Summer Research Fellowship Programme of India's Science Academies

A Study of Operators on Hilbert Space

Mainak Bhowmik

Indian Institute of Technology, Kanpur 208016

Dr. Jaydeb Sarkar

Indian Statistical Institute, Bengaluru 560059


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).


NLSNormed Linear Space
BOpen unit ball in the coressponding Banach space
HBTHahn Banach Theorem
UBPUniform Bounded Principle

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


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, lp and Lp spaces 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=n=1Yn\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 xtx_t in X s.t. dist(xtx_t,M) \geqt. (where dist(xtx_t,M) =inf {xm:mM\parallel x-m\parallel:m\in M} )​

Theorem: In any infinite dimen​sional NLS the closed unit ball {  x:x1}\{\;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 M=supx=1Ax . 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 A:=supx=1Ax.​​

Theorem: A:XY be a linear operator, where X, Y are NLS. A is bounded  A is continuous on XA is continuous at 0 .

The Hahn-Banach Theorem

Let X be a normed linear space. Let X0 be a subspace of it and let f0 be a linear functional on X0 such that f0(x)Cx for all xX0  and some C>0 . Then there exist a linear functional f on X such that f(x) =f0(x)   xX0  and  f(x)Cx  xX . Actually the theorem says that a linear functional on X0 can be extended to X without increasing its norm.

Corollaries: (i) Let X0X_0 be a subspace of a NLS and x1x_1be a vector such that dist (x1  ,X0x_1\;,X_0) =δ\delta>0. Then there exists a linear functional f on X such that f=1  ,f(x1)=δ  and  f(x)=0      xX0\parallel f\parallel=1\;,f(x_1)=\delta\;and\;f(x)=0\;\;\;\forall x\in X_0

(ii) For each non zero x0X  ,x_0\in X\;, there exists a linear functional f on X such that f=1  and  f(x0)=x0\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)|x=supfX,f=1|f(x)|. Hence we can seperate two point x1and x2in X,  fX* s.t. f=1 and f(x1)f(x2).

The Uniform Boundedness Principle

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

Application:- Let {fn}\{f_n\}be a bounded linear functionals on Banach space X. Suppose for each x , {fn(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:XY  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, A1A^{-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*.


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

ii) Conider the lp(N)  space  where  1p<l_p(N)\;space\;where\;1\leq p<\infty . If q be such that 1p+1q=1\frac1p+\frac1q=1 then the dual space (lp(N))(l_p(N))^\ast is isomorphic to lq(N)l_q(N). iii) similarly for Lp[0,1]  L_p\lbrack0,1\rbrack\; 1p<1\leq p<\infty . If q be such that 1p+1q=1\frac1p+\frac1q=1 then (Lp[0,1])  Lq[0,1](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 (X,τ)\left(X,\tau\right)be a topolgical space. Let A\mathcal A be family of maps from Y into X . The weak topology on Y generated by A\mathcal A i​s the smallest topology on Y for which all fA  f\in\mathcal A\;are continuous. The collection {kj=1  fj(Uj):Ujτ,  fjA,1jk,  k=1,2,.....}  \{{\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 {xnx_n} converges to xx in the norm or strong topology if xnx0  .  we  write  xnx.\parallel x_n-x\parallel\rightarrow0\;.\;we\;write\;x_n\rightarrow x.

We say that {xnx_n} converges to xxin the weak topology iff f(xn) converges to f(x) for all fX*. We denote thi​s convergence as xnxwx_n\overset w{\rightarrow x}.


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

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

iii) If {xn} in X is convergent in norm topology then it is bounded .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:XX    be  given  by  J(x)=Fx  ,xXJ:X\rightarrow X^{\ast\ast}\;\;be\;given\;by\;J(x)=F_x\;,\forall x\in X. whereFxX  defined  as  Fx(f)=f(x)  fXF_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 Jx=x\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*.



