n2N =�(xn)

1n=1 : xn 2 Xn, n2N, k(xn)k1 = sup

n2Nkxnk < 1

is a linear normed space with respect to the norm k · k1 , and Banach spaceif for all n 2 N (Xn, k · kn) is a Banach space.

7.1. NORMED LINEAR SPACES AND BANACH SPACES 129

Proposition 7.1.6. (Quotient spaces)Assume that X = (X, k · kX) is a normed linear space and that Y ⇢ X is aclosed subspace. Consider the quotient space

X/Y = {x+ Y : x 2 X}

(with usual addition and multiplication by scalars). For x 2 X put x =x+ Y 2 X/Y and define

kxkX/Y = infz2x

kzkX = infy2Y

kx+ ykX = dist(x, Y ).

Then k · kX/Y is norm on X/Y , and if X is a Banach space then X/Y is aBanach space.

Proof. For x1, x2 in X and � 2 K we compute

kx1 + x2kX/Y = infy2Y

kx1 + x2 + yk

= infy1,y22Y

kx1 + y1 + x2 + y2k

infy1,y22Y

�kx1 + y1k+ kx2 + y2k

�= kx1kX/Y + kx2kX/Y

and

k�x1kX/Y= inf

y2Yk�x1 + yk

= infy2Y

k�(x1 + y)k = |�| · infy2Y

kx1 + yk = |�| · kx1kX/Y .

Moreover, if kxkX/Y = 0, it follows that there is a sequence (yn) in Y ,for which limn!1 kx � ynk = 0, which implies, since Y is closed that x =limn!1 yn 2 Y and thus x = 0 (the zero element in X/Y ). This proves that(X/Y, k · kX/Y ) is a normed linear space.

Now assume that X is a Banach space. In order to show that X/Y iscomplete let xn 2 X with

Pn2N kxnkX/Y < 1. It follows that there are

yn 2 Y , n 2 N, so that1X

n=1

kxn + ynkX < 1,

and thus, since X is a Banach space,

x =1X

n=1

(xn + yn),

130 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS

exists in X and we observe that

���x�nX

j=1

xj

��� ���x�

nX

j=1

(xj + yj)���

1X

j=n+1

kxj + yjk !n!1 0,

which verifies that X/Y is complete.

We are now considering a linear map T : X ! Y between two normedlinear spaces X and Y . The following Proposition states conditions for T tobe continuous.

Proposition 7.1.7. If X and Y are two normed linear spaces, then fora linear map (we also say linear operator) T : X ! Y the following areequivalent:

a) T is continuous,

b) T is continuous at 0,

c) T is bounded, which means

kTk = supx2BX

kT (x)k = supx2X\{0}

���T (x)

kxk

��� < 1.

Proof. “(a) )(b) clear“(b) ) (c)” Assume that T is continuous at 0, then there is a � > 0 sothat for all x 2 X, kT (x)k < 1, whenever kxk � and thus, by linearitykT (x)k = 1�

1� for all kxk 1, which implies that kTk

1� .

“(c))(a)” Assume kTk = supx2BX kT (x)k < 1, and let (xn) ⇢ X withx = limn!1 xn. W.l.o.g we can assume that x 6= xn for all n 2 N

Then

kT (x)� T (xn)k = kT (x)� T (xn)k

= kx� xnk ·���T

⇣x� xx

kx� xnk

⌘��� kx� xnk · kTk !n!1 0.

Let X and Y be normed linear spaces. We put

L(X,Y ) =�T : X ! Y | linear and bounded

=�T : X ! Y | linear and continuous

.

7.2. LINEAR FUNCTIONALS 131

Often we call elements of L(X,Y ) bounded, linear operators.T 2 L(X,Y ) is called an isomorphic embedding if there is a c > 0 so that

kT (x)k � ckxk for all x 2 X.

and an isometric embedding if

kT (x)k = kxk for all x 2 X.

Isomorphic embeddings and isometric embeddings which are surjective arecalled isomorphisms, or onto isomorphisms, respectively isometries, or ontoisometries.

Proposition 7.1.8. Let X and Y be normed linear spaces.Then L(X,Y ) is a vector space (with respect to usual addition and mul-

tiplication of scalars of maps into vector spaces). and k · k as defined inProposition 7.1.7 (c) is a norm on L(X,Y ), and if Y is Banach space thenL(X,Y ) is also a Banach space.

Proof. Exercise.

7.2 Linear Functionals

Definition 7.2.1. For a normed linear space we put X⇤ = L(X,K). SinceK is a Banach space, by Proposition 7.1.8 X⇤ is a Banach space with thenorm

k · k : X⇤ ! [0,1), x⇤ 7! kx⇤k = supx2BX

|x⇤(x)|.

X⇤ is called the space of continuous, or bounded linear functionals on X.

