1 Inner Product Space - WordPress.com · All inner product spaces are normed spaces for the norm de ned by kxk= p hx;xi. Also, show that the triangle inequality for norm can be obtained

MAT641- MSC Mathematics, MNIT Jaipur

Functional Analysis-Inner Product Space and orthogonality

Dr. Ritu AgarwalDepartment of Mathematics,

Malaviya National Institute of Technology Jaipur

April 11, 2017

1 Inner Product Space

Inner product on linear space X over field K is defined as a function 〈.〉 : X ×X −→ Kwhich satisfies following axioms:For x, y, z ∈ X, α ∈ K,

i. 〈x+ y, z〉 = 〈x, z〉+ 〈y, z〉;

ii. 〈αx, y〉 = α〈x, y〉;

iii. 〈x, y〉 = 〈y, x〉;

iv. 〈x, x〉 ≥ 0, 〈x, x〉 = 0 iff x = 0.

An inner product defines norm as:

‖x‖ =√〈x, x〉

and the induced metric on X is given by

d(x, y) = ‖x− y‖ =√〈x− y, x− y〉

Further, following results can be derived from these axioms:

〈αx+ βy, z〉 = α〈x, z〉+ β〈y, z〉

〈x, αy〉 = α〈x, y〉〈x, αy + βz〉 = α〈x, y〉+ β〈x, z〉

α, β are scalars and bar denotes complex conjugation.

Remark 1.1. Inner product is linear w.r.to first factor and conjugate linear (semilin-ear) w.r.to second factor. Together, this property is called ‘Sesquilinear’ (means oneand half-linear).

Date: April 11, 2017

Email:[email protected] Web: drrituagarwal.wordpress.com

MAT641-Inner Product Space and orthogonality 2

Definition 1.2 (Hilbert space). A complete inner product space is known as HilbertSpace.

Example 1. Examples of Inner product spaces

i. Euclidean Space Rn, 〈x, y〉 = ξ1η1 + ...+ ξnηn, x = (ξ1, ..., ξn), y = (η1, ..., ηn) ∈ Rn.

ii. Unitary Space Cn, 〈x, y〉 = ξ1η1 + ...+ ξnηn, x = (ξ1, ..., ξn), y = (η1, ..., ηn) ∈ Cn.

iii. Function space L2[a, b], 〈x, y〉 =∫ ba x(t)y(t)dt, x(t) and y(t) are complex valued

functions (May be real valued also).

iv. l2 = {x = (ξ1, ξ2, ...) :∑∞

i=1 |ξi|2 <∞} is an IPS for 〈x, y〉 =∑∞

i=1 ξiηi.

2 Norm and Parallelogram equality

To prove triangle inequality for normed space, first we are required to prove the Schwarzinequality.

Theorem 2.1 (Schwarz inequality).

|〈x, y〉| ≤ ‖x‖‖y‖ (2.1)

where, equality holds if and only if {x, y} is linearly dependent set.

Proof. If y = 0, then (2.1) holds since 〈x, 0〉 = 0. Let y 6= 0. For every scalar α, wehave

0 ≤ ‖x− αy‖2 = 〈x− αy, x− αy〉= 〈x, x〉 − α〈x, y〉 − α[〈y, x〉 − α〈y, y〉]

(2.2)

Choosing α = 〈y,x〉〈y,y〉 so that the expression in brackets vanish. We get

0 ≤ 〈x, x〉 − 〈y, x〉〈y, y〉

〈x, y〉 = ‖x‖2 − |〈x, y〉|2

‖y‖2(Using〈y, x〉 = 〈x, y〉) (2.3)

Rearranging the terms, we get (2.1).Equality holds if ‖x− αy‖ = 0 i.e. x = αy, {x, y} are linearly dependent.

Theorem 2.2. All inner product spaces are normed spaces for the norm defined by‖x‖ =

√〈x, x〉. Also, show that the triangle inequality for norm can be obtained from

inner product as‖x+ y‖ ≤ ‖x‖+ ‖y‖ (2.4)

where, equality holds if and only if {x, y} is linearly dependent set x = cy, c ≥ 0 ory = 0.


Proof. (i) From axiom (iv) for inner product space, we have ‖x‖ =√〈x, x〉 ≥ 0 and

‖x‖ =√〈x, x〉 = 0 iff x = 0.

(ii) ‖αx‖ =√〈αx, αx〉 =

√αα〈x, x〉 =

√|α|2〈x, x〉 = |α| ‖x‖.

