A STUDY ON INNER PRODUCT SPACES

A SEMINAR REPORT

SUBMITTED IN PARTIAL FULFILLMENT OF

THE REQUIREMENT FOR THE DEGREE

OF

MASTER

OF

MATHEMATICS

BY

HIMANSHI

Roll No.: 23001553011

UNDER THE SUPERVISION OF

PROFESSOR SANJAY KUMAR

DEPARTMENT OF MATHEMATICS

DEENBANDHU CHHOTU RAM UNIVERSITY OF

SCIENCE & TECHNOLOGY,

SONIPAT, HARYANA,INDIA

SESSION: 2023-25

1
 CERTIFICATE

This is to certify that work embodies in the seminar report “INNER PRODUCT SPACE” is
original and has been carried in the Department of the Mathematics, Deenbandhu Chhotu
Ram University of Science and Technology, Murthal -131039 by HIMANSHI ( Roll no. –
23001553011) for the degree of Master of Science in Mathematics under my guidance and
supervision and has never been submitted, in part or full, for the award of any degree or
diploma to this or any other University/Institution.

Dr. Sanjay Kumar

Supervisor

Department of Mathematics

Deenbandhu Chhotu Ram University of Science and Technology

Murthal -131039 (Sonipat)

Haryana (INDIA)

2
 ACKNOWLEDGMENTS
This is to acknowledge all those without whom this seminar report would not have been a
reality. Firstly, I would acknowledge my debt of gratitude to my supervisor Dr. Sanjay
Kumar, Professor, DCRUST, Murthal -131039 who gave me immense support, dedicated
time towards my project. He also gave guidelines and told about the various sources from
where the information could be gathered.

Also this work would have been a distant dream for me without the love and support of my
family. I have no words to express my overwhelming sense of gratitude to my respected
parents for their love and blessings throughout my all academic pursuits.

Finally I would like to thank GOD for the spiritual power that endowed me during this work.

DATE : _________________. NAME : HIMANSHI

ROLL NO.: 23001553011

3
 INDEX
SNo. Topic Page No.

1. Notation 5

2. Introduction 6

3. Preliminaries 7-8
Inner product space
Hilbert space
Orthogonal complement
Orthogonal complement of set
Orthonormal set

4. Example of Inner product space 9-10

5. Result.: Every IPS is a metric space 11

6. Projection Theorem 12-13

7. Bessel’s Inequality 14-15

8. Result.: Every finite dimensional vector 16-17

space is an inner product space.

9. The Conjugate space H* 18-19

10. Riesz – Representation Thm. for Hilbert 19-20

space

11. Bibliography 21

4
 NOTATIONS

∀. For all

R. Set of real number

⊥ perpendicular

||𝑇|| norm of a bounded linear operator T

∈ belongs to

⊂ proper subset of

∑ sigma

𝜙 empty set

‖ || normal

Sup. Supremum

Inf. Infimum

Rn. n dimensional Euclidean space

C[a,b]. Set of all continuous ftn. Defined on [a,b]

N. Set of all positive number

Lp space consisting of measurable ftn. defined on

measure spaces X.
lp linear space of all convergent sequence

(X,d). Metric space

5
 INTRODUCTION

In linear algebra, an inner product space is a vector space with an additional

structure called an inner product space. This additional structure associates
each pair of vectors in the space with the scalar quantity known as the inner
product of the vectors. Inner product allow the rigorous introduction of
intuitive geometrical notions such as the length of a vector or the angle
between two vectors. They also provide the means of defining orthogonality
between vectors( zero inner product). Inner product spaces generalize
Euclidean spaces ( in which the inner product is the dot product, also known
as the scalar product) to vector space of any ( possibly infinite) dimensions,
and are studied in functional analysis. The first usage of the concept of a
vector space with an inner product is due to Giuseppe Peano, in 1898.

6
 Preliminaries
Inner Product Space
Let V be a vector space over a field F. Let 𝑎, 𝑏 ∈ 𝐹 and 𝑢,𝑣, 𝑤 ∈ 𝑉 be arbitrary.
The vector space V is called an inner product space if there exists a function
<,> : 𝑉 × 𝑉 → 𝐹satisfying the following properties:

i. < 𝑢, 𝑣 > = < ̅𝑣̅,̅𝑢̅ >, i.e complex conjugate of < 𝑣, 𝑢 >
ii. < 𝑢, 𝑢 > ≥ 0 𝑎𝑛𝑑 < 𝑢, 𝑢 ≥ 0 𝑖𝑓𝑓 𝑢 = 0
iii. < 𝑎𝑢 + 𝑏𝑣, 𝑤 > = 𝑎 < 𝑢, 𝑤 > +𝑏 < 𝑣, 𝑤 >
The function <,> satisfying i,ii,iii is called an inner product on V.

