NOTES ON SOME EXERCISES OF LECTURE 5, MODULE 2

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1 NOTES ON SOME EXERCISES OF LECTURE 5, MODULE 2 MARCO VITTURI Contents 1. Solution to exercise Solution to exercise Solution to exercise Solution to exercise Solution to exercise Solution to exercise 5-2 We want to prove that the following are equivalent: i) A is compact; ii) every net tx i u ipi Ă A has a cluster point. Notice that (i) is equivalent to the Finite Intersection Property (F.I.P.) Lemma 1. A is compact if and only if for every family of closed sets tf λ u λpλ such that F λ Ă A and Ş λpλ 0 F λ H for all finite subsets Λ 0 Ă Λ the intersection of all sets is non empty, i.e. Ş λpλ F λ H. Proof. Let A be compact and tf λ u λpλ a family of closed sets with Ş λpλ 0 F λ H whenever Λ 0 is a finite subset of Λ. Suppose by contradiction that Ş λpλ F λ H. Then the open sets A λ : AzF λ form an open covering of A, which by compactness has a finite open sub-covering ta λ u λpλ0. But then A ď č A λ Az F λ, λpλ 0 λpλ 0 i.e. Ş λpλ 0 F λ H, which is a contradiction. Now suppose that the Finite Intersection Property holds and take an open covering ta λ u λpλ of A. Define the closed sets F λ : AzF λ ; then Ş λpλ F λ H by assumption, and by the contrapositive of the Finite Intersection Property it follows there is a finite subset of indices Λ 0 s.t. Ş λpλ 0 F λ H. But this implies A Az č ď F λ A λ, λpλ 0 λpλ 0 proving A is compact. What we will prove is that F.I.P.ñ (ii) and viceversa then. First we prove a useful simple fact, namely that x being a cluster point of tx i u ipi is equivalent to x belonging to A i for all i P I, where A i : tx j s.t. j ě iu. Date: February 24,

2 2 MARCO VITTURI Remember indeed that x is a cluster point of a net if and only if P x nbhd of x, Dj P I s.t. j ě i and U x Q x j ; since x j P A i for j ě i, this can be reformulated equivalently P x nbhd of x, U x X A i H, which for any fixed i implies x P A i, so the two are equivalent. Proof of F.I.P. ñ(ii). Consider the family of closed sets ta i u ipi and let I 0 be a finite subset of I. Then the intersection Ş ipi 0 A i is non-empty since (by definition of nets) there exists k P I s.t. k ě i for all i P I 0, and therefore x k P A i for all i P I 0. By F.I.P. then we have that there exists x P Ş ipi A i. By the fact proved above, this x is a cluster point for the net tx i u ipi. Proof of (ii) ñ F.I.P.. Let tf λ u λpλ be a family of closed sets s.t. Ş λpλ 0 F λ H whenever Λ 0 is a finite subset of Λ. We introduce the following set of indices: I : tλ 0 finite subset of Λu; next we construct a net as follows: since for every Λ 0 P I it is Ş λpλ 0 F λ H, pick an element in the intersection and label it x Λ0. The collection tx Λ0 u Λ0PI is a net, because the set of indices I is a directed set with direction Ď: if Λ 0, Λ 1 P I, then Λ 0 Y Λ 1 is an element of I s.t. Λ 0, Λ 1 Ď Λ 0 Y Λ 1. Then by assumption this net has a cluster point x, i.e. as seen 0 P x nbhd of x, U x X tx Θ s.t. Θ P I, Θ Ě Λ 0 u H. We claim this cluster point proves that the intersection of all the F λ is non empty. For a fixed λ P Λ, choose Λ 0 tλu in the above, so x nbhd of x, U x X tx Θ s.t. Θ P I, Θ Q λu H; by definition of the x Θ we have that x Θ P Ť λ 1 PΘ F λ 1 Ă F λ, and therefore U x X F λ H for all nbhds of x, which implies x P F λ F λ. Since λ is arbitrary, x P Ş λpλ F λ, thus proving the Finite Intersection Property. 2. Solution to exercise 5-3 Let X, Y be normed spaces and ϕ : BpX, Y q Ñ R (not a priori continuous). We want to prove that the following are equivalent: i) φ can be written as ϕpt q f j pt x j q for an appropriate choice of x j P X and f j P Y ; ii) ϕ is continuous w.r.t. the weak operator topology; iii) ϕ is continuous w.r.t. the strong operator topology. Before proceeding on to the proof, remember that the weak operator topology is the one generated in BpX, Y q by the neighbourhoods of the form ts P BpX, Y q s.t. φ i psz i T z i q ă δ i u : N w pt, φ i, z i, δ i q, (1) 1 Here nbhd stands for neighbourhood.

