arXiv:1502.07205v2 [math-ph] 11 Aug 2015NoteonaFamilyofMonotoneQuantumRelativeEntropiesAndreas Deuchert, ChristianHainzl, Robert SeiringerAbstractGiven a convex function and two hermitian matrices A and B, Lewin and Sabinstudy in [1] the relative entropy dened by 1(A, B) = Tr [(A) (B) (B)(A B)].Amongst other things, they prove that the so-dened quantity is monotone if and only ifis operator monotone. The monotonicity is then used to properly dene 1(A, B) forbounded self-adjoint operators acting on an innite-dimensional Hilbert space by a limit-ing procedure. More precisely, for an increasing sequence of nite-dimensional projectionsPnn=1with Pn 1 strongly, the limit limn1(PnAPn, PnBPn) is shown to existand to be independent of the sequence of projections Pnn=1. The question whetherthis sequence converges to its obvious limit, namely Tr [(A) (B) (B)(A B)],has been left open. We answer this question in principle armatively and show thatlimn1(PnAPn, PnBPn) = Tr_(A) (B) dd(A + (1 )B) [=0. If the opera-torsAandBareregularenough,thatis(AB),(A) (B)and(B)(AB)aretrace-class,the identity Tr_(A) (B) dd(A + (1 )B) [=0=Tr [(A) (B) (B)(A B)]holds.MathematicsSubjectClassication(2010). 81Q99,46N50,47A99.Keywords. relativeentropy,operatormonotonicity.1I. INTRODUCTIONANDMAINRESULTSWe start with a quick review of the setting and the results of [1] that are of interest for us. Let C0([0, 1], R) beacontinuous,convexfunctionsuchthatiscontinouslydierentiableon(0, 1)andletAandBbetwohermitianmatriceswith0 A, B 1. LewinandSabindeneafamilyofrelativeentropiesofAwithrespecttoBbytheformula1(A, B) = Tr [(A) (B) (B)(AB)] . (1)Aslongas0 0.). When integrating the above expression,oneobtainsaprimitiveforwhichisoftheform(x) = ax +c b2_11_2x 1+ln (1 +(1 2x))2_d(). (11)Sincex ln(x)isanoperatorconvexfunction,thesameholdstruefor.Tokeepthemainargumentationstraight,werstprovetwotechnicalLemmata. Therstconcernstherelationbetweentheregularityofattheendpointsoftheinterval[0, 1]andthebehaviorofthemeasureinthevicinityof1and1.Lemma2. AssumeC0([0, 1], R)suchthat isoperatormonotoneon(0, 1). Then(1) = 0 = (1),_11/2ln(1 )d() < and_1/21ln(1 +)d() < . (12)6Ifinaddition C0([0, 1], R),thestrongerimplications_11/211 d() < and_1/2111 +d() < (13)hold. Incaseisnotcontinuousat1therstintegral inEq.(13)equals+andifitisnotcontinuousat0thisistrueforthesecondintegral.Proof. We start with the rst case, hence we assume that only is continuous on [0, 1]. Sincethelimitslimx0(x)andlimx1(x)existwecanconcludethat(1)=0=(1)holds. Wefurtherconcludethatthefollowinglimitexists(seeEq.(11)) >limx1_11/2ln(1 +(1 2x))2d() 14_11/2ln(1 )d(). (14)To come to the expression on the right hand side, we have applied Fatous Lemma. Doing thesame argumentation again, this time with the limit x 0, yields_1/21ln(1+)d() < .Ifalsoiscontinuouson[0, 1],wecomputelimx1(x) = a +b limx1_112x 11 (2x 1)d() a +_1111 d(), (15)where,as before,wehaveapplied Fatous Lemma. Thesameprocedure withlimx0(x)yields theother bound. Usingthemonotonicityof theintegrand, oneeasilyshows thatthe just discussedintegrals diverge to+incase is not continuous