Thus, vector space V with an inner product is called an inner product space .
Hare V is over a field F (where F = field of real or complex numbers).

# A real inner product space is called an Euclidean space , and a complex

inner product space is called unitary space.

Hilbert Space

Let X be a linear space over the complex field C .An inner product on X is a
function ( ): X×X → C which satisfies the following conditions.

(IP1) 𝛼𝑥 + 𝛽𝑦, 𝑧) = 𝛼( 𝑥, 𝑧) + 𝛽(𝑦, 𝑧) ∀ 𝑥, 𝑦, 𝑧 ∈ 𝑋 𝑎𝑛𝑑 𝛼, 𝛽 ∈ 𝐶 ( linearity in

the first variable).

(IP2) ̅(̅𝑥̅,̅𝑦̅) = ( 𝑦, 𝑥) where the bar denotes the complex conjugate.

(IP3) (𝑥, 𝑥) ≥ 0, (𝑥, 𝑥) = 0 ( positive definiteness).

A complex inner product space X is a linear space over C with an inner

product defined on it . We can also define inner product by replacing C by R in
the above definition. In that case , we get a real inner product space. Since the
theory of operators on a complex inner product space alone gives non – trivial

7
results in some important situations, we shall consider only complex inner
product space.

ORTHOGONAL COMPLEMENT
Definition :

Two vectors x and y in a Hilbert Space H are said to be orthogonal written as

𝑥⊥𝑦.

If (x,y) == 0 check this.

Orthogonal complement of a set

Let S be a non-empty set, then orthogonal complement of S denoted by 𝑆⊥ is
given by

𝑆⊥ = {𝑥: (𝑥, 𝑦) = 0 ∀ 𝑦 ∈ 𝑆}

Orthonormal Set

A subset S = {𝑢𝑖}of an inner product space V is said to be an orthonormal set if

1. < 𝑢𝑖, 𝑢𝑖 > = 1, 𝑖.𝑒 ||𝑢𝑖|| = 1 for all 𝑢𝑖 ∈ S.

2. < 𝑢𝑖, 𝑢𝑗 > = 0 𝑓𝑜𝑟 𝑖 ≠ 𝑗.

8
 EXAMPLE OF INNER PRODUCT SPACE:
EXAMPLE 01

Let u =(a1,a2,……..,an ) and v =( b1,b2 ,……..,bn ) be arbitrary members of Cn .

Define< 𝑢,𝑣 > = 𝑢. 𝑣= a1 𝑏̅ 1+a2 𝑏̅ 2+………….+an 𝑏̅ n.

Then Cn is an inner product space .

Solution:

i) < 𝑣̅̅,̅𝑢̅ > = ̅(̅𝑏̅1̅ 𝑎̅1̅ ̅+̅

̅ ̅𝑏̅2̅ ̅𝑎̅̅2̅+̅
̅ ̅⋯̅̅̅̅̅…̅̅+̅̅𝑏̅𝑛̅ ̅𝑎𝑛̅ ̅

= 𝑏̅1a1 + 𝑏̅ 2a2+ ………+ 𝑏̅ nan

= a1 𝑏̅1 + a2 𝑏̅2 + ………….+ an 𝑏̅ n

=< 𝑢, 𝑣 > ii) < 𝑢, 𝑢 > =a1 𝑎̅1+
a2 𝑎̅2+ ………+ an 𝑎̅n
= |a1|2+ |a2|2+………..+|an|2. ≥ 0
And < 𝑢, 𝑢 > = 0 𝑖𝑓𝑓 |a1|2+ |a2|2+………..+|an|2 = 0
Iff |a1 |=|a2 |= …..=|an | iff u= 0 iii) < 𝑎𝑢 + 𝑏𝑣, 𝑤 > =(aa1 + bb1 )
𝑐̅̅1+(aa2 + bb2 ) 𝑐̅̅2+ ……..+ (aan + bbn ) 𝑐̅̅n
Where w= (c1 ,c2 ,…cn )
= 𝑎(𝑎1 𝑐̅̅1 + 𝑎2𝑐̅̅2 + … + 𝑎𝑛 𝑐̅̅̅𝑛̅ ) + 𝑏(𝑏1 𝑐̅̅1 + 𝑏2 𝑐̅̅2 + ⋯̅̅̅ + 𝑏𝑛 𝑐̅̅̅𝑛̅
= 𝑎 < 𝑢, 𝑤 > + 𝑏 < 𝑣, 𝑤 >
Hence, Cn is an inner product space and the inner product defined is usually
referred as standard inner product on Cn.

EXAMPLE 02
If V be an inner product space then ,

9
(i) < 𝑎𝑢, 𝑣 > = 𝑎 < 𝑢, 𝑣 >
(ii) 𝑢, 𝑎𝑣 𝑢, 𝑣
(iii) 𝑢, 𝑏𝑣 𝑢, 𝑣 ̅ 𝑢, 𝑤 >
(iv) 𝟎, 𝑣
(v) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣 𝑉 =⟩ 𝑢 = 0 , 𝑤ℎ𝑒𝑟𝑒 𝑎,𝑏, 𝑐̅
𝐹 𝑎𝑛𝑑 𝑢,𝑣, 𝑤 𝑉 Proof:

(i) we know that for an inner product space, < 𝑎𝑢 + 𝑏𝑣, 𝑤

𝑢, 𝑤 > +𝑏 < 𝑣, 𝑤 >
Putting b= 0 in (1), we get
𝑢, 𝑤 𝑣, 𝑤 >
therefore ,<𝒂𝒖, 𝒗 𝒖, 𝒗
(ii) We have 𝑣,𝑎𝑢 ̅𝑎𝑢̅̅,̅𝑣̅ ̅𝑢̅,𝑣
̅ ̅>̅] = 𝑎̅ < ̅𝑢̅,̅𝑣̅ >

= 𝑣, 𝑢
<𝒗,𝒂𝒖 𝒗,𝒖 >
Interchanging v and u, we get
(iii) < 𝑢, 𝑎𝑣 > = 𝑎̅ < 𝑢,𝑣 >
< 𝑢, 𝑏𝑣 + 𝑐̅𝑤 > = < ̅𝑏𝑣̅̅̅+̅̅𝑐̅𝑤̅̅̅,̅𝑢̅ > = (̅𝑏̅<̅𝑣̅̅,̅𝑢̅>̅̅+̅̅𝑐̅̅<̅𝑤̅̅,̅𝑢̅>̅)
= 𝑏̅ < ̅𝑣̅,̅𝑢̅ > + 𝑐̅̅ < ̅𝑤,̅ ̅𝑢̅ >
= 𝒃̅̅ < 𝒖, 𝒗 > +𝒄̅̅ < 𝒖, 𝒘 >
< 𝟎, 𝑣 > = < 0. 𝟎, 𝑣 > = 0 < 𝟎, 𝑣 >
=0
(𝑣) < 𝑢,𝑣 > = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣 𝑉
< 𝑢, 𝑢 > = 0 [𝑏𝑦 𝑡𝑎𝑘𝑖𝑛𝑔 𝑣
𝒖=𝟎

10
 THEOREM
Every inner product space is a metric space.
Proof:
Let V(F) be an inner product space.
Let 𝛼 , 𝛽 ∈ 𝑉 be an arbitrary.
Define 𝑑(𝛼,𝛽) = ‖𝛼 − 𝛽‖ To
prove that d is a metric on V.
(iv) Since normal is always non – negative ,
∴ 𝑑(𝛼, 𝛽) ≥ 0
𝐴𝑙𝑠𝑜 𝑑(𝛼, 𝛽) = 0 𝑖𝑓𝑓 ‖𝛼 − 𝛽‖ = 0
i.e. < 𝛼 − 𝛽 , 𝛼 − 𝛽 > = 0 𝑖. 𝑒. 𝑖𝑓𝑓 𝛼 = 𝛽
∴ 𝑑(𝛼, 𝛽) = 0 𝑖𝑓𝑓 𝛼 = 𝛽
𝑑(𝛼, 𝛽) = ‖𝛼 − 𝛽‖ = ‖(−1)(𝛽 − 𝛼)‖ = |−1|.‖𝛽 − 𝛼‖
= 1. ‖𝛽 − 𝛼‖ = ‖𝛽 − 𝛼‖ = 𝑑 (𝛽 − 𝛼)
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 , 𝑑(𝛼, 𝛽) = 𝑑(𝛽, 𝛼)