Proposition 7.2.2. If X is a vector space over C, and f : X ! C is a linearfunctional on X. We can write f(x) =

132 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS

Proof. It is clear that

7.2. LINEAR FUNCTIONALS 133

Proof. We define

F =⇢(Z,F ) :

Z subspace of X, with Y ⇢ Z, F : Z ! R, linearF |Y = f and F (z) p(z) for all z 2 Z

�.

On F we consider the following order:

(Z1, F1) (Z2, F2) () Z1 ⇢ Z2 and F2|Z1 = F1.

We observe that F satisfies the assumption of Zorn’s Lemma. Indeed, ifL ⇢ F be linearly ordered subset, we put

Z0 =[�

Z : (Z,F ) 2 L for some linear F : Z ! R .

Since L is linearly ordered, Z0 is subspace (i.e., vectorspace) of X. Forz 2 Z0 we define F0(z) = F (z), where z 2 Z and (Z,F ) 2 L. Again sinceL is linearly ordered, F0 is well defined and since F0|Z , is linear for eachsubspace Z of X for which there is a linear F : Z ! R, with F (z) p(z),for z 2 Z, of X for which there is a linear F : Z ! R, with (Z,F ) 2 L, itfollows that F0 2 F , and that F0 is an upper bound of L.

By Zorn’s Lemma F has a maximal element (Z,F ). We need to showthat Z = X. We do this by assuming that Z ( X and x 2 X \Z, and provethat F can be extended to a linear functional

F̃ : Z + xR =�z + r · x : z2Z and r 2 R

! R,

with F̃ (z + rx) p(z + rx), for z 2 Z and r 2 R.

For z1, z2 2 Z it follows from the triangle inequality that

F (z1) + F (z2) = F (z1 + z2) p(z1 + z2) p(z1 � x) + p(z2 + x)

and thus

F (z1)� p(z1 � x) p(z2 + x)� F (z2).

Taking the sup over the right and the inf over the left side we obtain

supz2Z

F (z)� p(z � x) supz2Z

p(z + x)� F (z).

So we choose ↵ satisfying

(7.1) F (z1)� p(z1 � x) ↵ p(z2 + x)� F (z2),

134 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS

and define

F̃ (z + rx) = F (z) + r↵, whenever z 2 Z and r 2 R.

F̃ is linear and for z 2 Z and r > 0 we deduce from the properties of p andthe second inequality in (7.1)

F̃ (z + rx) = r�F (z/r) + ↵

�

r�F (z/r) + p((z/r) + x)� F (z/r)

�(using upper estimate of ↵)

= rp((z/r) + x) = p(z + rx).

And if r < 0 and letting s = �r, we deduce from the properties of p andthe second inequality in (7.1)

F̃ (z + rx) = s�F (z/s)� ↵

�

s�F (z/s)� (p((z/s)� x) + F (z/s))

�(using lower estimate of ↵)

= sp((z/s)� x)= p(z � sx) = p(z + rx).

Thus F̃ 2 F which is a contradiction, since we assumed that F was maximalin F

Note that if p is a seminorm on a vector spaceX, and f a linear functionalon X, with f(x) p(x), for all x 2 X, then �f(x) = f(�x) p(�x) = p(x)and thus |f(x)| p(x).

Theorem 7.2.5. (The Hahn-Banach Theorem, Complex Version)Let X be a complex vector space, p a seminorm on it, Y a subspace of X,and f a linear functional on Y , having the property that f(y) p(y), (andthus |f(y)| p(y)) for y 2 Y .

Then F can be extended to a linear functional on all of X with theproperty that F (x) p(x), for all x.

Proof. First apply the real version of the Hahn Banach Theorem 7.2.4 tothe real linear functional u =

7.2. LINEAR FUNCTIONALS 135

The following Theorem gathers some applications of the Hahn-BanachTheorem.

Theorem 7.2.6. Let X be a normed linear space.

1. If Y is a closed subspace of X and x 2 X \Y , there exists f 2 X⇤ suchthat f(x) 6= 0 and f |Y ⌘ 0. More precisely if � = infy2Y kx� yk > 0,we can choose f so that kfk = 1 and f(x) = �.

2. If x 2 X, x 6= 0, there exists f 2 X⇤, so that f(x) = kxk and kfk = 1.

3. The bounded linear functionals on X separate ponits of X.

4. If x 2 X define x̂ : X⇤ ! K, by x̂(x⇤) = x⇤(x). Then the map

c(·) : X ! X⇤⇤, x 7! x̂,

is an isometric embedding from X into X⇤⇤.

Proof. (1) For z = y + �x 2 Y + x ·K put g(z) = �� where � = infy2Y kx�yk > 0. Then f is linear f |Y ⌘ 0 and f(x) = �. Moreover, if � 6= 0, y 2 Y

|g(y + �x)| = |�|� |�| · ky · ��1 + xk = ky + �xk.