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

Example: Cn  C^n\;is a inner product space with the usual definition <x,y> <x,y>=j=1nxjyj    x,yCn<x,y>=\sum_{j=1}^nx_j\overline{y_j}\;\;\forall x,y\in C^nand it is a Hilbert space with the norm induced by the inner product. The space l2l_2with the inner product <x,y> <x,y>=xjyjj=1<x,y>=\overset\infty{\underset{j=1}{\sum x_j\overline{y_j}}}is also a Hilbert space. The space L2[a,b]  L_2\lbrack a,b\rbrack\;is a Hilbert space with the inner product <f,g> =abf(x)g(x)  dx\int_a^bf(x)\overline{g(x)\;}dx.

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

Parallelogram Law: x+y2+xy2=2(x2+y2).AAAdfdfcxvvvxvvddvddfdfgfdbggdf\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,yx,y>=14p=14ipx+ipy=\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: lpl_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 x+y2+xy2=21+2p  and      2(x2+y2)=4      so  21+2p=4  only  when  p=2.\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 Lp[a,b]L_p\lbrack a,b\rbrack is a Hilbert space only for p=2.

Some basic results on Orthogonal complement of a set

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


i) SS^\perpis a closed linear subspace of H.

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

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

iv) If S1S2  then  S2S1S_1\subset S_{2\;}then\;S_2^\perp\subset S_1^\perp. Also SSS\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 l2l_2space 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,.....)\inSS^\perpbut (0,0,0.....) ∉\not\inso SS=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 h0H\  K.h_0\in H\backslash\;K. Then there exists a unique vector hK  s.t.  h0h=dist(h0,K)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 h0h  K.h_0-h_\ast\;\perp K. Conversely if hK  s.t.  h0hK  then  h0h  =dist(h0,K).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 hH  and    Ph  the  unique  point  in  M  s.t.  hPh  Mh\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) P2=PP^2=P(ii) P=1\parallel P\parallel=1 (iii) Ker  P=M  and  Range  P=M.Ker\;P=M^\perp\;and\;Range\;P=M.

Proof: For any hH\in HP2(h)=P(Ph).  Since  PhM  and  PhP(Ph)  M  we  must  have  PhP(Ph)=0  i.e.,P2h=Ph  hH.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 uH  we  have  u2=Pu+(IdP)u2=Pu2+(IdP)u2Pu2u\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 M  \perp M\; <Pu, (Id-P)u>=0). So we get PuuP1\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=1supu=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=SS(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 h0H  s.t.  L(h)=h_0\in H\;s.t.\;L(h)=<h,h0h,h_0> for each h. Moreover L=h0\parallel L\parallel=\parallel h_0\parallel.

Proof: Let M=Ker  L.  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 MH.Then M{0}. so    f0H  s.t.  L(f0)=1.  Obviosly  f0M.so\;\exists\;f_0\in H\;s.t.\;L(f_0)=1.\;Obviosly\;f_0\in M^\perp. If hH  and  α=L(h)  then  L(hαf0)=0hαf0M.  Therefore  0=<h\in H\;and\;\alpha=L(h)\;then\;L(h-\alpha f_0)=0\Rightarrow h-\alpha f_0\in M.\;Therefore\;0=<hαf0,f0>h-\alpha f_0,f_0>. From this we get α =<h,α =<h,f0f02>\frac{f_0}{\parallel f_0\parallel^2}>. So our h0=f0f02.h_0=\frac{f_0}{\parallel f_0\parallel^2}. For uniqueness if h1  be  s.t.  h_{1\;}be\;s.t.\;L(h)=<h,h1h,h_1> then <h,h0h1h,h_0-h_1>. In particular <h0h1,  h0h1h_0-h_1,\;h_0-h_1>​ =0 so h0=h1h_0=h_1

Now L(h)hh0Lh0.But  L(h0)=h02  hence    L=h0\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 h0h_0is 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 TB(H)T\in\mathcal B(H). Let vHv\in H be fixed vector. Then the map   h\;h\mapsto​<T(h),vT(h),v> for every h in H, is a linear functional on H. Then by Riesz Representation theorem   !  h0H  s.t.  \exists\;!\;h_0\in H\;s.t.\;<T(h),vT(h),v> =<h,h0h,h_0> so we define T(v)=h0  for  each  vHT^\ast(v)=h_0\;for\;each\;v\in H. Then it is a linear operator on H s.t. <T(v),hT(v),h> = <v,T(h)v,T^\ast(h)> called Adjoint of T. Also there is another way to define adjoint of T :X\rightarrowY (X,Y are NLS) as T:YX  by  for  each  gY,  T(g)(x)=g(T(x))  xX.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(g)X.  T^\ast(g)\in X^\ast.\;(*) can be seen a​s <Tg,  xT^\ast g,\;x> ==<g,  Txg,\;Tx> where xX  and  gYx\in X\;and\;g\in Y^\ast