(v) Let 𝛼, 𝛽, 𝛾 ∈ 𝑉
Then 𝑑(𝛼, 𝛾) + 𝑑( 𝛾,𝛽) = ‖ 𝛼 − 𝛾‖ + ‖𝛾 − 𝛽‖
≥ ‖ 𝛼 − 𝛾 + 𝛾 − 𝛽‖
= ‖𝛼 − 𝛽‖ = 𝑑(𝛼,𝛽)
i.e. 𝑑(𝛼, 𝛾) + 𝑑(𝛾, 𝛽) ≥ 𝑑(𝛼,𝛽) Hence,
V is a metric space.

11
 Projection Theorem
If M is a closed linear subspace of Hilbert space H, then show that

𝑯 = 𝑴 ⊕ 𝑴⊥
, where 𝑴⊥ is the set of all vector orthogonal to M .

Proof:

Since 𝑀 𝑎𝑛𝑑 𝑀⊥ are orthogonal closed linear subspace of H. Let 𝑧 be a limit

point of 𝑀 + 𝑀⊥ .It is sufficient to show that 𝑍 ∈ 𝑀 + 𝑀⊥ .

Let < 𝑍𝑛 > be a sequence of points in 𝑀 + 𝑀⊥ such that 𝑍𝑛 → 𝑍.

Since 𝑀 + 𝑀⊥ , therefore 𝑀 ∩ 𝑀⊥ = {0} , so each 𝑍𝑛 can be uniquely written in

the form

𝑍𝑛 = 𝑥𝑛 + 𝑦𝑛 , 𝑤ℎ𝑒𝑟𝑒 𝑥𝑛 ∈ 𝑀 𝑎𝑛𝑑 𝑦𝑛 ∈ 𝑀⊥
For each 𝜖 > 0 there exists a positive integer N s.t.

‖𝑍𝑚 − 𝑍𝑛‖ < 𝜖 ∀ 𝑛, 𝑚 ≥ 𝑁(𝜖)

=⟩ ‖𝑍𝑚 − 𝑍𝑛‖2 < 𝜖2 ∀ 𝑛, 𝑚 ≥ 𝑁(𝜖)

=⟩ ‖(𝑥𝑚 + 𝑦𝑚) − (𝑥𝑛 + 𝑦𝑛)‖2 < 𝜖2

=⟩ ‖(𝑥𝑚 − 𝑥𝑛) + (𝑦𝑚 − 𝑦𝑛)‖2 ≤ 𝜖2

Or,

‖𝑥𝑚 − 𝑥𝑛‖2 + ‖𝑦𝑚 − 𝑦𝑛‖2 ≤ 𝜖2

=⟩ ‖(𝑥𝑚 − 𝑥𝑛‖2 < 𝜖2 , ‖𝑦𝑚 − 𝑦𝑛‖2 < 𝜖²

Or,

‖(𝑥𝑚 − 𝑥𝑛‖ < 𝜖, ‖(𝑦𝑚 − 𝑦𝑛‖ < 𝜖 ∀ 𝑛, 𝑚 ≥ 𝑁(𝜖)

Therefore < 𝑥𝑛 > 𝑎𝑛𝑑 < 𝑦𝑛 > are cauchy sequence in

12
𝑀 𝑎𝑛𝑑 𝑀⊥respectively. But 𝑀 𝑎𝑛𝑑 𝑀⊥ are complete being closed subspace of
Hilbert space H. So there exists vectors in x and y in M and 𝑀⊥ respectively s.t.

𝑥𝑛 → 𝑥 𝑎𝑛𝑑 𝑦𝑛 → 𝑦
Now,

𝑍 = lim 𝑧𝑛 = lim (𝑥𝑛 + 𝑦𝑛) = 𝑥 + 𝑦 ∈ 𝑀 + 𝑀⊥

𝑛→∞ 𝑛→∞

Therefore 𝑀 + 𝑀⊥ is closed subspace of H.

