Chapter 8 Bounded Linear Operators on a Hilbert Space

1 Chapter8 BoundedLinearOperatorsona HilbertSpaceIn thischapterwe describe someimportant classesof boundedlinearoperatorsonHilbertspaces,in cludingprojections,unitaryoperators,ands elf-adjoint alsoprove theRieszrepresentationtheorem,which characterizestheboundedlinearfunctionals on a Hilbertspace,anddiscussweakconvergencein beginby describingsomealgebraicpropertiesof a linearspaceXsuch thateveryx2 Xcanbe writtenuniquelyasx=y+zwithy2 Mandz2N, thenwe say thatX=M Nis thedirect sumofMandN, andwe callNacomplementarysubspaceofMinX. Thedecompositionx=y+zwithy2 Mandz2 Nis uniqueif andonlyifM\N=f0g. A givensubspaceMhasmany ,ifX=R3andMis a planethroughtheorigin,thenany linethroughtheoriginthatdoes notlieinMis a ,andthedimensionof a complementarysubspaceis N, thenwe de netheprojectionP:X!XofXontoMalongNbyP x=y, wherex=y+zwithy2 Mandz2N. Thisprojectionis Linear ,withranP=MandkerP=N, andsatis esP2=P.

2 Aswe willshow,thispropertycharacterizesprojec tions,so we make thefollowingde linearspaceXis a linearmapP:X!XsuchthatP2=P:( )Any projectionis associatedwitha a Linear Operatorson a HilbertSpace(a)IfP:X!Xis a projection,thenX= ranP kerP.(b)IfX=M N, whereMandNarelinearsubpacesofX, thenthereis aprojectionP:X!XwithranP=MandkerP= prove (a),we rstshow thatx2ranPif andonlyifx=P x. Ifx=P x, thenclearlyx2ranP. Ifx2ranP, thenx=P yforsomey2X, andsinceP2=P, it followsthatP x=P2y=P y= \kerPthenx=P xandP x= 0, so ranP\kerP=f0g. Ifx2X,thenwe havex=P x+ (x P x);whereP x2ranPand(x P x)2kerP, sinceP(x P x) =P x P2x=P x P x= 0:ThusX= ranP prove (b),we observe thatifX=M N, thenx2 Xhastheuniquedecompositionx=y+zwithy2 Mandz2N, andP x=yde nestherequiredprojection. WhenusingHilbertspaces,we areparticularlyinterestedin a closedsubspaceof a HilbertspaceH. Then, ,we haveH=M M?

3 We calltheprojectionofHontoMalongM?theortho gonalprojectionofHontoM. Ifx=y+zandx0=y0+z0, wherey; y02 Mandz; z02M?, thentheorthogonality ofMandM?impliesthathP x; x0i=hy; y0+z0i=hy; y0i=hy+z; y0i=hx; P x0i:( )Thisequationstatesthatanorthogonalproje ctionis self-adjoint ( ).Aswe willshow, theproperties( )and( ) thereforemake thefollowingde HilbertspaceHis a linearmapP:H!Hthatsatis esP2=P;hP x; yi=hx; P yiforallx; y2H:Anorthogonalprojectionis a nonzeroorthogonalprojection,thenkPk= HandP x6= 0, thentheuseof theCauchy-SchwarzinequalityimpliesthatkP xk=hP x; P xikP xk=hx; P2xikP xk=hx; P xikP xk kxk:ThereforekPk 1. IfP6= 0, thenthereis anx2 HwithP x6= 0, andkP(P x)k=kP xk, so thatkPk 1. Orthogonalprojections189 Thereis a one-to-onecorrespondencebetweenorthogona lprojectionsPandclosedsubspacesMofHsuch thatranP=M. Thekernelof theorthogonalprojectionis theorthogonalcomplement a Hilbertspace.

