# A Study of Operators on Hilbert Space

Mainak Bhowmik

Indian Institute of Technology, Kanpur 208016

Dr. Jaydeb Sarkar

Indian Statistical Institute, Bengaluru 560059

## Abstract

In this project work I have studied functional analysis from the beginning. Although I have learned general Banach Space theory I have given more importance on Hilbert Spaces and different types of operators on Hilbert spaces. The main goal of my project work is to study the Compact self-adjoint operators (especially the spectral theorem).

## Abbreviations

Abbreviation
 NLS Normed Linear Space bdd Bounded B Open unit ball in the coressponding Banach space HBT Hahn Banach Theorem UBP Uniform Bounded Principle

Unless stated I have considered X and Y to be Banach spaces.

## INTRODUCTION

In Linear Algebra we have mainly studied the Finite dimensional space, generalizing the 2-dimensional Euclidean geometry. But in functional analysis mostly we are concerned with infinite dimensional spaces like different sequence spaces, measure spaces etc. In finite dimensional case a linear operator is one-one iff it is onto. But in general normed linear space it is not true. Here we have to study the spectrum of an operator not only the eigenvalues. So there are lots of differences. That is why we can't simply generalize the Spectral theorem for normal operators from finite dimensional case to infinite dimensional. So we should know the version of spectral theorem in infinite dimensional Hilbert spaces. I have studied it for (a simple case) Compact self-adjoint operators only. But in advanced study we can learn it for normal operators.

## Banach Space

Definition: Suppose (X, ||.|| ) be a normed linear space (NLS). Let d(x,y) =|| x-y || for all x,y ϵ X. Then (X,d) is a metric space. If X is complete with this metric induced by the norm then the NLS is called a Banach space. For example C[0,1] with sup-norm, etc. are Banach spaces.

Hamel basis and Schauder basis: A linearly independent subset B of a vector space X, is aid to be a Hamel basis for X if every element of it is a linear combination of elements of B. A sequence {Yn} in a Banach space is said to be schauder basis for X if every element x has a unique representation x=$\sum_{n=1}^\infty Y_n$, where the series converges if the n-th partial sum converges in X.

Riesz's Lemma: Let M be a proper closed linear subspace of a normed linear space X. Then for each 0 < t < 1, there exists a unit vector $x_t$ in X s.t. dist($x_t$,M) $\geq$t. (where dist($x_t$,M) =inf {$\parallel x-m\parallel:m\in M$} )​

Theorem: In any infinite dimen​sional NLS the closed unit ball $\{\;x:\parallel x\parallel\leq1\}$ can not be compact.

We already know that closed unit ball in finite dim. NLS is compact. Therefore a NLS is finite dimensional if it is locally compact. This famous theorem was proved by Riesz.

Operator Norm : Let X and Y be two NLS and A is a linear map from X to Y. Let . If M is finite, we say that A is a bounded linear operator and A is called the Operator norm of A. The space of all bounded linear operators from X to Y is denoted by B(X,Y). This is a NLS with norm $\parallel A\parallel :=\underset{\parallel x\parallel =1}{sup}\parallel Ax\parallel .$​​

Theorem: be a linear operator, where X, Y are NLS.

## The Hahn-Banach Theorem

Let X be a normed linear space. Let ${X}_{0}$ be a subspace of it and let ${f}_{0}$ be a linear functional on ${X}_{0}$ such that . Then there exist a linear functional f on X such that . Actually the theorem says that a linear functional on ${X}_{0}$ can be extended to X without increasing its norm.

Corollaries: (i) Let $X_0$ be a subspace of a NLS and $x_1$be a vector such that dist ($x_1\;,X_0$) =$\delta$>0. Then there exists a linear functional f on X such that $\parallel f\parallel=1\;,f(x_1)=\delta\;and\;f(x)=0\;\;\;\forall x\in X_0$

(ii) For each non zero $x_0\in X\;,$ there exists a linear functional f on X such that $\parallel f\parallel=1\;and\;f(x_0)=\parallel x_0\parallel$. This shows that norm of x can be expressed as x=supfX,f=1|f(x)|. Hence we can seperate two point

## The Uniform Boundedness Principle

Let X be Banach space and let {$A_\alpha$} be a family of bounded linear operators from X to another normed linear space Y. Suppose for each xX   supα Aαx<. Then supα Aα<​.

Application:- Let $\{f_n\}$be a bounded linear functionals on Banach space X. Suppose for each x , $\{f_n(x)\}$converges to a limit f(x). Then f is a bounded (and hence continuous ) linear functional. (It is to be noted that in general the pointwise limit of continuous functions is not continuous).

## The Open Mapping Theorem

Let X and Y be two Banach Spaces and $A:X\rightarrow Y\;$be a bounded linear operator. If A is surjective then it is a open map.