(iii) Triangle inequality

‖x+ y‖2 = 〈x+ y, x+ y〉 = 〈x, x〉 + 〈x, y〉 + 〈y, x〉 + 〈y, y〉= ‖x‖2 + 〈x, y〉 + 〈y, x〉 + ‖y‖2

(2.5)

Using Schwarz inequality (2.1)

‖x+ y‖2 ≤ ‖x‖2 + 2|〈x, y〉|+ ‖y‖2 ≤ ‖x‖2 + 2‖x‖ ‖y‖+ ‖y‖2 = (‖x‖+ ‖y‖)2(2.6)

Taking square root on both sides, we get(2.4).

Further, equality holds in this derivation if and only if

〈x, y〉+ 〈y, x〉 = 2‖x‖ ‖y‖

i.e. 〈x, y〉+ 〈y, x〉 = 2<〈x, y〉 = 2‖x‖ ‖y‖. But Schwarz inequality says that‖x‖ ‖y‖ ≥ |〈x, y〉|, so that <〈x, y〉 ≥ |〈x, y〉|.

Since the real part of a complex number cannot exceed the absolute value, we musthave equality, which implies linear dependence from (2.1), say y = 0 or x = cy.With the equality sign, we have <〈x, y〉 = |〈x, y〉|, and hence imaginary part must bezero.

If x = cy, <(〈x, y〉) = <(〈cy, y〉) = <(c〈y, y〉) ≥ 0 ⇒ c > 0.

Remark 2.3. Every Hilbert space is Banach space. However converse is not necessarilytrue.

2.1 Parallelogram equality

A norm on an inner product space satisfies the important Parallelogram equality

‖x+ y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2) (2.7)

Proof.

LHS = ‖x+ y‖2 + ‖x− y‖2

= 〈x+ y, x+ y〉+ 〈x− y, x− y〉= 〈x, x〉+ 〈x, y〉+ 〈y, x〉+ 〈y, y〉+ 〈x, x〉 − 〈x, y〉+ 〈y, y〉 − 〈y, x〉= 2(〈x, x〉+ 〈y, y〉)= 2(‖x‖2 + ‖y‖2)

(2.8)


Examples of normed spaces which are not inner product spaces:

Example 2. Sequence space lp, p 6= 2Consider x = (1, 1, 0, 0, 0, ....) and y = (1,−1, 0, 0, 0, ...) ∈ lp and calculate

‖x‖ = ‖y‖ = 21/p, ‖x+ y‖ = ‖x− y‖ = 2

Parallelogram equality is not satisfied. Hence lp, p 6= 2 is a Banach space but notHilbert space.

Example 3. Space C[a,b] is not an inner product space for ‖x‖ = maxt∈J |x(t)|, J =[a, b].Consider x(t) = 1 and y(t) = (t − a)/(b − a). We have, ‖x‖ = 1, ‖y‖ = 1 and‖x(t) + y(t)‖ = ‖1 + t−a

b−a‖ = 2, ‖x(t)− y(t)‖ = ‖1− t−ab−a‖ = 1.

Thus, Parallelogram equality is not satisfied.

Theorem 2.4 (Polarization identity). The inner product in term of induced norm isgiven by

1. For a real IPS X,

〈x, y〉 =1

4(‖x+ y‖2 − ‖x− y‖2) (2.9)

2. For a complex IPS

<〈x, y〉 =1

4(‖x+ y‖2 − ‖x− y‖2) (2.10)

and

=〈x, y〉 =1

4(‖x+ iy‖2 − ‖x− iy‖2) (2.11)

Proof is left as an exercise to the reader.

3 Continuity in IPS

Theorem 3.1 (Continuity). If in an inner product space xn → x and yn → y, then〈xn, yn〉 → 〈x, y〉.

Proof. Consider

|〈xn, yn〉 − 〈x, y〉| = |〈xn, yn〉 − 〈xn, y〉+ 〈xn, y〉 − 〈x, y〉| = |〈xn, yn − y〉|+ |〈xn − x, y〉|

Applying Schwarz inequality

|〈xn, yn〉 − 〈x, y〉| ≤ ‖xn‖ ‖yn − y‖+ ‖xn − x‖ ‖y‖ → 0 as n→∞

Hence inner product is a continuous function.


Definition 3.2 (Isomorphism). An isomorphism T of an inner product space (X, 〈.〉)onto an IPS (X, 〈.〉0) over the same field K of scalars is a bijective linear operatorT : X → X which preserves the values of the inner products i.e. 〈Tx, Ty〉0 = 〈x, y〉.