4 (a)IfPis anorthogonalprojectiononH, thenranPis closed,andH= ranP kerPis theorthogonaldirectsumof ranPandkerP.(b)IfMis a closedsubspaceofH, thenthereis anorthogonalprojectionPonHwithranP=Mandk erP=M?. prove (a),supposethatPis anorthogonalprojectiononH. Then, ,we haveH= ranP kerP. Ifx=P y2ranPandz2kerP, thenhx; zi=hP y; zi=hy; P zi= 0;so ranP?kerP. Hence,we seethatHis theorthogonaldirectsumof ranPandkerP. It followsthatranP= (kerP)?, so ranPis prove (b),supposethatMis a closedsubspaceofH. M?. We de nea projectionP:H!HbyP x=y;wherex=y+zwithy2 Mandz2M?:ThenranP=M, andkerP=M?. Theorthogonality ofPwas shownin ( )above. IfPis anorthogonalprojectiononH, withrangeMandassociatedorthogonaldirects umH=M N, thenI Pis theorthogonalprojectionwithrangeNandasso ciatedorthogonaldirectsumH=N (R) is theorthogonaldirectsumof thespaceMofevenfunctionsandthespaceNof , respectively, aregiven byP f(x) =f(x) +f( x)2;Qf(x) =f(x) f( x)2:NotethatI P= a measurablesubsetofR|forexample,aninterva l |withcharacteristicfunction A(x) = 1ifx2A, (x) = A(x)f(x)190 Bounded Linear Operatorson a HilbertSpaceis anorthogonalprojectionofL2(R) onto thesubspaceof frequentlyencounteredcaseis thatof projectionsontoa one-dimensionalsubspaceof a HilbertspaceH.

5 For any vectoru2 Hwithkuk= 1, themapPude nedbyPux=hu; xiuprojectsa vectororthogonallyonto itscomponent in thedirectionu. Mathemati-ciansusethetensorproductnotati onu uto ,ontheotherhand,oftenusethe\bra-ket"nota tionintroducedby thisnotation,anelementxof a Hilbertspaceis denotedby a \bra"hxjor a \ket"jxi,andtheinnerproductofxandyis denotedbyhxjyi. Theorthogonalprojectionin thedirectionuis thendenotedbyjuihuj, so that(juihuj)jxi= , theorthogonalprojectionPuin thedirectionof a unitvectoruhastherankonematrixuuT. Thecomponent of a vectorxin thedirectionuisPux= (uTx) (Z), andu=en, whereen= ( k;n)1k= 1;andx= (xk), thenPenx= (T) is thespaceof 2 -periodicfunctionsandu= 1=p2 is theconstant functionwithnormone,thentheorthogonalpro jectionPumapsafunctionto itsmean:Puf=hfi, wherehfi=12 Z2 0f(x)dx:Thecorrespondingorthogonaldecomp osition,f(x) =hfi+f0(x);decomposesa functioninto a constant meanparthfianda a HilbertspaceAlinear functionalona complexHilbertspaceHis a linearmapfromHtoC.

6 Alinearfunctional'is Bounded ,or continuous,if thereexistsa constantMsuch thatj'(x)j Mkxkforallx2H:( )Thedualof a Hilbertspace191 Thenormof a boundedlinearfunctional'isk'k=supkxk=1j' (x)j:( )Ify2H, then'y(x) =hy; xi( )is a boundedlinearfunctionalonH, withk'yk= (T).Then,foreachn2Z, thefunctional'n:L2(T)!C,'n(f) =1p2 ZTf(x)e inxdx;thatmapsa functionto itsnth Fouriercoe cient is a havek'nk= 1 thefundamentalfactsaboutHilbertspacesis thatallboundedlinearfunctionalsareof theform( ). (Rieszrepresentation)If'is a boundedlinearfunctionalonaHilbertspaceH, thenthereis a uniquevectory2 Hsuch that'(x) =hy; xiforallx2H:( ) '= 0, theny= 0, so we supposethat'6= thatcase,ker'isa proper closedsubspaceofH, a nonzerovectorz2 Hsuch thatz?ker'. We de nea linearmapP:H!HbyP x='(x)'(z)z:ThenP2=P, so ranP kerP. Moreover,ranP=f zj 2Cg;kerP= ker';so thatranP?kerP. It followsthatPis anorthogonalprojection,andH=f zj 2Cg ker'is canthereforewritex2 Hasx= z+n; 2 Candn2ker':Takingtheinnerproductof thisdecompositionwithz, we get =hz; xikzk2;192 Bounded Linear Operatorson a HilbertSpaceandevaluating'onx= z+n, we ndthat'(x) = '(z):Theeliminationof fromtheseequations,anda rearrangement of theresult,yields'(x) =hy; xi;wherey='(z)kzk2z:Thus,everyboundedlin earfunctionalis givenby theinnerproductwitha have alreadyseenthat'y(x) =hy; xide nesa boundedlinearfunctionalonHforeveryy2H.

