Compact Operators on Hilbert Space - University of Minnesota

1 (February 18, 2012) Compact Operators on Hilbert SpacePaul garrett/Among all linear Operators on Hilbert spaces, thecompactones (defined below) are the simplest, and mostimitate the more familiar linear algebra of finite-dimensional operator theory. In addition, these are ofconsiderable practical value and importance. We prove a spectral theorem for self-adjoint Operators withminimal fuss. Thus, we donotinvoke broader discussions of properties of spectra. We only need theCauchy-Schwarz-Bunyakowsky inequalityand thedefinitionof self-adjoint Compact operator. It is true that variouspoints here admit great generalization, and receive definitive treatment only in such general setting. Compact Operators : definition Clever expression for the operator norm Spectral Theorem for self-adjoint Compact Operators : definitionA set in a topological Space is calledpre-compactif its closure is Compact . (Beware, sometimes this has amore restrictive meaning.) A linear operatorT:X Yfrom a pre- Hilbert spaceXto a Hilbert spaceYiscompactif it maps the unit ball inXto apre-compactset inY.

2 Equivalently,Tis Compact if and only ifit mapsboundedsequences inXto sequences inYwithconvergent Operators are the most amenable. Sources of such Operators will be considered elsewhere. For themoment we need only concentrate on the defining property, and its use in proof of the spectral expression for the operator normFirst is a little lemma, useful other places as well, which provides a necessary alternative expression for theuniform norm|T|=|T|unif= sup|x| 1|Tx|of a continuous linear operatorTfrom a Hilbert spaceXto itself. For present purposes, we say that a(continuous) linear operatorT:X Xisself-adjointif Tx,y = x,Ty for allx,y X.[ ] Lemma:ForTa self-adjoint (continuous linear) operator on a pre- Hilbert spaceX|T|= sup|x| 1| Tx,x |Proof:Letsbe that supremum. By the Cauchy-Schwarz-Bunyakowsky inequalitys |T|.For anyx,y X2| Tx,y + Ty,x |=| T(x+y),x+y T(x y),x y | | T(x+y),x+y |+| T(x y),x y | s|x+y|2+s|x y|2= 2s(|x|2+|y|2)1 Paul Garrett: Compact Operators on Hilbert Space (February 18, 2012)Lety=t Txwitht >0.

3 Using the self-adjointness ofT,| Tx,y + Ty,x |=| Tx,y |+| Ty,x |Dividing by 2,| Tx,y |+| Ty,x | s(|x|2+|y|2)Divide through bytand sett2=|Tx|/|x|to minimize the right-hand side. This gives| Tx,Tx |+| T2y,x | 2s|x||Tx|and2| Tx,Tx | 2s|x||Tx| 2s|x|2|T|The smallest non-negativesfor which this inequality is assured to hold for allx Xis|T| theoremLetTbe aself-adjoint compactoperator on a (non-zero) Hilbert spaceX. For complex , letX be the -eigenspaceX ={x X:Tx= x}ofTonX.[ ] Theorem: The completion of X is all ofX. That is, there is an orthonormal basis consisting ofeigenvectors. The only possibleaccumulation pointof the set of eigenvalues is 0, and ifXis infinite-dimensional itisanaccumulation point. The eigenspacesX arefinite-dimensional. All the eigenvalues arereal. One or the other of |T|is an eigenvalue ofTProof:The last assertion is the most crucial technical point. To prove it, we use the fact that for self-adjointTwe have|T|= sup|x| 1| Tx,x |And note that becauseTis self-adjoint any value Tx,x isreal.