One of the important consequence of this theorem is the Inverse Mapping theorem : If A is as above theorem and it is bijective then A being open map, $A^{-1}$is continuous and hence bounded linear operator.

## Dual spaces of some special Banach spaces

We know that the set of linear map from a vector space to it underleying scalar field is the Dual space of that vector space. But Here we will conider only the bounded linear functionals and this collection of bounded linear functionals on the Banach space X will be denoted by X*.

Examples:

i) for finite dimensional vector space X , X* $\cong$X.

ii) Conider the $l_p(N)\;space\;where\;1\leq p<\infty$ . If q be such that $\frac1p+\frac1q=1$ then the dual space $(l_p(N))^\ast$ is isomorphic to $l_q(N)$. iii) similarly for $L_p\lbrack0,1\rbrack\;$ $1\leq p<\infty$ . If q be such that $\frac1p+\frac1q=1$ then $(L_p\lbrack0,1\rbrack)^\ast\cong\;L_q\lbrack0,1\rbrack$

These dual spaces are very important. We shall use them when we shall discuss adjoint of an operator.

## The Weak and Weak* topology

We can define topolgies on a NLS other than the usual norm topology.

Weak topology : Let Y be a subset of X and $\left(X,\tau\right)$be a topolgical space. Let $\mathcal A$ be family of maps from Y into X . The weak topology on Y generated by $\mathcal A$ i​s the smallest topology on Y for which all $f\in\mathcal A\;$are continuous. The collection $\{{\bigcap\nolimits^k}_{j=1}\;f_j(U_j):U_j\in\tau,\;f_j\in\mathcal A,1\leq j\leq k,\;k=1,2,.....\}\;$ forms the bae for the weak topology on Y .

Let X be a Banach space and X* be its dual space. The weak topology generated by X* is called the weak topology on X.

Different types of convergence : A sequence {$x_n$} converges to $x$ in the norm or strong topology if $\parallel x_n-x\parallel\rightarrow0\;.\;we\;write\;x_n\rightarrow x.$

We say that {$x_n$} converges to $x$in the weak topology iff . We denote thi​s convergence as $x_n\overset w{\rightarrow x}$.

Notes:

i) The norm toplogy on X is stronger than the weak topology.

ii) The weak topology on X is a Haussdorff topology.

iii) If Also if the sequence converges weakly the it is bounded.

Weak* topology: Let X be a Banach space. Then dual of X* is again a Banach space X **(second dual of X). Let $J:X\rightarrow X^{\ast\ast}\;\;be\;given\;by\;J(x)=F_x\;,\forall x\in X$. where$F_x\in X^{\ast\ast}\;defined\;as\;F_x(f)=f(x)\;\forall f\in X^\ast$. Then one can show J is a linear map and $\parallel Jx\parallel=\parallel x\parallel$. Thus J is an isometric imbedding of X into X​**.

If the map J is surjective then X is iomorphic to X** via the map J and we say that X is reflexive.

The usual topology on X* is generated by its norm and its weak topology is the weak topology generated by X**. Now consider the weak topology on X* genetrated by the subspace X (considering as a subspace of X** via the natural imbedding ) of X** is called the weak* topology on X*.

## Concepts

Hilbert space is a Banach space with some special kind of norm. Let ( X ,<.>) be an inner product space. Now define $\parallel x\parallel:=\sqrt{}$$<$$x,x$$>^{1l2}$. Then $\parallel.\parallel$defines a norm on X. If X is a complete normed linear space then we say it is a Hilbert space.

Example: $C^n\;$is a inner product space with the usual definition <x,y> $=\sum_{j=1}^nx_j\overline{y_j}\;\;\forall x,y\in C^n$and it is a Hilbert space with the norm induced by the inner product. The space $l_2$with the inner product <x,y> $=\overset\infty{\underset{j=1}{\sum x_j\overline{y_j}}}$is also a Hilbert space. The space $L_2\lbrack a,b\rbrack\;$is a Hilbert space with the inner product <f,g> =$\int_a^bf(x)\overline{g(x)\;}dx$.

Cauchy Schwarz inequality: |<$x,y$> | $\leq\parallel x\parallel\parallel y\parallel$

Parallelogram Law: $\parallel x+y\parallel^2+\parallel x-y\parallel^2=2(\parallel x\parallel^2+\parallel y\parallel^2).AAAdfdfcxvvvxvvddvddfdfgfdbggdf$

Any inner product satifies these two. The norm and inner product is related by the Polariation Identity: <$x,y$>$=\frac14{\textstyle{\displaystyle\sum_{p=1}^4}i^p}\parallel x+i^py\parallel$. Now the question comes to our mind is that: When a Banach space is a Hilbert space? The anwer is the following

Remark: $l_p$ spaces is Hilbert space only for p=2. Take p to be other than 2. consider x=(1,0,0,.....) and y=(0,1,0,0,0,......)​ then x+y=(1,1,0,0,......) and x-y=(1,-1,0,0,........). Therefore $\parallel x+y\parallel^2+\parallel x-y\parallel^2=2^{1+\frac2p}\;and\;\;\;2(\parallel x\parallel^2+\parallel y\parallel^2)=4\;\;\;so\;2^{1+\frac2p}=4\;only\;when\;p=2.$ Hence the parallelogram law is satisfied ​only when p=2. Similaly one can show $L_p\lbrack a,b\rbrack$ is a Hilbert space only for p=2.

