Solutions to the Exercises of Lecture 13
Total Page:16
File Type:pdf, Size:1020Kb
Solutions to the Exercises of Lecture 13 Team Dresden d d Exercise 13.1. Let (Vn)n in L1(R ) and V 2 L1(R ). Characterise convergence of d Vn(m) ! V (m) in the strong operator topology of L(L2(R )) in terms of convergence of (Vn)n similar to as was done in Proposition 13.2.1. d Solution 13.1. Vn(m) ! V (m) in the strong operator topology of L(L2(R )) if and only d d if (Vn) is bounded in L1(R ) and Vn ! V in L1;loc(R ). d Proof. Let Vn(m) ! V (m) in the strong operator topology of L(L2(R )). Then (Vn(m))n d is bounded in L(L2;loc(R )) by the uniform boundedness principle and as in the proof of d d 1 d Proposition 13.2.1 (Vn) is bounded in L1(R ). So let K ⊂ R ) compact, then K 2 L2(R ) 1 1 d and Vn(m) K ! V (m) K in L2(K) and in L1(K). Hence Vn ! V in L1;loc(R ), since K is arbitrary. d d Assume Vn ! V in L1;loc(R ) and Vn;V bounded. Let f 2 L2(R ) with compact support. Then by the compactness of supp f we can interpret Vn;V as operators on some finite mea- d sure space; so we can apply Proposition 13.2.1 and derive kVn(m)f − V (m)fkL2(R ) ! 0. d By density we obtain Vn(m) ! V (m) in the strong operator topology of L(L2(R )). Exercise 13.2. Show that there exists an unbounded sequence (Vn)n in L1((0; 1)) and V 2 L1((0; 1)) with Vn ! V in L1((0; 1)). Solution 13.2. Let Vn = n · 1[1;n−2] and V = 0. n 1 1 n2 Then kVnkL1((0;1)) = n; so the sequence is unbounded in L1((0; 1)) and 1 kV − V k = ! 0; n L1((0;1)) n so Vn ! V in L1((0; 1)). 1 Exercise 13.3. Let (Ω; Σ; µ) be a finite measure space, (Vn)n a bounded sequence in L1(µ) and assume that Vn ! V in L1(µ) for some V 2 L1(µ). Show that there exists ν > 0 such that −1 −1 (@t,ν + Vn(m)) ! (@t,ν + V (m)) 1 in the strong operator topology of L L2,ν(R; L2(µ));Hν (R; L2(µ)) . Solution 13.3. First, we rewrite −1 −1 −1 (@t,ν + Vn(m)) = @t,ν @t,ν(@t,ν + Vn(m)) −1 = @t,ν Mn(@t,ν) −1 −1 where the functions Mn : dom(Mn) ! L(L2(µ)) are defined by Mn(z) = (1+z Vn(m)) for n 2 N. Now, if ν > supn2N jjVnjjL1(µ) we see that (Mn)n is a bounded sequence of material laws. By proposition 13.2.1(b) we obtain Vn(m) ! V (m) in the strong operator −1 topology of L(L2(µ)), which implies that (Mn(z))n converges to M(z) := (1 + z V (m)) in the strong operator topology of L(L2(µ)) for all z 2 CRe>ν. Then one has Mn(@t,ν) ! M(@t,ν) strongly by Thereom 13.1.1. and the claim follows. Exercise 13.4. Let D := S [n + 1=2; n + 1] and V := 1 (n·). For suitable ν > 0 n2Z n D compute the limit of −1 (@t,ν + Vn(m)) n in the weak operator topolgy of L2,ν R; L2((0; 1)) . Solution 13.4. Here is the graph of Vn. 1 n 1 n k By Theorem 13.2.3, Vn ! 2 in the weak* topology of L1(R ). Note that Vn = Vn for k 1 n every k 2 N. Hence, Vn ! 2 in the weak* topology of L1(R ) for every k 2 N. k 1 By Proposition 13.2.1, this implies that, Vn (m) ! 2 Id for every k 2 N in the weak operator topology of L2((0; 1)). Hence for ν > 1, Theorem 13.3.1 yields 1 X 1 (@ + V (m))−1 ! (−@−1)k @−1 t,ν n t,ν 2 t,ν k=0 1 1 X = @−1 (−@−1)k 2 t,ν t,ν k=0 1 = (@ + 1)−1: 2 t,ν 2 ∗ Exercise 13.5. Let H be a Hilbert space, c > 0 and c 6 Bn = Bn 2 L(H) for all n 2 N. Characterise, in terms of convergence of (Bn)n in a suitable sense, that −1 (@t,νBn) n converges in the weak operator topology. In the case of convergence, find its limit and a sufficient condition for which there exists a B 2 L(H) such that −1 −1 (@t,νBn) ! (@t,νB) in the weak operator topology. Solution 13.5. We first proof the