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Appendix 1 Operators on Hilbert Space Appendix 1 Operators on Hilbert Space We collect in this appendix necessary information on linear operators on Hilbert space. We give here almost no proofs and we give references for more detailed information. 1. Singular Values and Operator Ideals Let T be a bounded linear operator from a Hilbert space 11. to a Hilbert space K. The singular values Sn (T), n E Z+, of T are defined by sn(T) ~f inf{IIT - KII: K: 11. --t K, rankK::; n}, where rankK stands for the rank of K. Clearly, the sequence {sn(T)}n?:o is nonincreasing and its limit soo(T) ~f lim sn(T) n-+oo is equal to the essential norm of T, which is by definition IITlle = inf{IIT - KII: K E C(li, K)}, where C(1i, K) is the space of compact operators from 11. to K. It is easy to see that if Tl and T2 are operators from 11. to K, then sm+n(T1 + T 2 ) ::; sn(Td + sm(T2 ), m, nE Z+. If A, T, and B are bounded linear Hilbert space operators such that the product ATB makes sense, then it can easily be seen that 706 Appendix 1. Operators on Hilbert Space If T is a self-adjoint operator and D is a nonnegative rank one operator, and T # = T + D, then sn(T#) ~ sn(T) ~ sn+l(T#) ~ sn+l(T) for any nE 1:+. The operator T is compact if and only if lim sn(T) = O. If T is a n-HXl compact operator from 1l to K, it admits a Schmidt expansion Tx = L sn(T)(x, fn)gn, x E 1l, n::::O where {fn}n::::o is an orthonormal sequence in 1l and {gn}n::::O is an or­ thonormal sequence in K. Note that the sum is finite if and only if T has finite rank. If 0 < p < 00, we denote by Bp(1l, K) the Schatten-von Neumann class of operators T from H to K such that If this does not lead to a confusion, we write T E Bp instead of T E Bp(1l, K). If 1 ::; p < 00, the space Bp = Bp(H, K) is a Banach space with norm 11 . lisp' We refer the reader to Gohberg and Krein [2] or Simon [2] for the proof of the triangle inequality in Bp. Together with C(H,K) we use the notation Boo(H,K) for the space of compact operators from 1l to K. If p < 1, the space Bp is not a Banach space. However, the following triangle inequality is very useful for p < 1. Theorem ALL Let 0 < p < 1 and A, B E Bp. Then (ALl) IIA + BII~ p ::; IIAII~ p + IIBII~ p . ([)J. Moreover, IIA + BII~ p = IIAII~ p + IIBII~ p if and only if A* B = AB* = Since it is not easy to find the proof of this theorem in monographs, we give a proof here that is due to A.B. Aleksandrov (private communication); apparently, it is published here for the first time. We refer the reader to Rotfel'd [1] and C.A. McCarthy [1] for other proofs. Suppose that A = ( ~~ ) and S = ( :~ ) are distinct vectors. Suppose also that IAol > IA11 and Isol ~ IS11. Consider the function f on (0,00) defined by f(q) = IIAII£q -llslleq. (A1.2) We need the following fact whose proof is an easy exercise. Lemma Al.2. The function f can have at most one zero on (0, (0). If it has a zero, it changes sign at the zero. Appendix 1. Operators on HiIbert Space 707 Lemma AI.3. Let 0 < p < 1 and let A and B be rank one opera­ tors. Suppose that IIAII ~ IIBII and IIA + Blit ~ IIAII~p + IIBII~p. Then so{A + B) = IIAII and sl{A + B) = IIBII and, in particular, IIA + BII~p = IIAII~ p +IIBII~·p Proof. Assume that IIAII ~ IIBII. Let A = Ao{·,XO)Yo, B = Al{·,Xl)Yl, where xo, Xl E 1l, Yo, Yl E /C, IIxoll = IIxIII = IIyoII = IIYllI = 1, AO = IIAII, Al = IIBII· Let So = so{A + B) and SI = sl{A + B). Put By the assumptions, IIslllP ~ IIAIIlP. By the triangle inequality in 81. IIslll1 ::; IIAIIl1. Assume now that A f. s. It follows now from Lemma A1.2 that the function f defined by (A1.2) has one zero on [p, 1]. Clearly, f{q) < 0 for any q > 1. It follows that AO > so. Since rankA = 1, it follows that SI ::; Al. We have obtained a contradiction.