We can therefore apply Theorems 7.2.5 and 7.2.6 for p(·) = k · k and find anextension f 2 X⇤ of g.(2) We apply (1) to x and Y = {0}(3) If x 6= y are in X apply (2) to x� y.(4) It for x 2 X, kx̂k = supx⇤2X⇤,kx⇤k1 |x̂(x⇤)| = supx⇤2X⇤,kx⇤k1 |x⇤(x)| kxk Thus c(·) is well defined, it is also easy to see that it is linear, andmoreover, the above chain of inequalities show that c(·) 2 L(X,X⇤⇤) withkc(·)kL(X,X⇤⇤) 1.

By (2) there is for every x 2 X an x⇤ 2 SX⇤ so that x̂(x⇤) = x⇤(x) = kxk,which implies that c(·) is an isometry.

X⇤⇤ is a Banach space, and if X is a Banach space and its isometric

image under the map X 3 x 7! x̂ 2 X⇤⇤. Usually we identify X with itsimage under c(·), and think of X being a subspace of X⇤⇤.

We call a Banach space reflexive if that map is surjective.

Definition 7.2.7. Let X and Y be two Banach spaces and let T : X ! Ybe linear and bounded.

136 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS

Then

T⇤ : Y ⇤ ! X⇤, y⇤ 7! y⇤�T, i.e., T ⇤(y⇤)(x) = y⇤(T (x)), for y⇤ 2 Y ⇤ and x 2 X

is called the adjoint of T .

Proposition 7.2.8. Let X and Y be two Banach spaces and let T : X ! Ybe linear and bounded. Then the adjoint T ⇤ : Y ⇤ ! X⇤ is also linear andbounded and kTk = kT ⇤k.

If T is an (onto) isomorphism (onto isometry) then T ⇤ is an (onto)isomorphism (onto isometry) .

Proof. Exercise.

Now we will now formulate and prove the Geometric Version of the HahnBanach Theorem

Definition 7.2.9. Let K be a subset of a vector space X. A point x0 2 Kis called an internal point of K if for each x 2 X, there is some �0 > 0 forwhich x0 + � · x 2 K whenever |�| �0.

Let K ⇢ X for which 0 is an internal point of K. The gauge functionalfor K is the map:

pK : X ! [0,1), pK(x) = inf{� > 0 : x 2 � ·K} for x 2 X.

Proposition 7.2.10. Let K be a convex set of a vector space X that con-tains 0 as internal point.

Then the gauge function pK for K is sublinear and positively homoge-nous.

Moreover:

{x 2 X : pK(x) < 1} ⇢ K ⇢ {x 2 X : pK(x) 1}

Proof. If u, v 2 x and " > 0 is arbitrary, we find 0 < �u < pK(u) + " and0 < �v < pK(v) + ", so that u/�u 2 K and v/�v 2 K and thus

u+ v

�u + �v=

�u

�u + �v

u

�u+

�v

�u + �v

v

�v2 K,

which implies that pK(u + v) �u + �v pK(u) + pK(v) + 2", and, since," > 0 is arbitrary, pK(u+ v) pK(u) + pK(v).

Finally for � > 0 and v 2 x

pK(�v) = infnr > 0 :

�v

r2 K

o= � inf

nr

�:�v

r2 K

o= �pK(v).

7.2. LINEAR FUNCTIONALS 137

To prove the first “⇢” in the “Moreover” part, assume that x 2 X andpK(x) < 1. Then there is 0 < � < 1 so that x 2 �K and thus (x/�) 2 Ksince K is convex and since 0 2 K we deduce that

x = � · x�+ (1� �) · 0 2 K

.The second “⇢” is clear since for x 2 K it follows that 1 2 {� > 0 : x 2

K/�}.

Lemma 7.2.11. (The Hyperplane Separation Lemma) Let K1 and K2 betwo nonempty disjoint convex subsets of a linear space X, one of which hasan internal point. Then there is a nonzero linear functional F : X ! R forwhich

supx2K1

F (x) infx2K2

F (x).

Proof. Let x1 be an internal point of K1 and x2 an arbitrary point of K1and put

z = x2 � x1 and K = K1 �K2 + z = (K1 � x1)� (K2 � x2).

Then 0 is an internal point of K and z 62 K. Indeed if z = y1 � y2 + z, forsome y1 2 K1, and y2 2 K2 and thus 0 = y1 � y2 which is contradictionsince K1 \K2 = ;.

Let p = pK , be the gauge function for K. Since z 62 K it follows fromProposition 7.2.10 that pK(z) � 1. Let Y = span(z) = zR, and definef(rz) = r. It follows that f(rz) = r rpK(z) = pK(rz).

By the Hahn -Banach theorem we can extend f to a linear functional onX so that f(x) pK(x) for all x 2 X.

Now let x 2 K1 y 2 K2, then x� y+ z 2 K and thus pK(x� y+ z) 1and

F (x)� F (y) + F (z) = F (x� y + z) pK(x� y + z) 1

Since F (z) = 1 it follows that F (x) F (y).Since x 2 K1 y 2 K2, where arbitrary it follows that

supx2K1

F (x) infx2K2

F (x).