7 To prove thatthereis a uniqueyinHassociatedwitha givenlinearfunctional,supposethat'y1='y2 . Then'y1(y) ='y2(y) wheny=y1 y2,which impliesthatky1 y2k2= 0, soy1=y2. ThemapJ:H ! H givenbyJy='ythereforeidenti esa HilbertspaceHwithitsdualspaceH . Thenormof'yis equalto thenormofy( ),soJis an isometry. In thecaseof complexHilbertspaces,Jis antilinear,ratherthanlinear,because' y= 'y. Thus,Hilbertspacesareself-dual, meaningthatHandH areisomorphicas Banach spaces,andanti-isomorphicas anin nite-dimensionalBanachspace,such as anLp-spacewithp6= 2 orC([a; b]), is in quantummechanics,theobservablesof a systemarerepresentedby a spaceAof linearoperatorsona HilbertspaceH. Astate!of a quantummechanicalsystemis a linearfunctional!onthespaceAof observableswiththefollowingtwo properties:!(A A) 0forallA2A;( )!(I) = 1:( )Thenumber!(A) is theexpectedvalueof theobservableAwhenthesystemisin thestate!

8 Condition( )is calledpositivity, andcondition( )is be speci c,supposethatH=CnandAis thespaceof alln a Hilbertspacewiththeinnerproductgiven byhA; Bi= trA B:BytheRieszrepresentationtheorem,foreac h state!thereis a unique 2 Asuchthat!(A) = tr AforallA2A:Theadjointof an operator193 Theconditions( )and( )translateinto 0, andtr = 1, theRieszrepresentationtheoremis given in ,wherewe useit to prove theexistenceanduniquenessof weaksolutionsof Laplace' of anoperatorAnimportant consequenceof theRieszrepresentationtheoremis theexistenceoftheadjointof a boundedoperatorona ningproperty of theadjointA 2B(H) of anoperatorA2B(H) is thathx; Ayi=hA x; yiforallx; y2H:( )TheuniquenessofA nitionimpliesthat(A ) =A;(AB) =B A :To prove thatA exists,we have to show thatforeveryx2H, thereis a vectorz2H, dependinglinearlyonx, such thathz; yi=hx; Ayiforally2H:( )For xedx, themap'xde nedby'x(y) =hx; Ayiis a boundedlinearfunctionalonH, withk'xk kAkkxk.

9 BytheRieszrepresen-tationtheorem,thereis a uniquez2 Hsuch that'x(y) =hz; yi. Thiszsatis es( ),so we setA x=z. Thelinearity ofA followsfromtheuniquenessin theRieszrepresentationtheoremandthelinea rity of theadjoint of a linearmaponRnwithmatrixAisAT, sincex (Ay) = ATx y:In component notation,we havenXi=1xi0@nXj=1aijyj1A=nXj=1 nXi=1aijxi!yj:Thematrixof theadjoint of a linearmaponCnwithcomplexmatrixAis theHermitianconjugatematrix,A =AT:194 Bounded Linear Operatorson a andleftshiftoperatorsonthesequencespace` 2(N), de nedbyS(x1; x2; x3; : : :) = (0; x1; x2; x3; : : :);T(x1; x2; x3; : : :) = (x2; x3; x4; : : :):ThenT=S , sincehx; Syi=x2y1+x3y2+x4y3+: : :=hT x; :L2([0;1])!L2([0;1])be anintegraloperatorof theformKf(x) =Z10k(x; y)f(y)dy;wherek: [0;1] [0;1]!C. Thentheadjoint operatorK f(x) =Z10k(y; x)f(y)dyis theintegraloperatorwiththecomplexconjuga te, plays a crucialrolein studyingthesolvability of a linearequationAx=y;( )whereA:H !

10 His a Hbe any solutionof thehomogeneousadjoint equation,A z= 0:We take theinnerproductof ( )withz. Theinnerproductontheleft-handsidevanishe sbecausehAx;zi=hx; A zi= 0:Hence,a necessaryconditionfora solutionxof ( )to existis thathy; zi= 0forallz2kerA , meaningthaty2(kerA )?. Thisconditiononyis notalwayssu cient toguaranteethesolvability of ( );themostwe cansay forgeneralboundedoperatorsis :H!His a boundedlinearoperator,thenranA= (kerA )?;kerA= (ranA )?:( ) , thereis ay2 Hsuch thatx=Ay. For anyz2kerA , wethenhavehx; zi=hAy; zi=hy; A zi= 0:Theadjointof an operator195 Thisproves thatranA (kerA )?. Since(kerA )?is closed,it followsthatranA (kerA )?. Ontheotherhand,ifx2(ranA)?, thenforally2 Hwe have0 =hAy; xi=hy; A xi:ThereforeA x= 0. Thismeansthat(ranA)? kerA . Bytakingtheorthogonalcomplement of thisrelation,we get(kerA )? (ranA)??=ranA;which proves the rstpartof ( ).