Example : Consider the l2  spacel_{2\;}spaceand T:l2  l2    given  by  T(x1,x2,......)=(0,x1,x2,......)  (right  shift  operator)T:l_2\;\rightarrow l_{2\;}\;given\;by\;T(x_1,x_2,......)=(0,x_1,x_2,......)\;(right\;shift\;operator). Now ​<T(v),hT(v),h> = <v,T(h)v,T^\ast(h)> where v,hl2v,h\in l_2.<T(v),h> = <(0,v1,v2,v3,.......0,v_1,v_2,v_3,.......),(h1,h2,h3,.....h_1,h_2,h_3,.....)> =<(v1,v2,....),(h2,h3,....)(v_1,v_2,....),(h_2,h_3,....)> = <v,T(h)v,T^\ast(h)>. Therefore T(h1,h2,h3,....)=(h2,h3,....)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 {xn}\{x_n\} be an orthonormal sequence in Hilbert space H and h in H. k=1<xk,h>2  h2k=1<xk,h>2  h2

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

Definition: An orthonormal sequence {xn}\{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>xkh=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 {e1,e2,.......en}\{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=k=1nPh=\sum_{k=1}^n<h,ekh,e_k>.eke_k for each h in H.​

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

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

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

a) E is a basis.

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

c) span closure of E is H.

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

e) h2=\parallel h\parallel^2=∑{ |<h,e>|2{}^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 ef2=e2+f2=2\parallel e-f\parallel^2=\parallel e\parallel^2+\parallel f\parallel^2=2. Hence the collection {B(e,2 ):eE} 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)|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 B=A\parallel B\parallel=\parallel A\parallel.

Outline of the proof: See for each fixed y let  fy(x)=B(x,y)let\;f_y(x)=B(x,y). This is a bounded linear functional. Hence by Riesz representation there exists z such that fy(x)=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 A=\parallel A\parallel= A=supx=1Ax

The strong operator topology: We say that a sequence {An}  in  B(X,Y)\{A_n\}\;in\;B(X,Y) converges strongly to A if for each xX  ,  Anx  converges  to  Ax  i.e.,AnxAx0  for  each  x.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 Tx  :B(X,Y)  Y  by  AAxT_x\;:B(X,Y)\;\rightarrow Y\;by\;A\rightarrow Ax where x varies over X.

The weak operator topology: We say that a sequence {An}  in  B(X,Y)\{A_n\}\;in\;B(X,Y) converges in weak operator topology if f(Anx)  f(Ax)    fY,xX.  Denoted  by  An  wA.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 Tx,f  :B(X,Y)  C  (where  C  is  the  field  of  complex  no.)  defined  as  Tx,f(A)=f(Ax)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 An  w  A  iff    A_n\;\xrightarrow w\;A\;iff\;\;<Anx,yA_nx,y> → <Ax,yAx,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 l2spacel_2space. An=1nId.  A_n=\frac1nId.\;Then clearly the sequence of operators converges to zero operator in the norm topology. Now consider {ej}\{e_j\}: orthonormal basis for l2l_2. Let PnP_nbe the orthogonal projection onto the linear span of {e1,e2,.....en}\{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 IdPn=1    n  as  IdPn\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 {e1,e2,.....en}\{e_1,e_2,.....e_n\}.

Another example: Consider the right shift operator S on l2l_2. One can l2l_2see {Sn}  \{S_n\}\;converges to zero in the weak operator topology. But Snx=x\parallel S^nx\parallel=\parallel x\parallel. Therefore it does not converges in strong operator topology.​​​

Theorem: Let {An}\{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. An  AwA_n\overset w{\rightarrow\;A}.