## Some basic results on Orthogonal complement of a set

Let $S$ be any subset of a Hilbert space H. Orthogonal complement of S is the set $\{x\in H\;:\;x\perp y\;for\;all\;y\in S\}$ denoted by $S^\perp$

Results:

i) $S^\perp$is a closed linear subspace of H.

ii) $S\cap S^\perp\subset\{0\}$

iii)$\{0\}^\perp=H\;and\;H^\perp=\{0\}$

iv) If $S_1\subset S_{2\;}then\;S_2^\perp\subset S_1^\perp$. Also $S\subset S^{\perp\perp}$

In finite dimensional inner product space the above containments are actually equalities. But in general we can give example for which the containments are proper. Consider the $l_2$space S=linear span {(1,0,0,....),(1,1/2,0,0,....),(1,1/2,1/3,0,0,....),.....(1,1/2,...,1/n,0,0,...),......}. Then (0,0,.....)$\in$$S^\perp$but (0,0,0.....) $\not\in$so $S\cap S^\perp=\varnothing$. Using this example we can show that the containment in the last part of (iv)​ may be proper.

## Orthogonal projection

Theorem: Let K be a nonempty closed convex subset of Hilbert space H and $h_0\in H\backslash\;K.$ Then there exists a unique vector $h_\ast\in K\;s.t.\;\parallel h_0-h_\ast\parallel=dist(h_0,K)$.

The theorem says that each point in H has a unique best approximation from any given closed convex set K. This is not true in general Banach space. The result becomes more powerful when K is a closed linear subspace of H. In that case $h_0-h_\ast\;\perp K.$ Conversely if $h_\ast\in K\;s.t.\;h_0-h_\ast\perp K\;then\;\parallel h_0-h_\ast\parallel\;=dist(h_0,K).$

Theorem: Let M be a closed linear subspace of a Hilbert space H and $h\in H\;and\;\;Ph\;the\;unique\;point\;in\;M\;s.t.\;h-Ph\;\perp M$. Then P is a linear transformation on H with the following properties (i) $P^2=P$(ii) $\parallel P\parallel=1$ (iii) $Ker\;P=M^\perp\;and\;Range\;P=M.$

Proof: For any h$\in H$$P^2(h)=P(Ph).\;Since\;Ph\in M\;and\;Ph-P(Ph)\;\perp M\;we\;must\;have\;Ph-P(Ph)=0\;i.e.,P^2h=Ph\;\forall h\in H.$ Now for any $u\in H\;we\;have\;\parallel u\parallel^2=\parallel Pu+(Id-P)u\parallel^2=\parallel Pu\parallel^2+\parallel(Id-P)u\parallel^2\geqslant\parallel Pu\parallel^2$(Here Id is the identity operator on H and since u-Pu $\perp M\;$ <Pu, (Id-P)u>=0). So we get $\parallel Pu\parallel\leqslant\parallel u\parallel\Rightarrow\parallel P\parallel\leqslant1$. Also if u be a nonzero vector in M then Pu=u. Thereforesupu=1Pu=1P=1. The last part is quite obvious.

P is called an Orthogonal projection of H onto the subspace M along orthogonal complement of M. We can also show that H have the Direct sum decomposition $H=S\oplus {S}^{\perp }$(also called othogonal decomposition of H).

## Concept of Adjoint of an operator

Before going to Adjoint we should know the famous Riesz Representation Theorem:

Theorem: ​Suppose H be a Hilbert space and L be a bounded linear functional on it. Then there exists a unique vector $h_0\in H\;s.t.\;L(h)=$<$h,h_0$> for each h. Moreover $\parallel L\parallel=\parallel h_0\parallel$.

Proof: Let $M=Ker\;L.\;$Since L is bounded (and hence continuous) M is closed. If M=H then L is nothing but the zero functional. Now let . $so\;\exists\;f_0\in H\;s.t.\;L(f_0)=1.\;Obviosly\;f_0\in M^\perp.$ If $h\in H\;and\;\alpha=L(h)\;then\;L(h-\alpha f_0)=0\Rightarrow h-\alpha f_0\in M.\;Therefore\;0=<$$h-\alpha f_0,f_0>$. From this we get α =<h,$\frac{f_0}{\parallel f_0\parallel^2}>$. So our $h_0=\frac{f_0}{\parallel f_0\parallel^2}.$ For uniqueness if $h_{1\;}be\;s.t.\;$L(h)=<$h,h_1$> then <$h,h_0-h_1$>. In particular <$h_0-h_1,\;h_0-h_1$>​ =0 so $h_0=h_1$