Exercise 3.3. Let Y be a subspace of a Hilbert space H.

1. Y is complete iff Y is closed in H.

2. If Y is a finite dimensional subspace then Y is complete.

3. Every subspace of a separable space is separable.

Theorem 3.4 (Minimizing Vector). Let X be an inner product space and Y 6= φ is aconvex subset which is complete (in the metric induced by the inner product) subset.Then for every given x ∈ X, there is an unique y ∈ Y such that

δ = infy′∈Y‖x− y′‖ = ‖x− y‖.

Proof: (Existence of y): Given δ = infy′∈Y ‖x− y′‖. By definition of infimum, thereexists a sequence (yn) in Y such that

δn = ‖x− yn‖ δn → δ as n→∞

We will show that (yn) is a Cauchy sequence in Y.Let x− yn = vn. Then

‖vn + vm‖ = ‖2x− (yn + ym)‖ = 2

∥∥∥∥x− (yn + ym)

2

∥∥∥∥But yn, ym ∈ Y , (yn + ym)/2 ∈ Y as Y is convex. Hence

‖vn + vm‖ ≥ 2δ. (3.1)

Now ‖yn− ym‖2 = ‖vn− vm‖2 = 2‖vn‖2 + 2‖vm‖2−‖vn− vm‖2 (by Parallelogram law).⇒ ‖yn− ym‖2 = 2‖x− yn‖2 + 2‖x− ym‖2−‖vn− vm‖2 ≤ 2δ2n + 2δ2m− 4δ2 (using (3.1)).As n,m→∞, δn → δ and δm → δ. Thus‖yn − ym‖ → 0 as n,m → ∞ i.e. (yn) is a Cauchy sequence. Since Y is complete,(yn)→ y ∈ Y . Thus existence of y is guaranteed.

(Uniqueness): Let y0 and y1 ∈ Y such that

δ = ‖x− y0‖ = ‖x− y1‖

. Now ‖y1− y0‖ = ‖(y1− x)− (y0− x)‖ ≤ 2‖y1− x‖2 + 2‖y0− x‖2−‖y1− x+ y0− x‖2(by Parallelogram law)⇒ ‖y1 − y0‖ ≤ 2δ2 + 2δ2 − 4δ2 → 0.But ‖y1 − y0‖ ≥ 0, we have ‖y1 − y0‖ = 0 i.e. y1 = y0.


4 Orthogonality

Definition 4.1 (Orthogonality). An element x of an inner product space X is said tobe orthogonal to an element y ∈ X if

〈x, y〉 = 0

and we write x⊥y.

Theorem 4.2 (Orthogonality). Let X be an inner product space and Y 6= φ is a convexcomplete (in the metric induced by the inner product) subset. Then for a fixed x ∈ X,z = x− y is orthogonal to Y where

‖x− y‖ = δ = infy′∈Y‖x− y′‖, y ∈ Y.

Proof: Let z ⊥ Y is false. Then there exist a y′ ∈ Y such that

〈z, y′〉 = β 6= 0

Clearly, y′ 6= 0 since otherwise 〈z, y′〉 = 0.Consider the norm

‖z − αy′‖2 = 〈z − αy′, z − αy′〉 = 〈z, z〉 − α〈z, y′〉 − α(〈z, y′〉 − α〈y′, y′〉)

Choose α = 〈z,y′〉〈y′,y′〉 = β

‖y′‖2 so that

‖z − αy′‖2 = ‖z‖2 − ββ

‖y′‖2< ‖z‖2 = δ2

We claim, that ‖z − αy′‖2〈δ2 can never happen. Since

‖z − αy′‖ = ‖x− y − αy′‖ = ‖x− αy1‖ ≥ δ

y1 = y+αy′ and δ = infy∈Y ‖x−y‖. This shows contradiction. Hence β = 0⇒ 〈z, y〉 = 0i.e. (x− y) ⊥ Y .

4.1 Orthogonal Compliments and Direct sums

Definition 4.3 (Direct Sum). A vector space X is said to be direct sum of two sub-spaces Y and Z of X, written X = Y ⊕ Z, if each x ∈ X has a unique representationx = y + z.Z is called algebraic complement of Y in X and vice versa and Y, Z is called a comple-mentary pair of subspaces in X.