at 0and/or 1,respectively.In order to obtain a handy formula for the operatordd(A+ (1 )B) [=0, we explicitlycomputethedirectionalderivative.Lemma3. AssumeC0([0, 1], R)suchthatisoperatormonotoneon(0, 1). Ifiscontinuouson[0, 1]orifitisdiscontinuousat0and/or1andBhasnoeigenvalueatthesepointsthendd (, (A+ (1 )B))=0= a (, (AB)) b2_11__, 2(AB)_(16)2_0_,11 +(1 2B) +t(AB)11 +(1 2B) +t_dt_d(),wherea,bandaredenedbyEq. (10). Incaseof therst scenario(continuous on[0, 1]),thederivativeistakenforall hwhileinthesecondscenarioitistakenonlyfor7allinadensesetD h. Explicitly,thesetDisgivenbyD = >01( < B< 1 )hincase0and1arepointsof discontinuityofandbytheobviousgeneralizationwhenisdiscontinuous only at one of these points. This accounts for the fact that the limiting operatormaybeunbounded. IncasehasdiscontinuitiesandBhaseigenvaluesatatleastoneofthesepoints,wehavetotreattheabovelimitwithbeingoneoftheeigenvectors tothejustmentionedeigenvaluesseparately. Wedistinguishbetweentwocases. If (A B),=0wehave lim0_,_(A+ (1 B)) (B)__ = . (17)If(A B)= 0instead,thelimitinEq.(17)equalszero.Proof. UsingEq.(11),onecaneasilychecktheidentitydd (, (A+ (1 )B) )=0= a (, (AB)) b2lim0_11__, 2(AB)_(18)+_, ln (1 +(1 2(B +(AB)))) ln (1 +(1 2B))2__d().Thesecondtermis just thedierencequotient deningthedirectional derivativeof thesecondterm inEq.(11). Letushaveacloserlook at thetermwiththe logarithms. We usetheformulaln(x)=_0_11+t1x+t_dtandapplytheresolventidentityonce, toseethatitcanbewrittenas_, ln (1 +(1 2(B +(AB)))) ln (1 +(1 2B))2_ = (19)2_0_,11 +(1 2B) +t(AB)11 +(1 2(B +(AB))) +t_dt.Inordertoexplicitlycomputethelimit0, itneedstobeinterchangedinarststepwiththeintegral over andinasecondstepwiththeintegral over t. Thesecondstepwill followeasilyfromtheestimatesusedtoshowtherststepsince(1, 1)isthenxedwhichimpliesthatall resolventsareuniformlybounded. Wethereforefocusontheinterchange of the limit 0withthe integral over . Inorder tobe able toapplydominatedconvergence, wehavetondapositivefunctiong L1()with_, 2(AB)_2_0(, R(B)(AB)R(B +(AB))) dt g() (20)8forall atleast inadensesubset of h(ThecasewhereBhaseigenvalues atpointsofdiscontinuityofwillbetreatedattheend.). Toshortenthewriting,wehaveintroducedthenotationR(B) = (1 +(1 2B) +t)1.Let us rst investigate thebehavior of our integrandfor (1 +, 1 ). We writeR(B) =11+t1+t(1 2B)R(B)(andthesameforR(B+ (A B))) andevaluatethecontributionofthersttermwhichreads2_011 +t (, (AB))11 +tdt = 2 (, (AB)) . (21)Itcancelsthersttermundertheintegral ontherighthandsideof Eq. (18). Thethreeremainingtermshavenosingularityandcanbeboundedbyaconstant.In the vicinityof = 1and = 1 the situationis a littledierentand one needsto arguemorecarefully. Wewilldistinguishthreecasesdependingon theregularityofat 0and1and on the spectrum of B. First let us assume that is not continuous at 0 and 1 and thatBhas no eigenvaluesat these points. Let D=1( < B< 1 ) h and dene D = >0D.DuetoourassumptionsonB,thesetDisdenseinh. For D,weinvestigate_0_,11 +(1 2B) +t(AB)11 +(1 2(A+ (1 )B)) +t_dt (22)_0____11 +(1 2B) +t____|AB|____11 +(1 2(A+ (1 )B)) +t____dtwhichistherelevantcontributionfromEq.