Now $\left|L(h)\right|\leq\parallel h\parallel\parallel h_0\parallel\Rightarrow\parallel L\parallel\leqslant\parallel h_0\parallel.But\;\left|L(h_0)\right|=\parallel h_0\parallel^2\;hence\;\;\parallel L\parallel=\parallel h_0\parallel$.(proved)

The remark here is that $h_0$is in orthogonal complement of Null space of L. We can prove it for finite dim. IPS more easily. Suppose H be a Hilbert space and $T\in\mathcal B(H)$. Let $v\in H$ be fixed vector. Then the map $\;h\mapsto$​<$T(h),v$> for every h in H, is a linear functional on H. Then by Riesz Representation theorem $\exists\;!\;h_0\in H\;s.t.\;$<$T(h),v$> =<$h,h_0$> so we define $T^\ast(v)=h_0\;for\;each\;v\in H$. Then it is a linear operator on H s.t. <$T(v),h$> = <$v,T^\ast(h)$> called Adjoint of T. Also there is another way to define adjoint of T :X$\rightarrow$Y (X,Y are NLS) as $T^\ast:Y^\ast\rightarrow X^\ast\;by\;for\;each\;g\in Y^\ast,\;T^\ast(g)(x)=g(T(x))\;\forall x\in X.$.....(*) . Look $T^\ast(g)\in X^\ast.\;$(*) can be seen a​s <$T^\ast g,\;x$> $=$<$g,\;Tx$> where $x\in X\;and\;g\in Y^\ast$

Example : Consider the $l_{2\;}space$and $T:l_2\;\rightarrow l_{2\;}\;given\;by\;T(x_1,x_2,......)=(0,x_1,x_2,......)\;(right\;shift\;operator)$. Now ​<$T(v),h$> = <$v,T^\ast(h)$> where $v,h\in l_2$.<T(v),h> = <($0,v_1,v_2,v_3,.......$),($h_1,h_2,h_3,.....$)> =<$(v_1,v_2,....),(h_2,h_3,....)$> = <$v,T^\ast(h)$>. Therefore $T^\ast(h_1,h_2,h_3,....)=(h_2,h_3,....)$ which is the left Shift operator.

Recall that Orthogonal Projection operator P, one can easily prove that P=P*.

## Orthonormal Basis for Hilbert space

We know the terms orthogonal set and orthonormal set from Linear Algebra. So we can discuss Bessel's Inequality: For $\{x_n\}$ be an orthonormal sequence in Hilbert space H and h in H. k=1<xk,h>2  h2

Theorem: Let $\{x_n\}$ be an orthonormal sequence in Hilbert space H and h in H. Then the series k=1<xk,h>xk converges strongly in H and the vector h-k=1<xk,h>xkis orhogonal to each $x_k$.

Definition: An orthonormal sequence $\{x_n\}$ in a Hilbert space H is called an Orthonormal basis for H if for any h in H, it can be expressed as h=k=1<xk,h>xk.

Proposition: If E is an orthonormal set in H then there is a basis for H. (The proof uses Zorn's Lemma)

In Hilbert space we can use Gram-Schmidt orthogonalization process to get orthonormal set from an linealy independent set.

Proposition: Let $\{e_1,e_2,.......e_n\}$ be an orthonormal set in H and M be the closed linear span of the orthonormal set. If P be the orthogonal projection of H onto M then $Ph=\sum_{k=1}^n$<$h,e_k$>.$e_k$ for each h in H.​

Proposition: If E is an orthonormal set and $h\in H\;then$<h,e>≠0 for at most countable no. of vectors in E.

Proof: For n≥1 let $E_n=${e:|<h,e>| ≥ 1/n }. By Bessel's Inequality $E_n$ is finite. But Un=1En={e: <h,e> ≠0}​ .

Theorem: Let E be an orthonormal set in H . Then TFAE:

a) E is a basis.

b) If $h\in H\;and\;h\perp H\;then\;h=0$.

c) span closure of E is H.

d) If $g,h\in H\;then\;$<g,h>=∑{<g,e><e,h>: e ϵ E}.

e) $\parallel h\parallel^2=$∑{ |<h,e>|${}^2$: e is in H}

Remark : The last equality is called PARSEVAL'S IDENTITY.

## Separable Hilbert Space

A metric space is said to Separable if it has countable dense set. Now will give a necessary and sufficient condition for separebility of a Hilbert space.

Theorem: If H is an infinite dimensional Hilbert space then H is separable iff dim.H=|N| .