Definition 4.4 (Anihilator). An orthogonal compliment of Y 6= φ in an IPS X is theset Y ⊥ = {x ∈ X : x ⊥ Y }, called anihilator.


Theorem 4.5. (i) Y ⊥ is a vector subspace.(ii) Y ⊥ is a closed subspace.(iii) Y ⊂ Y ⊥⊥.

Main concern is for a Hilbert space H, a closed subspace Y and its ’OrthogonalCompliment’

Y ⊥ = {z ∈ H|z ⊥ Y }

Theorem 4.6 (Projection theorem). Let Y be any closed convex subspace of a Hilbertspace H. Then H = Y ⊕ Y ⊥.

Proof: Since H is complete and Y is closed, Y is complete. By Theorem 4.2, forevery x ∈ H, there is a y ∈ Y such that

x = y + z z ∈ Z = Y ⊥

To prove uniqueness, assume that x = y + z = y′ + z′, where y, y′ ∈ Y and z, z′ ∈ Z.Since y−y′ ∈ Y and z− z′ ∈ Z, we see that y−y′ ∈ Y ∩Y ⊥ = {0}. This implies y = y′

and z = z′. Hence proved.

Definition 4.7 (Orthogonal Projection). Define a mapping P : H → Y i.e. Px = y. Pis called ‘Orthogonal Projection’ of H onto Y. Obviously P is a bounded linear operator.It maps H onto Y, Y onto itself and Z = Y ⊥ onto {0}.P is idempotent, i.e. P 2 = P .

Lemma 4.8 (Null space). The orthogonal compliment Y ⊥ of a closed subspace Y of aHilbert space H is the null space N(P ) of the orthogonal projection P of H onto Y .

Proof. Since Y is a closed subspace of H, H = Y ⊕ Y ⊥. Therefore any x ∈ H can bewritten as x = y+z where y ∈ Y and z ∈ Y ⊥. If P : H → Y is the projection mapping,Px = y, x ∈ H. In particular let x ∈ Y ⊥ then x = 0 + z = z since 〈x, y〉 = 0. HencePz = 0. z is arbitrary element of Y ⊥. Hence the theorem.

Lemma 4.9 (Closed subspace). If Y is a closed subspace of a Hilbert space H, thenY = Y ⊥⊥

Proof. To prove Y = Y ⊥⊥, we need to show that Y ⊂ Y ⊥⊥ and Y ⊃ Y ⊥⊥.Let x ∈ Y ⇒ x ⊥ Y ⊥ ⇒ x ∈ Y ⊥⊥. That is Y ⊂ Y ⊥⊥.

To show Y ⊃ Y ⊥⊥, let x ∈ Y ⊥⊥. Since Y ⊥⊥ is a vector space, y ∈ Y ⊂ Y ⊥⊥,z = x− y ∈ Y ⊥⊥. Hence z ⊥ Y ⊥ but z ∈ Y ⊥.Hence z = 0 so that x = y i.e. x ∈ Y .Since x is arbitrary, Y ⊃ Y ⊥⊥.

Lemma 4.10 (Dense Set (3.3-7)). For any subset M 6= φ of a Hilbert space H, the spanof M is dense in H if and only if M⊥ = {0}.


Proof. Let x ∈M⊥ and assume V = spanM to be dense in H. Then x ∈ V = H. Thus,there exists a sequence (xn) in V such that xn → x. Since x ∈ M⊥ and M⊥ ⊥ V , wehave 〈xn, x〉 = 0. The continuity of the inner product implies that 〈xn, x〉 → 〈x, x〉 =‖x‖2 = 0, so that x = 0. Since x ∈M⊥ is arbitrary, this shows that M⊥ = {0}.

Conversely, suppose that M⊥ = {0}. If x ⊥ V=Span M, then x ⊥M so that x ∈M⊥

and x = 0. Hence V ⊥ = {0}. H being Hilbert space, can be written as H = Y ⊕Y ⊥, Yis closed. Noting that V is subspace of H, taking Y = V , we thus obtain H = V .

5 Orthonormal sets and sequences

An orthogonal set M in an inner product space X is a subset M ⊂ X whose elementsare pairwise orthogonal.

An orthonormal set M in an inner product space X is a subset M ⊂ X is an orthog-onal set in X whose elements have norm 1. That is for all x, y ∈M

〈x, y〉 =

{0 x 6= y

1 x = y(5.1)

If an orthogonal or orthonormal set M is countable, we can arrange it in a sequence(xn) and call it orthogonal or orthonormal sequence.