• Corollary AI.4. If A and B are rank one operators, then IIA + BII~p ::; IIAII~p + IIBllt· Lemma AI.5. Suppose that A and B are rank one operators such that IIAII ~ IIBII. The following are equivalent: (i) so(A + B) = IIAII and sl(A + B) = IIBlli (ii) A*B = (()) and AB* = (()). Proof. It is sufficient to prove that (i) implies (ii). Again, let Ax = AO(X, xo)Yo, Bx = Al (x, XdYb X E 1l, where xo, Xl E 1l, Yo, Yl E /C, IIxoll = IIXlII = IIYoII = IIYllI = 1, AO = IIAII, Al = IIBjj. Clearly, we may assume that Al > o. We have . sl(A + B) ::; Al sup j(~I~I)1 = Al(l-j(xO,Xl)12)1/2. x..Lxo X It follows that {xo, xd = O. To prove that (yo, Yl) = 0, we can apply the above reasoning to the operators A * and B* .• Corollary AI.6. Let A and B be rank one operators. Then IIA + BII~ p = IIAII~ p + IIBII~ p if and only if A* B = (()) and AB* = (()). Lemma AI. 7. Let A : 1-l -t /C be an operator of rank at most Nand 0< p < 1. Then jjAllt = inf {f)AJII~p: A = tAj; rankAj ::; I}. J=l J=l (A1.3) Proof. Clearly, it is sufficient to consider on the right-hand side of (A1.3) only those Aj that satisfy Ker A C Ker Aj and Range Aj C Range A. The 708 Appendix 1. Operators on Hilbert Space set of such operators is finite-dimensional, and so the infimum on the right­ hand side is attained. N Let A = L: Aj be a representation that minimizes the infimum in (A1.3). j=l Clearly, it is sufficient to show that AjAk = (]) and AjAi; = (]). Suppose that AjAk =1= (]) or AjAi; =1= (]) for some j and k. By Lemma A1.5, one can represent Aj + Ak as the sum of rank one operators B1 and B2 such that N which contradicts the fact that the representation A = L: Aj realizes the j=l infimum in (A1.3) .• Proof of Theorem ALL Let us prove (ALl). Clearly, we can assume that A and B are finite rank operators. Let where the Aj and Bk are of rank one. Then N M A+B = LAj + LBk j=l k=l and by Corollary AI.4, N M IIA + BII~p ::; L IIAj lI~p + L IIBk II~p = IIAII~p + IIBII~p' j=l k=l It is obvious that the conditions A * B = (]) and AB* = (]) imply that IIA + BliPBp = IilAII PBp + liB liPBp . Suppose now that IIA + BII~ p = IIAII~ p + IIBII~ p. Consider the Schmidt expansions of A and B: A = LAj(-,Xj)Yj, B = Lftk(-,Uk)Vk. j k Then IIA+BII~p = LIAjIP+ LlftklP . j k It is sufficient to show Xj .1 Uk and Yj .1 Vk for all j and k. If this is not true for some jo and ko, then by Lemma Al.5, we can represent the operator Aj(',Xjo)Yjo + ftk("Uko)Vko as the sum of Al and B1 such that IIAj(-'Xjo)Yjo, + ftk("Uko)Vkoll~ p = IIA1 + B111~ p < IAjolP + IftkolP. Appendix 1. Operators on Hilbert Space 709 Hence, IIA + BII~p < IIAl + Blll~p + L IAjlP + L lJ-ljlP jho k=f.ko < IAjolP + lJ-lkol P + L IAjlP + L lJ-ljlP j=f.jo k=f.ko IIAII~ p + IIBII~ p , which contradicts the assumption .• Operators of class SI are also called nuclear operators or operators of trace class. The last term reflects the fact that on the space SI of operators ona Hilbert space H we can introduce the important functional trace. Suppose that {ej h::::o is an orthonormal basis in H. Put traceT = L(Tej, ej), T E SI· j::::O Then trace is a linear functional on SI and ItraceTI :::; IITlls,. Moreover, trace T does not depend on the choice of the orthonormal basis {ej} j::::O· If 1 < p < 00, the dual space S; = Sp(H, lC)* can be identified with the space Sp' = Sp' (H, lC) with respect to the pairing (T,R)=traceTR*, TESp, RESp'. (A1.4) With respect to the same pairing we can identify Si with the space B of bounded linear operators and we can identify the dual space C* to the space of compact operators with the space SI. The space S2 is also called the Hilbert~Schmidt class. It is a Hilbert space with respect to the inner product (A1.4). If H = L2(X, J-l) and lC = L2(y, v), an operator T E B(H, lC) belongs to S2 if and only if there exists a function k E L2(X X y, J-l x v) such that (Tf)(y) = Ix k(x,y)f(x)dJ-l(x). Consider now the ideals Sq,W' 1:::; q < 00, of operators T such that It is easy to see that Sp c Sq,w for any p < 00 and any q E [1,00).
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