Proof: Let E be a orthonormal basis for H. If e, f be two basis vectors then $\parallel e-f\parallel^2=\parallel e\parallel^2+\parallel f\parallel^2=2$. Hence the collection is of pairwise disjoint open balls. Now suppose H is separable and D be a countable dense set. If E is not countable then each of the ball must contain atleast one element of the countable dense set D. Then D will not be countable, which is a contradiction. Therefore E is countable. Conversely if H has countable dimension then it has a countable orthonormal basis say, E. Now take D to be the collection of rational (for complex no. take those no. whose real and imaginary parts are rational) linear combinations of elements of D. Obviously this is a countable ​set and the closure of D is H. Therefore D is the required countable dense set. Hence the proof.

## Forms on Hilbert space

A complex valued function B(.,.) on H x H is called a sesquilinear form if it is linear in its first component and conjugate linear in second component. Its norm is defined as follows:B=supx=y=1|B(x,y)|. If this no. is finite we call it bounded.

Theorem: For every bounded sesquilinear form B(.,.) on H x H there corresponds a unique linear oprator A on H such that B(x,y)=<x,Ay> and $\parallel B\parallel=\parallel A\parallel$.

Outline of the proof: See for each fixed y $let\;f_y(x)=B(x,y)$. This is a bounded linear functional. Hence by Riesz representation there exists z such that $f_y(x)=$<x,z> for every x in H. Put z=Ay. This is our required linear operator on H.

Note that inner product is a particular type of sesquilinear form. Also we have seen quadratic form earlier. ​​

## Topologies on operators

Previously we have studied strong and weak topologies on Banach spaces. But now we shall use the adjectives strong and weak on the Banach space B(X,Y) in a different sense here.

The norm topology: It is the usual topology on B(X,Y) coming from the operator norm defined as $\parallel A\parallel=$ $\parallel A\parallel =\underset{\parallel x\parallel =1}{sup}\parallel Ax\parallel$

The strong operator topology: We say that a sequence $\{A_n\}\;in\;B(X,Y)$ converges strongly to A if for each $x\in X\;,\;A_nx\;converges\;to\;Ax\;i.e.,\parallel A_nx-Ax\parallel\rightarrow0\;for\;each\;x.$ The associated topology is called the strong operator topology. It is the weak topology generated by family of maps $T_x\;:B(X,Y)\;\rightarrow Y\;by\;A\rightarrow Ax$ where x varies over X.

The weak operator topology: We say that a sequence $\{A_n\}\;in\;B(X,Y)$ converges in weak operator topology if $f(A_nx)\rightarrow\;f(Ax)\;\;\forall f\in Y^\ast,x\in X.\;Denoted\;by\;A_n\;\xrightarrow wA.$ This is the weak topology generated by the family of maps $T_{x,f}\;:B(X,Y)\;\rightarrow C\;(where\;C\;is\;the\;field\;of\;complex\;no.)\;defined\;as\;T_{x,f}(A)=f(Ax)$ where x varies over X and f varies over Y*. If X and Y are Hilbert spaces using Riesz representation one can show $A_n\;\xrightarrow w\;A\;iff\;\;$<$A_nx,y$> → <$Ax,y$> for x in X and y in Y.

Note: Clearly the convergence in norm implies convergence in strong operator topology and convergence in strong operator topology implies convergence in weak operator topology. But the converse implications are not true. Counter example: Consider X and Y to be $l_2space$. $A_n=\frac1nId.\;$Then clearly the sequence of operators converges to zero operator in the norm topology. Now consider $\{e_j\}$: orthonormal basis for $l_2$. Let $P_n$be the orthogonal projection onto the linear span of $\{e_1,e_2,.....e_n\}$.Then The sequence of operators converges to Id in strong operator topology but it does not converge to Id in norm topology as $\parallel Id-P_n\parallel=1\;\;\forall n\;as\;Id-P_n$ is the orthogonal projection onto orthogonal complement of the space linear span of $\{e_1,e_2,.....e_n\}$.

Another example: Consider the right shift operator S on $l_2$. One can $l_2$see $\{S_n\}\;$converges to zero in the weak operator topology. But $\parallel S^nx\parallel=\parallel x\parallel$. Therefore it does not converges in strong operator topology.​​​

Theorem: Let $\{A_n\}$be a sequence of operators on a Hilbert space H. Suppose the sequence is weakly cauchy. Then there exists an operator A s.t. $A_n\overset w{\rightarrow\;A}$.

Proof: For each x and y let. Then we get a sesquilinear form B(. , .) Now |<$A_nx,\;y$>| $\leq\parallel A_n\parallel\parallel x\parallel\parallel y\parallel$.....(*). As $for\;each\;x,y\;$<$A_nx,\;y$> is bounded for each fixed x considering $A_nx\;as\;linear\;functional\;on\;H\;acting\;as\;(A_nx)(y)=$<$y,A_nx$>. Then by UBP for each x $\underset{n}{sup}\parallel {A}_{n}x\parallel$Again by UBP $\underset{n}{sup}\parallel {A}_{n}x\parallel$. Therefore B is bounded and so by Riesz representation there exists A such that B(x,y)=<Ax,y>. Threfore $A_n\overset w{\rightarrow\;A}$.