For orthogonal elements x, y, we have 〈x, y〉 = 0 so that parallelogram equality leadsto

‖x+ y‖2 = ‖x‖2 + ‖y‖2 (5.2)

which is Pythogorean relation.More generally, if {x1, x2, ...xn} is an orthogonal set then

‖x1 + ...+ xn‖2 = ‖x1‖2 + ...+ ‖xn‖2 (5.3)

Lemma 5.1 (Linear independence). An orthonormal set is linearly independent.

Proof. Let {e1, e2, ..., en} be orthonormal. Consider the equation

α1e1 + α2e2 + ...+ αnen = 0

Taking inner product with ej, 1 ≤ j ≤ n, we get

〈n∑k=1

αkek, ej〉 =n∑k=1

αk〈ek, ej〉 = αj〈ej, ej〉 = αj = 0

This proves the linear independence for any finite orthonormal set.

Example 4. In the space R3 the basis (1,0,0), (0,1,0), (0,0,1) is an orthonormal basis.


Example 5. In Example 1(iv), e1, .., en, ... is the orthonormal sequence in l2 where

en = (0, 0, ..., 1︸︷︷︸nth position

, 0, ...)

Example 6. Let X be the IPS of all real valued continuous functions on [0, 2π] withinner product defined by

〈x, y〉 =

∫ 2π

0

x(t) y(t)dt (5.4)

An orthogonal sequences in X are un = cosnt and vn = sinnt. That is 〈un, um〉 = 0 forn 6= m. Also 〈vn, vm〉 = 0 for n 6= m. If we write e0 = u0/

√2π and en = cosnt√

π, en

sinnt√π

,

we get an orthonormal sequence of functions {e0, en, en}.

5.1 Advantage of orthonormal sequence

Expressing elements of IP space X in terms of orthonormal basis actually makes thedetermination of coefficients very easy as compared to arbitrary linearly independentset. Let e1, .., en, ... is the orthonormal sequence in X. Then x ∈ X can be written as

x =n∑k=1

αkek

〈x, ej〉 =n∑k=1

αk〈ek, ej〉 = αj

With these coefficients, we can write

x =n∑k=1

〈x, ek〉ek

Theorem 5.2 (Bessel inequality). Let (ek) be an orthonormal sequence in an innerproduct space X. Then for every x ∈ X

∞∑k=1

|〈x, ek〉|2 ≤ ‖x‖2 (5.5)

The inner product 〈x, ek〉 in (5.5) are called Fourier Coefficients of x with respectto orthonormal sequence (ek).

Proof. Let y =∑n

k=1 |αkek where n is fixed. Here αk = 〈y, ek〉 as (ek) is orthonormalsequence. We claim that for a particular choice of αk = 〈x, ek〉, k = 1, 2, ..., n, we shallobtain a y ∈ Y such that z = x− y ⊥ Y . By orthonormality,

‖y‖2 = 〈n∑k=1

〈x, ek〉ek,n∑j=1

〈x, ej〉ej〉 =n∑k=1

|〈x, ek〉|2


Using this, we can show that z ⊥ y ∈ Y as

〈z, y〉 = 〈x−y, y〉 = 〈x, y〉−〈y, y〉 =

⟨x,

n∑k=1

〈x, ek〉ek

⟩−‖y‖2 =

n∑k=1

〈x, ek〉〈x, ek〉−n∑k=1

|〈x, ek〉|2 = 0

Using the Pythagorean relation

‖x‖2 = ‖y‖2 + ‖z‖2

here, we get

‖z‖2 = ‖x‖2 − ‖y‖2 = ‖x‖2 −n∑k=1

|〈x, ek〉|2

Since ‖z‖2 ≥ 0, for n ∈ N, we have

‖x‖2 ≥n∑k=1

|〈x, ek〉|2

Gram-Schmidt orthogonalization process

6 Total orthonormal sets

Definition 6.1. A total set (or fundamental set) in a normed space X is a subsetM ⊂ X whose span is dense in X. Accordingly an orthonormal set (or sequence orfamily) in an IPS X which is total in X is called a total orthonormal set (or sequenceor family) in X.

Remark 6.2. M is total in X if and only if span M = X.

Remark 6.3. A total orthonormal family in X is sometimes called an orthonormal basisfor X but not in sense of algebra except for the finite dimensional spaces.

Lemma 6.4. In every Hilbert space H 6= {0}, there exists a total orthonormal set.