Proof: For each x and y let. Then we get a sesquilinear form B(. , .) Now |<Anx,  yA_nx,\;y>| Anxy\leq\parallel A_n\parallel\parallel x\parallel\parallel y\parallel.....(*). As for  each  x,y  <Anx,y>for\;each\;x,y\;<A_nx,y><Anx,  yA_nx,\;y> is bounded for each fixed x considering Anx  as  linear  functional  on  H  acting  as  (Anx)(y)=A_nx\;as\;linear\;functional\;on\;H\;acting\;as\;(A_nx)(y)=<y,Anxy,A_nx>. Then by UBP for each x supnAnxAgain by UBP supnAnx. Therefore B is bounded and so by Riesz representation there exists A such that B(x,y)=<Ax,y>. Threfore An  AwA_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 A1A^{-1}is also bdd.

Theorem: ​If IA<1\parallel I-A\parallel<1 then AA is invertible and A1  =I  +(IA)  +(IA)2  +.........A^{-1}\;=I\;+(I-A)\;+(I-A)^2\;+..........(Neumann series)

Proof: If  Sn=k=0n(IA)k      then  Sn+mSnj=n+1n+mIAj0    as  n,m.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 TSn=I(IAn)n+1TS_n=I-(I-A_n)^{n+1}. By continuity of operator multiplication AT=I  .AT=I\;. Similarly we can show TA=I.  Therefore  T=A1.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 : A=A\parallel A\parallel=\parallel A^\ast\parallel.

Proof: If  fY  and  f=1If\;f\in Y\ast\;and\;\parallel f\parallel=1then A*f=supx=1A(f)(x)supx=1Ax. Thus A*A. Now by Hann Banach Theorem there exists a linear functional f on Y such that f=1\parallel f\parallel=1and f(Ax)=Ax.f(Ax)=\parallel Ax.\parallel Therefore Ax=f(Ax)=(A*f)(x)  A*fx=A*x.This implies AA*.

Theorem: AB(H)  AA=A2A\in B(H)\;\Rightarrow\parallel A^\ast A\parallel=\parallel A\parallel^2

Proof: Since ABAB  we  have  AAAA=A2.\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 Ax2=\parallel Ax\parallel^2=<Ax,Ax> =<A*Ax,x> ≤ AAxxAAx2\parallel A^\ast Ax\parallel\parallel x\parallel\leqslant\parallel A^\ast A\parallel\parallel x\parallel^2. Therefore A2AA.\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 A=A\parallel A\parallel=\parallel A^\ast\parallel iii) (AB)* = B*A*. iv) I*=I and (A1)=(A)1(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

Self-adjoint operator

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: Ax,x=x,Ax=x,Ax=Ax,x\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 Ah,hR\left\langle Ah,h\right\rangle\in\mathbb{R}. If αC  then  A(h+αg),  h+αg  =Ah,h+α  Ah,g  +αAg,  h  +α2Ag,g  R.  As  Ah,h  and  Ag,gR  we  must  have  α  Ah,g  +αAg,  h=α  h,Ag  +αg,Ah=α  Ah,g  +αAg,  h  \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 α=1  and  α=i\alpha=1\;and\;\alpha=i.Then we get Ag,h+Ah,g=Ah,g+Ag,h    and    iAg,hiAh,g=iAh,g+iAg,h  \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 Ag,h=Ag,h.  Therefore  A=A.\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    A=sup  {Ah,  h:h=1}.A=A^\ast\;\Rightarrow\;\parallel A\parallel=sup\;\{\left|\left\langle Ah,\;h\right\rangle\right|:\parallel h\parallel=1\}.