## Inverse and its existence

Let A is a bdd linear operator on X. If A is bijective then by Inverse mapping theorem $A^{-1}$is also bdd.

Theorem: ​If $\parallel I-A\parallel<1$ then $A$ is invertible and $A^{-1}\;=I\;+(I-A)\;+(I-A)^2\;+.........$.(Neumann series)

Proof: $If\;S_n=\sum_{k=0}^n(I-A)^k\;\;\;then\;\parallel S_{n+m}-S_n\parallel\leq\sum_{j=n+1}^{n+m}\parallel I-A\parallel^j\rightarrow0\;\;as\;n,m\rightarrow\infty.$Let T be the sum of the series .Then $TS_n=I-(I-A_n)^{n+1}$. By continuity of operator multiplication $AT=I\;.$ Similarly we can show $TA=I.\;Therefore\;T=A^{-1}.$

Proposition : The set of bdd invertible operators is open in B(X). In finite dimensional space X it is dense.

## Some results on norm of adjoint operator

We have already defined Adjoint of an operator from X to Y.

Theorem : $\parallel A\parallel=\parallel A^\ast\parallel$.

Proof: $If\;f\in Y\ast\;and\;\parallel f\parallel=1$then $\parallel {A}^{*}f\parallel =\underset{\parallel x\parallel =1}{sup}\left|A\left(f\right)\left(x\right)\right|\le \underset{\parallel x\parallel =1}{sup}\parallel Ax\parallel$. Thus $\parallel {A}^{*}\parallel \le \parallel A\parallel .$ Now by Hann Banach Theorem there exists a linear functional f on Y such that $\parallel f\parallel=1$and $f(Ax)=\parallel Ax.\parallel$ Therefore This implies $\parallel A\parallel ⩽\parallel {A}^{*}\parallel .$

Theorem: $A\in B(H)\;\Rightarrow\parallel A^\ast A\parallel=\parallel A\parallel^2$

Proof: Since $\parallel AB\parallel\leq\parallel A\parallel\parallel B\parallel\;we\;have\;\parallel A^\ast A\parallel\leq\parallel A^\ast\parallel\parallel A\parallel=\parallel A\parallel^2.$ Again $\parallel Ax\parallel^2=$<Ax,Ax> =<A*Ax,x> ≤ $\parallel A^\ast Ax\parallel\parallel x\parallel\leqslant\parallel A^\ast A\parallel\parallel x\parallel^2$. Therefore $\parallel A\parallel^2\leq\parallel A^\ast A\parallel.$ Hence the result.

Properties of the map A→A* on B(H) : i) it is conjugate linear. ii) It is isometric as $\parallel A\parallel=\parallel A^\ast\parallel$ iii) (AB)* = B*A*. iv) I*=I and $(A^{-1})^\ast=(A^\ast)^{-1}$​, provided A is invertible.

Also the map A→A* from B(H) to B(H*) is continuous in the norm topology.

## Some special operators on Hilbert space

A be an operator on H is called self adjoint if A=A*.

Proposition: If A is self adjoint then <Ax,x> is real for each x in H. Conversely if H is a complex Hilbert space then for each x in H <Ax,x> is real implies A is self adjoint.

Outline of the proof: $\left\langle Ax,x\right\rangle=\left\langle x,A^\ast x\right\rangle=\left\langle x,Ax\right\rangle=\overline{\left\langle Ax,x\right\rangle}$ For the con​verse let $\left\langle Ah,h\right\rangle\in\mathbb{R}$. If $\alpha\in\mathbb{C}\;then\;\left\langle A(h+\alpha g),\;h+\alpha g\right\rangle\;=\left\langle Ah,h\right\rangle+\overline\alpha\;\left\langle Ah,g\right\rangle\;+\alpha\left\langle Ag,\;h\right\rangle\;+\left|\alpha\right|^2\left\langle Ag,g\right\rangle\;\in\mathbb{R}.\;As\;\left\langle Ah,h\right\rangle\;and\;\left\langle Ag,g\right\rangle\in\mathbb{R}\;we\;must\;have\;\overline\alpha\;\left\langle Ah,g\right\rangle\;+\alpha\left\langle Ag,\;h\right\rangle=\overline\alpha\;\left\langle h,Ag\right\rangle\;+\alpha\left\langle g,Ah\right\rangle=\overline\alpha\;\left\langle A^\ast h,g\right\rangle\;+\alpha\left\langle A^\ast g,\;h\right\rangle\;$. Now take $\alpha=1\;and\;\alpha=i$.Then we get $\left\langle Ag,h\right\rangle+\left\langle Ah,g\right\rangle=\left\langle A^\ast h,g\right\rangle+\left\langle A^\ast g,h\right\rangle\;\;and\;\;i\left\langle Ag,h\right\rangle-i\left\langle Ah,g\right\rangle=-i\left\langle A^\ast h,g\right\rangle+i\left\langle A^\ast g,h\right\rangle\;$. Now little calculation shows $\left\langle Ag,h\right\rangle=\left\langle A^\ast g,h\right\rangle.\;Therefore\;A=A^\ast.$

Note that the converse may not be true for real Hilbert space.