Moreover since 𝑀 ⊥ 𝑀⊥ , 𝑠𝑜 𝑀 ∩ 𝑀⊥ = {0}

It remains to prove that 𝐻 = 𝑀 + 𝑀⊥

So 𝑀 + 𝑀⊥ is a proper closed linear subspace of H and hence there exists a

non zero vector 𝑍0 𝑠. 𝑡.𝑍0 ⊥ (𝑀 + 𝑀⊥ )

Which is possible only when 𝑍0 ⊥ 𝑀 𝑎𝑛𝑑 𝑍0 ⊥ 𝑀⊥.

i.e. when 𝑍0 ∈ 𝑀⊥ 𝑎𝑛𝑑 𝑍0 ∈ (𝑀⊥)⊥

But 𝑀⊥ ∩ (𝑀⊥)⊥ = {0}.𝑆𝑜 𝑍0 = 0

A contradiction because 𝑍0 ≠ 0

∴ 𝐻 = 𝑀 + 𝑀⊥ 𝑎𝑛𝑑 𝑀 ∩ 𝑀⊥ = {0}

∴ 𝐻 = 𝑀 ⊕ 𝑀⊥ .
Hence proved.

13
 BESSEL’s INEQUALITY
Theorem:

If < 𝒆𝒊 > is an orthogonal set in Hilbert space H, then

|𝒙, 𝒆𝒊|𝟐 ≤ ||𝒙||𝟐

For every vector 𝒙 ∈ 𝑯 .

Proof:

Let S = { 𝑒𝑖, (𝑥, 𝑒𝑖) ≠ 0}

If S is empty, then we define ∑|𝑥, 𝑒𝑖|2 to be the number zero and the result is
obvious in this case. We now assume that S is non- empty, then by the
theorem (which state that if < 𝑒𝑖 > is an orthonormal set in a Hilbert space H.
Then the set S= {𝑒𝑖, (𝑥, 𝑒𝑖) = 0} is an either empty or countable) it must be
finite or countably infinite. If S is , then it can be written in the form

𝑆 = { 𝑒1, 𝑒2, … . ,𝑒𝑛} for some positive integer n. In this case,we define
∑|𝑥, 𝑒𝑖|2 𝑡𝑜 𝑏𝑒

| . Then inequality to be proved now reduces to

|𝑥, 𝑒𝑖|2 ≤ ‖𝑥‖2.

Now consider the case

𝑆 = [𝑒𝑖 ∶ (𝑥, 𝑒𝑖) ≠ 0] is countably infinite.

Let the vectors in S be arranged in a definite order.

𝑆 = [𝑒1,𝑒2, 𝑒3, … … … … , 𝑒𝑛, … … … … … … … ]

14
By the theory of absolutely convergent series, if |𝑥, 𝑒𝑖|2 converges, then
every series obtained from it by rearranging it’s terms and also converges and
all such series have the same sum . We therefore, define

∑|𝑥, 𝑒𝑖|2 𝑡𝑜 𝑏𝑒 ∑ 𝑖 =∞1 |𝑥, 𝑒𝑖|2 and it follows that ∑ 𝑖 =∞1 |𝑥, 𝑒𝑖|2 is a non
negative extended real number which depends only on S and not on the
arrangement of its vectors . We now observe that

∑|𝑥, 𝑒𝑖| |𝑥, 𝑒𝑖|2

= lim ∑ |𝑥, 𝑒𝑖|

𝑖 =∞1 2

𝑛→∞ ≤ lim‖𝑥‖2 =

‖𝑥 ‖ 2
𝑛→∞ Hence

∑|𝑥, 𝑒𝑖|2 ≤ ||𝑥||2 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑥 ∈ 𝐻 .

15
 THEOREM

Every finite dimensional vector space is an inner product space.

Proof:

Let V(F) be a finite dimensional vector space over a field F (real or complex )
such that din V = n.

Let S = {𝑢1, 𝑢2, 𝑢3, … … … . , 𝑢𝑛} be basis of V.

Then each vector of V can be expressed as a linear combination of the vector

of S.

Let 𝑢, 𝑣, 𝑤 ∈ 𝑉 be an arbitrary.

Then 𝑢 𝑎𝑖𝑢𝑖, 𝑣 𝑏𝑖𝑢𝑖, 𝑤 𝑐̅𝑖𝑢𝑖 𝑤ℎ𝑒𝑟𝑒 𝑎𝑖, 𝑏𝑖, 𝑐̅𝑖 ∈ 𝐹 .