4 Then choose a sequence{xn}so that|xn| 1 and| Tx,x | |T|. Then, replacing it by a subsequence if necessary, the sequence Tx,x of realnumbers has a limit = |T|.Then0 |Txn xn|2= Txn xn,Txn xn =|Txn|2 2 Txn,xn + 2|xn|2 2 2 Txn,xn + 2 The right-hand side goes to 0. Invoking thecompactnessofT, we can replacexnby a subsequence so asto be able to assume without loss of generality thatTxnconverges to some vectory. Then the previousinequality shows that xnconverges toy. For = 0, we have|T|= 0, soT= 0. For 6= 0, xn yimpliesxn 1yThus, lettingx= 1y, we haveTx= x2 Paul Garrett: Compact Operators on Hilbert Space (February 18, 2012)andxis the desired eigenvector with eigenvalue |T|.///Now we use a sort of induction. LetYbe the completion of the sum of all the eigenspaces. ThenYisT-stable. LetZ=Y . We claim thatZis alsoT-stable, and that on the Hilbert spaceZthe (restrictionof)Tis a Compact operator. Indeed, forz Zandy Y, we have Tz,y = z,Ty = 0which proves stabilityeasily.

5 And the unit ball inZis certainly a subset of the unit ballBinX, so ismapped byTto a pre- Compact setTB ZinX. SinceZisclosedinX, the intersectionTB ZofZwiththe pre- Compact setTBis pre- Compact . This proves thatTrestricted toZ=Y is still Compact . Theself-adjoint-ness is the restriction ofTtoZ. By construction,T1has no eigenvalues onZ, since any such eigenvaluewould also be an eigenvalue ofTonZ. But unlessZ={0}this would contradict the previous argumentwhich showed that |T1|is an eigenvalue on anon-zeroHilbert Space . Thus, it must be that the completionof the sum of the eigenspaces is all proceeding, note that in an infinite-dimensional Hilbert spaceYa ballBof positive radiusr >0isnotpre- Compact . Indeed, let{e1,e2,..}be a Hilbert Space basis. Then{re1,re2, }is a sequencewithnoconvergent subsequence, because all these points are distancer 2 prove that the eigenspaces are finite-dimensional, and that there are only finitely-many eigenvalues with| |> for given >0, letBbe the unit ball inY= | |> X Then the image ofBbyTcontains the ball of radius inY.

6 SinceTis Compact , this ball must bepre- Compact , so it must be thatYis finite-dimensional. Thus, since the dimensions of theX are positiveintegers, there can be only finitely-many of them with| |> , and each must be finite-dimensional. It followsthat the only possible accumulation point of the set of eigenvalues is 0. Further, ifXis infinite-dimensional,0mustbe an accumulation , we prove that all eigenvalues arereal. Indeed, letx X . Then x,x = Tx,x = x,Tx = Tx,x which implies that is real ifx6= of Compact Operators [ ] Proposition:Uniform-norm limits of Compact Operators on Banach spaces are :LetTn Tin uniform operator norm, where theTnare Compact . Given >0, letnbe sufficientlylarge such that|Tn T|< /2. SinceTn(B) is pre- Compact , there are finitely many pointsy1,..,ytsuchthat for anyx Bthere isisuch that|Tnx yi|< /2. By the triangle inequality|Tx yi| |Tx Tnx|+|Tnx yi|< This proves thatT(B) is covered by finitely many balls of radius .///[ ] Remark:The -finiteness hypothesis in the following theorem is necessary to make Fubini s theoremwork as Garrett: Compact Operators on Hilbert Space (February 18, 2012)[ ] Theorem:( Hilbert -Schmidt)LetX, andY, be -finite measure spaces.

7 LetK L2(X Y, ).Then the operatorT:L2(X, ) L2(Y, )defined byTf(y) = XK(x,y)f(x)d (x)is a Compact :We grant ourselves that, for orthonormal bases forL2(X) and forL2(Y), the collection offunctions (x) (y) is an orthonormal basis forL2(X Y). This plausible result is non-trivial, needingFubini s theorem and the -finiteness. Thus,K(x,y) = ijcij i(x) j(y)with complexcij, where we should not initially presume that the index set is countable. The square-integrability asserts that ij|cij|2=|K|2L2(X Y)< In particular, this implies that the indexing sets can be taken to be countable, since an uncountable sum ofpositive reals cannot converge. Then, givenf L2(X), the imageTfis inL2(Y), sinceTf(y) = ijcij f, i j(y)whoseL2(Y) norm is easily estimated by|Tf|22 ij|cij|2| f, i |2| j|22 |f|22 ij|cij|2| i|22| j|22=|f|22 ij|cij|2=|f|22 |K|2L2(X Y)We claim that we can writeK(x,y) = i i(x)T i(y)Indeed, the inner product inL2(X Y) of the right-hand side against any i(x) j(y) agrees with the innerproduct of the latter againstK(x,y).