Proposition: $A=A^\ast\;\Rightarrow\;\parallel A\parallel=sup\;\{\left|\left\langle Ah,\;h\right\rangle\right|:\parallel h\parallel=1\}.$

Proof: M=$sup\{\left|\left\langle Ah,\;h\right\rangle\right|:\parallel h\parallel=1\}.$ If $\parallel h\parallel=1\;then\;\left|\left\langle Ah,\;h\right\rangle\right|\leqslant\parallel A\parallel;\;hence\;M\leqslant\parallel A\parallel.$ If $\parallel h\parallel=\parallel g\parallel=1\;then\;\left\langle A(h\pm g),\;h\pm g\right\rangle\;=\;\left\langle Ah,h\right\rangle\pm\left\langle Ah,\;g\right\rangle\pm\left\langle g,A^\ast h\right\rangle+\left\langle Ag,\;g\right\rangle.$As A is self adjoint this implies $\left\langle A(h\pm g),\;(h\pm g)\right\rangle\;=\left\langle Ah,h\right\rangle\pm2Re\left\langle Ah,g\right\rangle+\left\langle Ag,g\right\rangle.\;subtracting\;these\;two\;equation\;4Re\;\left\langle Ah,g\right\rangle=\left\langle A(h+g),h+g\right\rangle-\left\langle A(h-g),h-g\right\rangle.Put\;h=g=f\;then\;\left|\left\langle Af,f\right\rangle\right|\leq M\parallel f\parallel^2$Again by Parallelogram law $\begin{array}{l}4Re\left\langle Ah,g\right\rangle\;\leqslant M(\parallel h+g\parallel^2+\parallel h-g\parallel^2)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=2M(\parallel h\parallel^2+\parallel g\parallel^2)=4M\end{array}$ Suppose $\left\langle Ah,g\right\rangle=e^{i\vartheta}\left|\left\langle Ah,g\right\rangle\right|.\;Replace\;h\;by\;e^{-i\vartheta}h\;then\;\left|\left\langle Ah,g\right\rangle\right|\leqslant M\;if\;h,g\;are\;unit\;vector.$ Taking supremum over all g and when $\parallel h\parallel=1\;we\;get\;\parallel Ah\parallel\leq M.\;Therefore\;\parallel A\parallel\leqslant M.$

Cor: If A is self adjoint and $\left\langle Ah,\;h\right\rangle=0\;\;\forall h\;then\;A=0.$

Proposition: If H is a complex Hilbert space abd A is bdd operator ​such that $\left\langle Ah,h\right\rangle=0\;\forall h\in H\;then\;A=0.$

## Positive operator

A self adjoint operator on H is called positive if $\left\langle Ah,h\right\rangle\geqslant0\;\;\;\forall h$ denoted as $A\geqslant0$. If the inequality is strict then we call it as strictly positive operator denoted as $A>0$

Result: For any operator $A,\;AA^\ast\;\;and\;A^\ast A\;are\;positive\;operator.$

Proof: $\left\langle AA^\ast x,x\right\rangle=\left\langle A^\ast x,A^\ast x\right\rangle=\parallel A^\ast x\parallel^2\;\geq0\;\;and\;\;\left\langle A^\ast Ax,x\right\rangle=\parallel Ax\parallel^2\geq0.$ Obviously both of the operator are are self adjoint.

## Normal operator

A linear operator A is called Normal if $AA^\ast=A^\ast A.$

Lemma: A is​ normal iff $\parallel Ah\parallel=\parallel A^\ast h\parallel\;\;\forall h.$

Proof: $\parallel Ah\parallel^2=\parallel A^\ast h\parallel^2\;\Leftrightarrow\left\langle Ah,Ah\right\rangle=\left\langle A^\ast h,A^\ast h\right\rangle\;\Leftrightarrow\left\langle A^\ast Ah,h\right\rangle=\left\langle AA^\ast h,h\right\rangle\;\Leftrightarrow\left\langle(A^\ast A-AA^\ast)h\;,h\right\rangle=0\;\;\;\;\forall h.\;Therefore\;\;A^\ast A-AA^\ast=0.\;$

Lemma: A is normal then $\parallel A^2\parallel=\parallel A\parallel^2.$

Proof: By previous lemma $\parallel A(Ax)\parallel=\parallel A^\ast(Ax)\parallel\;\;\forall x.\;Therefore\;\parallel A^2\parallel=\parallel A^\ast A\parallel.Again\;\parallel A^\ast A\parallel=\parallel A\parallel^2.$