following lemma. Lemma. Let H be a Hilbert space, (Bn)n be a sequence in L(H) such that Re Bn > c −1 (n 2 N); supn kBnk 6 d for some c; d > 0 and Bn ! C in L(H) in the weak operator topology. Then C is invertible in L(H): −1 −1 Proof. We know that Bn is invertible in L(H) and kBn k 6 1=c: Let y 2 H; x := Bn y (then kyk = kBnxk 6 dkxk): We have c RehB−1y; yi = Rehx; B xi ckxk2 kyk2 n n > > d2 c 2 for all n 2 N: This implies that RehCy; yi > d2 kyk (for all y 2 H), and therefore C is −1 2 invertible (with kC k 6 d =c). Let H be a Hilbert space, ν > 0 and (Bn)n be a sequence in L(H) such that Re Bn > c −1 for some c > 0. Then ((@t,νBn) )n converges in the weak operator topology if and only −1 if (Bn )n converges in the weak operator topology. Moreover, if sup kBnk 6 d for some −1 d > 0 and Bn ! C 2 L(H) in the weak operator topology, then C is invertible and −1 −1 −1 (@t,νBn) ! (@t,νB) (where B := C ). −1 −1 Proof. We know that Bn is invertible in L(H) and kBn k 6 1=c: Let Mn(z) := (zBn) for z 2 CRe>ν: Then (Mn)n is a bounded sequence in M(H; ν): The equivalence of con- −1 −1 vergence of ((@t,νBn) )n and (Bn )n in the weak operator topology now follows from Theorem 13.1.1. −1 Now assume sup kBnk 6 d for some d > 0 and Bn ! C 2 L(H) in the weak oper- ator topology. Then by the above Lemma we obtain that C is invertible. Let B = C−1: −1 −1 Applying Theorem 13.1.1 gives (@t,νBn) ! (@t,νB) in the weak operator topology. Exercise 13.6. Let H be a Hilbert space. Show that BL(H) := fT 2 L(H); kT k 6 1g is a compact subset under the weak operator topology. If, in addition, H is separable, show that BL(H) is also metrisable under the weak operator topology. Solution 13.6. The proof will be given in two parts: (a) for the compactness, (b) for the metrisability. The proof of (a) is remindful of the proof of the Banach–Alaoglu theorem. (a) The weak operator topology τw is defined as the initial topology with respect to the family of mappings (L(H) 3 A 7! hx; Ayi 2 K)x;y2H , i.e., τw is the coarsest topology on L(H) making all the mappings A 7! hx; Ayi 2 K continuous. Equivalently, τw is defined by the set fpx;y; x; y 2 Hg of seminorms, where px;y(A) := jhx; Ayij (A 2 L(H); x; y 2 H): 3 Let S denote the set of sesquilinear mappings b: H × H ! K. Then S is a closed subset of KH×H (with the product topology). Indeed, the mappings H×H ∗ 'λ,x,y;z : K ! K; b 7! b(λx + y; z) − λ b(x; z) − b(y; z); H×H λ,x,y;z : K ! K; b 7! b(x; λy + z) − λb(x; y) − b(x; z) are continuous. (The reason is, that the canonical projections KH×H 3 b 7! b(x; y) 2 K are continuous, for all (x; y) 2 H × H.) Therefore \ −1 \ −1 S = 'λ,x,y;z(0) \ λ,x,y;z(0) λ2K;x;y;z2H λ2K;x;y;z2H is a closed subset of KH×H . H×H We define the mapping η : L(H) ! K by η(A) := (hx; Ayi)(x;y)2H×H ; then obvi- ously η(L(H)) ⊆ S. (Note that τw is the initial topology with respect to the mapping η.) Q H×H The set C := (x;y)2H×H BK[0; kxkkyk] is a compact subset of K , by Tikhonov’s theorem; hence the closed subset C \ S of C is compact as well. We show that η(BL(H)) = C \ S.‘⊆’ follows from jhx; Ayij 6 kAkkxkkyk 6 kxkkyk (A 2 BL(H); x; y 2 H). ‘⊇’: b 2 C \ S means that b is sesquilinear and jb(x; y)j 6 kxkkyk (x; y 2 H). This implies that there exists A 2 L(H) with kAk 6 1 such that b(x; y) = hx; Ayi (x; y 2 H), i.e., b = η(A). As both topologies, τw on BL(H) and the topology of C \S, are induced by the product H×H topology on K , it follows that η :(BL(H); τw) ! C \ S is a homeomorphism; hence (BL(H); τw) is compact.