8 In particular, with the coefficientscijfrom above, we see thatT i= jcij jSince ij|cij|2converges,limi|T i|2= limi|cij|2= 0In fact, for the same reason,limn i>n|T i|2= limn i>n|cij|2= 0 This fact is essential just the kernelKbyKn(x,y) = 1 i n i(x)T i(y)4 Paul Garrett: Compact Operators on Hilbert Space (February 18, 2012)These give the obvious finite-rank operatorsTnf(y) = XKn(x,y)f(x)dxGranting that anyn-dimensional subspace of a Hilbert Space is isomorphic toCn, with all open balls pre- Compact , these Operators are Compact . We claim that they approachTin operator norm. Indeed, letg= ici ibe inL2(X). Then(T Tn)g(y) = i>nbiT i(y)and by the triangle inequality and Cauchy-Schwarz-Bunyakowsky inequality|(T Tn)g(y)| i>n|bi|2|T i|2 ( i>n|bi|2)1/2( i>n|T i|22)1/2 |g|2 ( i>n|T i|22)1/2As observed in the previous paragraph,limn i>n|T i|22= 0 Thus,|T Tn| [ ] Remark:Given the -finiteness of the measure spaces, the argument above is correct whetherKismeasurable with respect to the product sigma-algebra or only with respect to of finite-rank operatorsA continuous linear operator is offinite rankif its image is finite-dimensional.

9 Note that a finite-rankoperator iscompact, since all balls are pre- Compact in a finite-dimensional Banach Space .[ ] Theorem:A Compact operatorT:X YwithYa Hilbert Space is a uniform operator norm limitof finite rank Operators , and :The converse is a special case of the previous theorem that operator norm limits of Compact operatorsare Compact (even in Banach spaces), since finite-rank Operators are the closed unit ball inX. SinceT(B) is pre- Compact it is totally bounded, so for given >0coverT(B) by open balls of radius centered at pointsy1,..,yn. Letpbe the orthogonal projection to thefinite-dimensional subspaceFspanned by theyiand defineT =p T. Note that for anyy Yand for anyyi|p(y) yi| |y yi|sincey=p(y) +y withy orthogonal to allyi. ForxinXwith|x| 1, by construction there isyisuchthat|Tx yi|< . Then|Tx T x| |Tx yi|+|T x yi|< + Thus,T Tin operator norm as [ ] Remark:The conclusion of the previous theorem is known to be false in Banach spaces, although theonly example known to this author (Per Enflo,Acta Math.)

10 , vol. 130, 1973) is rather complicated. Certainlythe previous argument using orthogonal projections cannot be employed.[ ] Remark:In the proof above that Hilbert -Schmidt Operators are Compact , we needed the fact thatfinite-dimensional subspaces of Hilbert spaces are linearly homeomorphic toCnwith its usual topology. In5 Paul Garrett: Compact Operators on Hilbert Space (February 18, 2012)fact, it is true thatanyfinite dimensional topological vector Space is linearly homeomorphic toCn. Thatis, we need not assume that the Space is a Hilbert Space , a Banach Space , a Frechet Space , locally convex,or anything else. However, the general argument is most reasonably a by-product of the development ofthe general theory of topological vector spaces, and is best delayed until we do that. Thus, we give moreelementary and immediate proofs that apply to Hilbert and Banach spaces, despite the fact that thesehypotheses are needlessly strong.[ ] Lemma:LetWbe a finite-dimensional subspace of a pre- Hilbert spaceV.