(20). Thepartoftheintegralovertfromsay1toiseasytocontrol. Onejustboundstheresolventsinoperatornormby1/t. Aftertheevaluation of the integral, we end up with a constant. To bound the other part of the integralover t (the one from 0 to 1), we use the fact that D which implies that___11+(12B)+t___ 1/foran>0thatdependson. Ontheotherhand___11+(12(A+(1)B))+t___ 11++tforcloseto1. Puttingthistogether,weobtain_0_____11 +(1 2B) +t________AB________11 +(1 2(A+ (1 )B)) +t_____dt (23) |AB|_1_1011 + +tdt +C_ C() (ln(1 +) + 1) .A similarbound canbe obtainedfor closeto1. Therethefunction ln(1 ) entersthe9analysis. Hence,thereexistsaconstantC()dependingonsuchthat_, 2(AB)_2_0_, R(B)(AB)R(B+(AB))_dt(24) C() (ln(1 [[) + 1) .BecauseofLemma2, theboundallowsustotakethelimitinsidetheintegral andprovestheclaiminthissituation.Nearly the same argumentation goes through when B has spectrum at 0 and/or 1 and if iscontinuousat thesepoints. Byboundingboth resolventslikewedidwiththesecondinthepreviousstep,that is|R(B)| (1 [[ +t)1and the samewith|R(A+ (1 )B)|,oneobtains_, 2(AB)_2_0_, R(B)(AB)R(B +(AB))_dtC1 [[. (25)AgainduetoLemma2, thisisenoughtointerchangethelimitandtheintegral. Thecasewhereisdiscontinuousonlyatonepointistreatedintheobviousway.For thelastcasewehavetoassumethatisnot continuousat 0and/or 1andthatBhasaneigenvalueatatleastoneofthesepoints. Weonlyinvestigatetherelevantcontribution.LetbetheeigenvectorofBtotheeigenvalue0forexample(theothercasesgothesameway). Wewillshowthatlim0_1/211_0(, R(B)(AB)R(B +(AB))) dtd() = , (26)if (AB) ,=0andthat the above limit equals zeroincase (AB) =0. UsingR(B+(AB)) = R(B)+2R(B+(AB))(AB)R(B),the integrand can be writtenas1(, R(B)(AB)R(B)) 2(, R(B)(AB)R(B +(AB))(A B)R(B))=_11 + +t_2_1(, A) 2(, AR(B +(AB))A)_. (27)Letusrstassumethat(A B),= 0whichimpliesthat(, A)> 0. Sincethefunctiont 1tis operator convexontheinterval (0, ), see[4, Exercise V.2.11], weknowthatR(A + (1 )B) R(A) + (1 )R(B). Ifweapplythisinequalityontherighthand10sideofEq.(27)anddiscardallpositivetermsinordertoobtainalowerbound,wend_11 + +t_2_1(, A) 2(, AR(B +(AB))A)_(28)_11 + +t_2_1(, A) 22(, AR(A)A) 2(, AR(B)A)_.Therighthandsideofthisequation, viewedasafunctionof,iscertainlymonotoneandsowecanusemonotoneconvergencetoshowthatlim0_1/21_0_11 + +t_2_1(, A) 22(, AR(A)A) (29)2(, AR(B)A)_dtd()= (, A)_1/211_0_11 + + t_2dtd() (, A)_1/2111 +d() = .The last equality is achieved with the help of Lemma 2. Now assume that (AB)= 0 whichmeans that A= 0. Hence, [A+ (1 )B] = 0 and (A+(1 )B)= (0). Sincethisexpressionisaconstantthederivativewithrespecttovanishes. AsimilarargumentcanbedonewhenBhas1asaneigenvalue. ThisconcludestheproofofLemma3.Beforewecometothemainpartoftheproof,wehavetoarguehowthetraceontherighthand side of Eq. (7) can be dened. Let us for the moment assume that B has no eigenvaluesatpointsofdiscontinuityofwith(A B),=0onthecorrespondingeigenspaces. ThenbyLemma3,wecandenethequadraticformq(, ) =lim0_,_(A) (B) (B +(AB)) (B)__(30)on the densesetD h (ThesetDhas been denedin Lemma3.). The operator convexityof implies that(B+(AB))(B)(A) (B) holds for all 0