## Unitary operator

An operator U is unitary if $U^\ast U=UU^\ast=I.$

Proposition : U is surjective on H and preserves the inner product iff it carries orthonormal basis of H to an orthonormal basis. (This is same as saying U is unitary)

Isometry: A linear operator is said to be isometry if it preserves the norm i.e, $\forall x\;\;\parallel Ux\parallel=\parallel x\parallel.$

Obviously every unitary operator is an isometry. Although the converse is true for finite dimensional case but it may not be true in infinite dimensional Hilbert space. For example the right shift operator S on $l_2$is an isometry but it is not an unitary operator. $S(x_1,x_2,....)=(0,x_1,x_2,....)\;and\;S^\ast(x_1,x_2,x_3,,....)=(x_2,x_3,,.....).\;Then\;S^\ast S(x_1,x_2,....)=(x_1,x_2,....\;)\;\Rightarrow S^\ast S=I.$

Lemma: An operator A on H is an isometry iff $A^\ast A=I$

Proof: $\parallel Ax\parallel^2=\parallel x\parallel^2\;\Leftrightarrow\left\langle Ax\;,\;Ax\right\rangle\;=\left\langle x,x\right\rangle\;\Leftrightarrow\left\langle A^\ast Ax\;,\;x\right\rangle=\left\langle x,x\right\rangle\;\Leftrightarrow\left\langle(A^\ast A-I)x,x\right\rangle=0\;\;\forall x\;\Leftrightarrow A^\ast A=I.$

Properties of Orthogonal Projection

a)An Idempotent operator is on H is an orthogonal projection is iff it is self adjoint. b) Every orthogonal projection is a positive operator.

## Invariant subspace of an operator

Let A be an operator on H . A subspace M is said to be Invariant under A if $A(M)\;\subseteq M\;.$If both $M\;and\;M^\perp$are invariant under A we call M is a reducing subspace for A.

Proposition: A closed subspace $M$ is invariant under $A$ iff $M^\perp$is invariant under $A^\ast$. Thus ​$M$​is a reducing subspace for$A$ iff $M$ is invariant under both ​​$A$​and $A^\ast$.

## THE SPECTRUM OF AN OPERATOR

In finite dimensional case spectrum is nothing but the eigenvalues. But in infinite dimensional case there are so many complications.​

Definition: The spectrum of a linear operator on a Banach space X​ is the set $\sigma(A)=$ $\{\lambda\in\mathbb{C}\;:\;A-\lambda I\;is\;not\;invertible\}.$

Definition: The reolvent set of A is the set $\rho(A)=\{\lambda\in\mathbb{C}\;:\;A-\lambda I\;is\;invertible\;\}$. The operator $R_\lambda(A)=(A-\lambda I)^{-1}$is called the resolvent of A at $\lambda.$

Proposition: If A is a bdd linear operator on X then $\rho(A)\;$ is nonempty open bdd subset of $\mathbb{C}.$

Proof: If $\parallel A\parallel<\left|\lambda\right|\;then\;\parallel\frac A\lambda\parallel<1\;.\;So\;I-\frac A\lambda\;$i​s invertible which imples $A-\lambda I\;is\;invertible\;and\;\;(A-\lambda I)^{-1}=\frac{-1}\lambda\sum_{n=0}^\infty(\frac A\lambda)^n\;\;for\;\parallel A\parallel<\left|\lambda\right|$. Thus resolvent set is nonempty. $A-\lambda I\;=(A-\lambda_0I)\;\lbrack I-(\lambda-\lambda_0)(A-\lambda_0I)^{-1}\rbrack\;.\;So\;if\;\left|\lambda-\lambda_0\right|<\frac{\;1}{\parallel(A-\lambda_0I)^{-1}\parallel.}$then $(A-\lambda I)\;is\;\;invertible.$ Hence it is easy to see that $\rho(A)$is open. ​

Cor: $\sigma(A)\;=(\rho(A))^c.\;\;\sigma(A)\;$is closed and bounded.

Proof: If $\parallel A\parallel<\left|\lambda\right|\;then\;(A-\lambda I)\;is\;invertible.$​So $\sigma(A)\subseteq\{\lambda\;:\;\left|\lambda\right|\leq\parallel A\parallel\}$and hence bounded.

Theorem: $\sigma(T)\;is\;nonempty.$

Proof: Let us suppose that $\sigma(T)$be empty. So $\rho \left(T\right)=\mathrm{ℂ}$. Then by Neumann series we can say that $\parallel R_\lambda(T)\parallel\rightarrow0\;as\;\left|\lambda\right|\rightarrow\infty.$Now for each . . By linearity and continuity of f and continuity of the inversion map and since Hence $\phi(z)$is analytic and so it is an entire function. But Therefore $\phi(z)$