Define < 𝑢, 𝑣 𝑎𝑖 𝑏̅𝑖

Let us now check that < , > satisfies all the axioms of an inner product space.

(𝑖) < ̅𝑢̅,̅𝑣̅ > = < 𝑣, 𝑢 >

< ̅𝑢̅,̅𝑣̅ > = (̅̅∑̅𝑖̅=̅𝑛̅1̅̅𝑎̅𝑖̅ 𝑏̅̅𝑖) = ∑ 𝑖 =𝑛 1 ̅𝑎̅𝑖̅ ̅𝑏̅̅𝑖 = ∑ 𝑖 =𝑛 1 𝑎̅𝑖𝑏𝑖 = ∑ 𝑖 =𝑛 1 𝑏𝑖 𝑎̅𝑖

= < 𝑣, 𝑢 >

(𝑖𝑖) < 𝑢, 𝑢 > = 0

𝑖 =𝑛 1 𝑛 2≥ 0 [∴ |𝑎𝑖| ≥ 0 𝑓𝑜𝑟 𝑒𝑎𝑐̅ℎ 𝑖

< 𝑢, 𝑢 > = ∑ 𝑎𝑖 𝑎̅𝑖 = ∑ 𝑖 = 1 |𝑎𝑖|

16
< 𝑢, 𝑢 > = 0 𝑖𝑓𝑓 𝑢 = 0

𝑁𝑜𝑤 , < 𝑢, 𝑢 > = 0 𝑖𝑓𝑓

|𝑎1|2 + |𝑎2|2 + ⋯̅̅̅ … … … … . . +|𝑎𝑛|2 = 0

|𝑎𝑖| = 0 ∀ 𝑖

𝑎𝑖 = 0 ∀ 𝑖
𝑛 𝑛
𝑢 = ∑ 𝑖 = 1 𝑎𝑖𝑢𝑖 = ∑ 𝑖 = 1 𝑎𝑖𝑢𝑖 =⟩

u=0 .

(𝑖𝑖𝑖) < 𝑎𝑢 + 𝑏𝑣, 𝑤 > = 𝑎 < 𝑢, 𝑤 > + 𝑏 < 𝑣, 𝑤 > ∀ 𝑎, 𝑏 ∈ 𝐹

< 𝑎𝑢 + 𝑏𝑣, 𝑤

𝑎𝑎𝑖 𝑐̅̅𝑖 𝑏𝑏𝑖𝑐̅̅𝑖

𝑎𝑖𝑐̅̅̅𝑖̅ 𝑏𝑖 𝑐̅̅𝑖
= 𝑎 < 𝑢, 𝑣 > + 𝑏 < 𝑣,𝑤 >
From (i),(ii) and (iii) it follows that <,> is an inner product on V(F) and hence
the finite dimensional space V(F) is an inner product space.

17
 THE CONJUGATE SPACE 𝑯∗

Let H be a Hilbert space and H* , it’s conjugate space .

Let y be a fixed vector in H , Define a function 𝑓𝑥 on H by

𝑓𝑦(𝑥) = (𝑥, 𝑦) ∀ 𝑥 ∈ 𝐻

We assert that 𝑓𝑦 is linear, for 𝑓𝑦(𝑥1 + 𝑥2) = (𝑥1 + 𝑥2, 𝑦) ∀ 𝑥1, 𝑥2 ∈ 𝐻

= (𝑥1, 𝑦) + (𝑥2, 𝑦)

= 𝑓𝑦(𝑥1) + 𝑓𝑦(𝑥2)

And 𝑓𝑦(𝛼𝑥) = (𝛼𝑥, 𝑦)

= 𝛼(𝑥, 𝑦) = 𝛼 (𝑓𝑦(𝑥))

𝐴𝑙𝑠𝑜 |𝑓𝑦(𝑥)| = |𝑥, 𝑦| ≤ ||𝑥|| ||𝑦|| (By Schwartz’s Inequality)

Which proves that |𝑓𝑦(𝑥)| ≤ ||𝑦||

Which implies that 𝑓𝑦 is a cont. Thus 𝑓𝑦 is linear and cont. mapping and
hence is a linear functional on H . On the other hand if y = 0, then