Proof: M=sup{Ah,  h:h=1}.sup\{\left|\left\langle Ah,\;h\right\rangle\right|:\parallel h\parallel=1\}. If h=1  then  Ah,  hA;  hence  MA.\parallel h\parallel=1\;then\;\left|\left\langle Ah,\;h\right\rangle\right|\leqslant\parallel A\parallel;\;hence\;M\leqslant\parallel A\parallel. If h=g=1  then  A(h±g),  h±g  =  Ah,h±Ah,  g±g,Ah+Ag,  g.\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 A(h±g),  (h±g)  =Ah,h±2ReAh,g+Ag,g.  subtracting  these  two  equation  4Re  Ah,g=A(h+g),h+gA(hg),hg.Put  h=g=f  then  Af,fMf2\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^2Again by Parallelogram law 4ReAh,g  M(h+g2+hg2)                                              =2M(h2+g2)=4M\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 Ah,g=eiϑAh,g.  Replace  h  by  eiϑh  then  Ah,gM  if  h,g  are  unit  vector.\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 h=1  we  get  AhM.  Therefore  AM.\parallel h\parallel=1\;we\;get\;\parallel Ah\parallel\leq M.\;Therefore\;\parallel A\parallel\leqslant M.

Cor: If A is self adjoint and Ah,  h=0    h  then  A=0.\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 Ah,h=0  hH  then  A=0.\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 Ah,h0      h\left\langle Ah,h\right\rangle\geqslant0\;\;\;\forall h denoted as A0A\geqslant0. If the inequality is strict then we call it as strictly positive operator denoted as A>0A>0

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

Proof: AAx,x=Ax,Ax=Ax2  0    and    AAx,x=Ax20.\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=AA.AA^\ast=A^\ast A.

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

Proof: Ah2=Ah2  Ah,Ah=Ah,Ah  AAh,h=AAh,h  (AAAA)h  ,h=0        h.  Therefore    AAAA=0.  \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 A2=A2.\parallel A^2\parallel=\parallel A\parallel^2.

Proof: By previous lemma A(Ax)=A(Ax)    x.  Therefore  A2=AA.Again  AA=A2.\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 UU=UU=I.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, x    Ux=x.\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 l2l_2is an isometry but it is not an unitary operator. S(x1,x2,....)=(0,x1,x2,....)  and  S(x1,x2,x3,,....)=(x2,x3,,.....).  Then  SS(x1,x2,....)=(x1,x2,....  )  SS=I.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 AA=IA^\ast A=I

Proof: Ax2=x2  Ax  ,  Ax  =x,x  AAx  ,  x=x,x  (AAI)x,x=0    x  AA=I.\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)  M  .A(M)\;\subseteq M\;.If both M  and  MM\;and\;M^\perpare invariant under A we call M is a reducing subspace for A.

Proposition: A closed subspace MM is invariant under AA iff MM^\perpis invariant under AA^\ast. Thus ​MM​is a reducing subspace forAA iff MM is invariant under both ​​AA​and AA^\ast.


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 σ(A)=\sigma(A)= {λC  :  AλI  is  not  invertible}.\{\lambda\in\mathbb{C}\;:\;A-\lambda I\;is\;not\;invertible\}.

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

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

Proof: If A<λ  then  Aλ<1  .  So  IAλ  \parallel A\parallel<\left|\lambda\right|\;then\;\parallel\frac A\lambda\parallel<1\;.\;So\;I-\frac A\lambda\;i​s invertible which imples AλI  is  invertible  and    (AλI)1=1λn=0(Aλ)n    for  A<λA-\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. AλI  =(Aλ0I)  [I(λλ0)(Aλ0I)1]  .  So  if  λλ0<  1(Aλ0I)1.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 (AλI)  is    invertible.(A-\lambda I)\;is\;\;invertible. Hence it is easy to see that ρ(A)\rho(A)is open. ​

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

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

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

Proof: Let us suppose that σ(T)\sigma(T)be empty. So ρ(T)=. Then by Neumann series we can say that Rλ(T)0  as  λ.\parallel R_\lambda(T)\parallel\rightarrow0\;as\;\left|\lambda\right|\rightarrow\infty.Now for each f(B(H,H))*  define ϕ: by ϕ(z)=f(Rz(T)) for z. Consider z0ρ(T). Now limzz0φ(z)-φ(z0)z-z0 = limzz0f((A-zI)-1-(A-z0I)-1z-z0) =f((A-z0I)-2) . By linearity and continuity of f and continuity of the inversion map and since (A-zI)-1 - (A-z0I)-1=(z-z0)(A-z0I)-1(A-zI)-1.Hence ϕ(z)\phi(z)is analytic and so it is an entire function. But ϕ(z)0 as z.Therefore ϕ(z)\phi(z)