3 NOTES ON SOME EXERCISES OF LECTURE 5, MODULE 2 3 for some choices of φ i P Y, z i P X and δ i ą 0; analogously, the strong operator topology is the one generated in BpX, Y q by the neighbourhoods of the form ts P BpX, Y q s.t. }Sz i T z i } Y ă δ i u : for some choices of z i P X and δ i ą 0. N s pt, z i, δ i q, (2) Proof of (i) ñ (ii). ϕ is continuous in the weak operator topology means that the inverse image of an open set of R is a set of operators in BpX, Y q that is open in the weak operator topology. In other words, if I is an open interval of R we have to prove that every T P ϕ 1 piq ts P BpX, Y q s.t. ϕpsq P Iu has a neighbourhood of the form (1) that is entirely contained in ϕ 1 piq. It suffices to consider I p r, rq of course. Fix T P ϕ 1 pp r, rqq, i.e. ϕpt q ă r r ϕpt q and let ε : 2. Then consider the following (weak)neighbourhood, mč N : N w pt, f j, x j, ε m q; we claim that N Ă ϕ 1 pp r, rqq. Indeed, let S P N, then we have that, since S P N w pt, f j, x j, ε mq for all j, ε ϕps T q ď f j psx j T x j q ď m ď ε, and therefore by triangular inequality ϕpsq ď ϕps T q ` ϕpt q ď ε ` ϕpt q ă r by choice of ε. Therefore S P ϕ 1 pp r, rqq too, or equivalently N Ă ϕ 1 pp r, rqq. Proof of (ii) ñ (iii). Assume that ϕ is continuous in the weak operator topology. Then, as just seen, this means that for any r ą 0 and for any T P ϕ 1 pp r, rqq we can find a (weak)neighbourhood N w of T of the form (1) s.t. N w Ă ϕ 1 pp r, rqq. We claim that we can find a (strong)neighbourhood N s of T of the form (2) s.t. N s Ă N w, thus proving continuity w.r.t. the strong operator topology. Let T P ϕ 1 pp r, rqq and let N w be given by then we define δ i N s : N s pt, z i, q }φ i } Y N w pt, φ i, z i, δ i q; ts P BpX, Y q s.t. }Sz i T z i } Y ă δ i u. }φ i } Y Let S P N s, then for any i 1,..., k φ i psx i T x i q ď }φ i } Y }Sx i T x i } Y ă }φ i } Y }φ i } Y thus proving that S P N w too and showing the claim holds. δ i δ i, Proof of (iii) ñ (i). By corollary 5.6 in the notes, ϕ is continuous w.r.t. the strong operator topology if and only if there exist x 1,..., x m s.t. ϕpt q ď C }T x j } Y,