Proof: For a finite dimensional space, this is clear. For an infinite dimensionalseparable space H, it follows from Gram-Schmidt orthogonalization process. For anon-separable H, a non-constructive proof follows from Zorn’s lemma.

Definition 6.5 (Hilbert Dimension). All total orthonormal sets in a given Hilbertspace H 6= {0} have the same cardinality. The latter is called the ‘Hilbert dimension’or orthogonal dimension of H.

Theorem 6.6 (Totality). Let M be a subset of IPS X. Then


(a) If M is total in X, then there does not exist a non-zero x ∈ X which is orthogonalto every element of M. Briefly

x ⊥M ⇒ x = 0 (6.1)

(b) If X is complete, (6.1) is sufficient for the totality of M in X.

Proof. (a) Let H be completion of X [1, Theorem 3.2-3]. Then X is regarded as a densesubspace of H. By assumption, M is total in X so that span M is dense in X and hencedense in H. Lemma 4.10, orthogonal compliment of M in H is {0}. Therefore, if x ∈ Xand x ⊥M then x = 0.

(b) If X is a Hilbert space and M satisfies the given condition M⊥ = {0}, by Lemma4.10, the span of M is dense in X i.e. M is total in X.

Remark 6.7. The completeness of X in (b) is essential. If X is not complete, there maynot exist an orthonormal set M ⊂ X such that M is total in X. An example was givenby J. Diximier (1953).

Theorem 6.8 (Parseval relation). An orthonormal set M in a Hilbert space H is totalin H if and only if for all x ∈ H, the following Parseval relation holds

‖x‖2 =∑k=1

|〈x, ek〉|2 (6.2)

Proof. If M is not total, there is a non-zero x ⊥M in H. Since x ⊥M , in (6.2), we have〈x, ek〉 = 0 for all k, so that the RHS in (6.2) is zero whereas ‖x‖2 6= 0. This showsthat (6.2) does not hold.

Conversely assume that M is total in H. Consider any x ∈ H and its non-zero Fouriercoefficients arranged in a sequence 〈x, e1〉, 〈x, e2〉, .... We now define y as

y =∑k

〈x, ek〉ek, (6.3)

k may be finite or infinite as convergence follows from [1, Theorem 3.5-2].Let us show that x − y ⊥ M . Using the orthonormality, for every ej in (6.3), we

have〈x− y, ej〉 = 〈x, ej〉 −

∑k

〈x, ek〉〈ek, ej〉 = 〈x, ej〉 − 〈x, ej〉 = 0.

For every v ∈M not in (6.3), we have 〈x, v〉 = 0 and 〈ek, v〉 = 0 so that

〈x− y, v〉 = 〈x, v〉 −∑k

〈x, ek〉〈ek, v〉 = 0− 0 = 0.

Hence x− y ⊥ M i.e. x− y ∈ M⊥. Since M is total in H, M⊥ = {0}. This impliesx− y = 0 or x = y. Hence

‖x‖2 =

⟨∑k

〈x, ek〉ek,∑k

〈x, ek〉ek

⟩=∑k

〈x, ek〉〈x, ek〉 (6.4)

This completes the proof.


Theorem 6.9. Let H be a Hilbert space. Then

(a) If H is separable, every orthonormal set is countable.

(b) If H contains an orthonormal sequence which is total in H, then H is separable.

Proof. (a) Let H be separable

References

[1] Erwin Kreyszig, Introductory Functional Analysis With Applications, Wiley India Edition,1989.

[2] Walter Rudin, Functional Analysis, Tata Mc-Graw Hill, New Delhi.

[3] Balmohan Vishnu Limaye, Functional analysis, New Age International, 1996.

[4] Aldric Loughman Brown and A. Page, Elements of functional analysis, Van Nostrand-Reinhold, 1970.

[5] J.B. Conway, A Course in Functional Analysis, 2nd Edition, Springer, Berlin, 1990.

Contents

1 Inner Product Space 1

2 Norm and Parallelogram equality 22.1 Parallelogram equality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 Continuity in IPS 4

4 Orthogonality 64.1 Orthogonal Compliments and Direct sums . . . . . . . . . . . . . . . . . . . . . . . . . . 6

5 Orthonormal sets and sequences 85.1 Advantage of orthonormal sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

6 Total orthonormal sets 10

1 Inner Product Space - WordPress.com · All inner product spaces are normed spaces for the norm de ned by kxk= p hx;xi. Also, show that the triangle inequality for norm can be obtained

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