𝑓𝑦(𝑥) = (𝑥, 0) = 0 =⟩ ||𝑓𝑦|| ≤ ||𝑦||

If y ≠ 0 , then

||𝑓𝑦|| = sup {|𝑓𝑦(𝑥)|;||𝑥|| = 1

𝑦
≥ |𝑓𝑦 (| |𝑦||)|

18
 𝑦
≥ |( ,𝑦)|
||𝑦|
Hence,

||𝑓𝑦𝑦|| = ||𝑦||

Thus for each 𝑦 ∈ 𝐻. There is a linear functional 𝑓𝑦 ∈ 𝐻∗ such that

||𝑓𝑦|| = ||𝑦||

Hence the mapping 𝑦 → 𝑓𝑦is a normal preserving mapping of H into 𝐻∗.

RIESZ- REPRESENTATION THEOREM FOR HILBERT SPACE:

THEOREM:
Let H be a Hilbert space and let f be an arbitrary functional in 𝑯∗. Then
there exists a unique vector y in H such that 𝒇(𝒙) = (𝒙, 𝒚) 𝒇𝒐𝒓 𝒆𝒗𝒆𝒓𝒚 𝒙 ∈
𝑯.
Proof:
We shall show first that if such a y exists, then it is necessarily unique.

Let 𝑦′ be another vector in H such that 𝑓(𝑥) = (𝑥,𝑦′). Then clearly (𝑥, 𝑦) = (𝑥,
𝑦′) i.e.(𝑥, 𝑦 − 𝑦′) = 0 ∀ 𝑥 ∈ 𝐻. Since zero is the only vector .

Orthogonal to every vector, this implies that 𝑦 − 𝑦′ =

0 𝑤ℎ𝑖𝑐̅ℎ 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑡 𝑦′ = 𝑦.
Now we turn to the existence of such vector y. If 𝑓=0 , then it clearly suffice to
choose y = 0. We may therefore assume that 𝑓 ≠ 0. The null space

𝑀 = {𝑥 ∈ 𝐻 ; 𝑓(𝑥) = 0} is thus a proper closed linear subspace of H and

therefore there exists a non – zero vector 𝑦0 ∈ 𝐻 which is orthogonal to M. We
show that if a is a suitably chosen scalar, then the vector 𝑦 = 𝛼𝑦0 meets our
requirements. If 𝑥 ∈ 𝑀, then whatever values of 𝛼 may be, we have

19
𝑓(𝑥) = (𝑥, 𝛼𝑦0) = 0

We now choose 𝑥 = 𝑦0. Then we must have

2
𝑓(𝑦0) = (𝑦0, 𝛼𝑦0) = 𝛼̅(𝑦0, 𝑦0) = 𝛼̅||𝑦0|| Therefore, it

follows that the vector

̅𝑓̅(̅𝑦̅0̅)
𝛼= 2 .𝑦0
||𝑦||
Satisfies the required condition for each

𝑥 ∈ 𝑀 𝑎𝑛𝑑 𝑓𝑜𝑟 𝑥 = 𝑦0. Each x in H can be written in the form

𝑥 = 𝑚 + 𝛽𝑦0 ,𝑚 ∈ 𝑀. For this all that necessary is to choose 𝛽

In such a way that 𝑓(𝑥 − 𝛽𝑦0) = 𝑓(𝑥) − 𝛽𝑓(𝑦0) = 0 and this is justified by
putting
𝑓(𝑥)
𝛽=
𝑓(𝑦0)

Now we show that the conclusion of the theorem holds for each x in H .

For this, we have

𝑓(𝑥) = 𝑓(𝑚 + 𝛽𝑦0)

= 𝑓(𝑚) + 𝛽𝑓(𝑦0)

= (𝑚,𝑦) + 𝛽(𝑦0,𝑦) = (𝑚 + 𝛽𝑦0,𝑦) = (𝑥,𝑦)

20
 BIBLIOGRAPHY
1. Somsundaram .D, A first course in functional analysis , Narosa
publishing House ,2006
2. Kreyszig Erwin , Introductory functional Analysis with Application, John
Wiley ,1989.
3. Royden H.L., Real Analysis, MacMillan Publishing Co.,Inc, New York,4th
Edition,1993.
4. Pundir S.K., Functional Analysis ,CBS publication, 2016

21