4 4 MARCO VITTURI for some constant C ą 0. Let Y 0 : SpantpT x 1,..., T x m q for T P BpX, Y qu; this is a subspace of Y m in general. Define the linear functional this functional belongs to Y 0, since2 ΦpT x 1,..., T x m q : ϕpt q; ΦpT x 1,..., T x m q ď C }T x j } Y. Therefore we can extend it by Hahn-Banach theorem to a continuous functional Φ defined on all of Y m and coinciding with Φ on the subspace Y 0. Notice by linearity Φpy 1,..., y m q Φp0,..., 0, y j, 0,..., 0q : f j py j q, and therefore ϕpt q ΦpT x 1,..., T x m q ΦpT x 1,..., T x m q for any T P BpX, Y q, thus proving the claim. 3. Solution to exercise 5-7 f j pt x j q We want to show the existence of a sequence s.t. 0 is a cluster point of it in the weak topology, but no subsequence converges to 0 (always in the weak topology). This is radically different from the case of metric spaces, where the two are equivalent instead. Before we start, let s clarify the terminology. A point x P X is a cluster point of A in the weak topology if and only if it belongs to the closure of A in the weak topology, denoted by A w. Equivalently, for every (weak)neighbourhood N w of x it must be N w X A H. Since (weak)neighbourhoods are generated by neighbourhoods of the form ty P X s.t. φ j px yq ă δ j u : N w px, φ j, δ j q for some choice of φ j P X, δ j ą 0, it suffices to consider only these last ones. Moreover, a sequence tx i u i converges to x in the weak topology if and only if for every (weak)neighbourhood N w of x there is an i 0 s.t. for all i ě i 0 it is x i P N w. Equivalently, for all (weak)nbhds of the form N w px, φ j, δ j q (3) it holds that for i ě i 0 (for some i 0 sufficiently large) φ j px x i q ă δ j for all j 1,..., m. Heuristically, we are probing the distance of x and x i through a limited set of lenses (the finite collection of φ j s). The setup is the following. Consider H a Hilbert space with orthonormal basis te n u npn, and define the sequence Let A tx m,n s.t. m, n P Nu. x m,n : e m ` me n. 2 The sum gives a norm for Y m, as can be easily verified.

5 NOTES ON SOME EXERCISES OF LECTURE 5, MODULE 2 5 Proof that 0 P A w. As shown in the introduction, we have to show that for every (weak)neighbourhood N w of 0 it is x m,n P N w for some indices m, n. It suffices to consider nbhds of the form (3), thus fix one of these, say N w N w p0, φ j, δ j q, where φ j P H. By Riesz representation theorem there exist z j s.t. φ j pxq xz j, xy for all x P H, so we want to show there exist m, n s.t. xz j, x m,n y ă δ j (4) for all j 1,..., k. Indeed, let z j ř npn αpjq n e n be the expansion of z j in terms of vectors of the basis, then since z j 2 ř n αpjq n xz j, x m,n y α pjq m 2 ă 8 it is αn pjq Ñ 0. Then ` mαn pjq ď αm pjq ` m αn pjq, which can be made smaller than δ j ą 0 by taking m large enough that αm pjq ă δ j {2 and n large enough that αn pjq ă δ j {2m. Since there are only a finite number of js, we can find m and n large enough that (4) is satisfied for all j. Proof that no subsequence converges weakly to 0. Suppose by contradiction that there exists a subsequence tx mj,n j u jpn that converges weakly to 0, i.e. for every (weak)neighbourhood N w we have x mj,n j P N w eventually 3. We will choose a bad neighbourhood and show that this is not possible. First of all, we can see that there is no M ą 0 such that m j ă M for all j. Indeed, suppose this is the case, then take Mÿ z : and consider the neighbourhood N w p0, xz, y, 1{2q. We would have k 0 e k xz, x mj,n j y 1 ` m j xz, e nj y ě 1, so that x mj,n j R N w p0, xz, y, 1{2q, which contradicts our assumption. Therefore m j is unbounded, and by taking a further subsequence we can assume that m j is monotone increasing. With a symmetric reasoning we see that n j is unbounded too, and therefore we can assume that n j is monotone increasing too. With a small abuse of notation, we denote this final subsequence again by tx mj,n j u jpn. Now we choose the bad neighbourhood mentioned before. Let z 1 : ÿ 1 e nj. m jpn j Notice this is well defined as an element of H indeed, because }z 1 } 2 xz 1, z 1 y ÿ 1 m 2 ď ÿ 1 jpn j n 2 ă 8. npn Now consider the neighbourhood we see that for any j N w : N w p0, xz 1, y, 1{2q; xz 1, x mj,n j y ě 1 m j m j xe nj, e nj y ě 1, which again contradicts our assumption on the subsequence. 3 i.e. for all j except a finite number of them.

6 6 MARCO VITTURI 4. Solution to exercise 5-8 Remember that by σpx, X q, the weak-topology on X, we denote the topology of X generated by neighbourhoods of x P X of the form ty P X s.t. φ j py xq ă δ j u : N px, φ j, δ j q for some choice of φ j P X and δ j ą 0. Analogously, by σpx, Xq, the weak*-topology on X, we denote the topology of X generated by neighbourhoods of φ P X of the form tψ P X s.t. ψpx j q φpx j q ă δ j u : N pφ, x j, δ j q for some choice of x j P X and δ j ą 0. Notice that although similar the second one is not completely specular to the first one, i.e. it s not the same as σpx, X q because in general X can only be identified as a (proper) subspace 4 of X ; in particular σpx, Xq is a coarser topology than σpx, X q in general. Let B X denote the closed unit ball of the space X w.r.t. its metric. We want to show that in a reflexive 5 (and therefore Banach) space the image of B X under any continuous linear operator T is closed. Notice that by continuity we always have in the other direction that T 1 pb X q is closed. To do this, notice that Banach-Alaoglu theorem says that B X is weak*-compact for any normed space X, so in particular B X is weak**-compact in X, i.e. compact w.r.t. the σppx q, X q topology, but since X X this is the same topology as σpx, X q if we identify X with X. In other words, B X is weakly compact in X, i.e. compact w.r.t. σpx, X q. Now, observe that by exercise 5-4 (not included in these notes), a linear operator T between normed spaces X and Y is norm continuous 6 if and only if it is weakly continuous 7. Therefore T is weakly continuous, and since continuous functions map compact sets to compact sets, it follows that T pb X q is weakly-compact, i.e. compact w.r.t. σpx, X q. Since X is Banach in particular it is a Hausdorff space 8, and compact sets are closed in a Hausdorff space. Therefore T pb X q is (weakly)- closed too, or T pb X q T pb X q w. Now, Theorem 5.14 in the notes tells us that a convex set K in a normed space is closed if and only if it is weakly closed. Since T pb X q is obviously convex, then this implies T pb X q T pb X q w T pb X q, 4 Think of X L 1, then pl 1 q L 8 and pl 1q pl 8 q ŋ L 1. 5 i.e. X X. 6 i.e. there is C ą 0 s.t. }T x}y ď C}x} X for all x. 7 i.e. for every neighbourhood in Y of the form NY w Ş k ty P Y s.t. φ jpy y 0 q ă δ j u for some choice of φ j P Y, δ j ą 0, the inverse image T 1 pny w q contains a neighbourhood in X of the form NX w Ş m l 1 tx P X s.t. ψ lpx x 0 q ă ε l u for some ψ l P X, ε l ą 0. Equivalently, by Proposition 5.5 in the notes, T is weak continuous if and only if for all φ P Y there exist C C φ ą 0, ψ 1,..., ψ m P X s.t. φpt xq ď C ψ j P X. 8 i.e. for every x y there are Ux nbhd of x and W y nbhd of y s.t. U x X W y H. I m making a big fuss here, but this is essentially trivial - moreover, all spaces we consider in the course are Hausdorff, as non-hausdorff spaces are terribly ill-behaved.

7 NOTES ON SOME EXERCISES OF LECTURE 5, MODULE 2 7 i.e. T pb X q is closed. 5. Solution to exercise 5-10 The natural embedding of a Banach space X into X is given by the map ι : X ãñ X defined as ιpxq X Q φ ÞÑ φpxq, or rιpxqspφq φpxq. We are asked to prove that the image ιpb X q of the unit ball B X is σpx, X q-dense in the unit ball B X of X. Notice that writing Y X, we have σpx, X q σpy, Y q, i.e. the weak*-topology on Y. Just to be clear, to be dense in such a topology means that for every neighbourhood N of Ξ 0 P B X of the form N tξ P B X s.t. Ξpφ j q Ξ 0 pφ j q ă δ j u for some choice of φ j P X, δ j ą 0, there is x P B X s.t. ιpxq P N, i.e. for all j 1,..., k. φ j pxq Ξ 0 pφ j q rιpxqspφ j q Ξ 0 pφ j q ă δ j In this solution we don t use explicitely the definition of denseness though - it would be clumsy. We keep the discussion high-level. Suppose by contradiction that 9 K : ιpb X q B X (in particular, it must be K Ĺ B X ). Take then Ξ P B X zk. Since K is a σpx, X q-closed and convex set, by Theorem 5.8 in the notes (one of the versions of Hahn-Banach) there exists a σpx, X q-continuous linear functional f : X Ñ R s.t. fpξq ą 1 ą P K. Now, let s consider what it means to be σpx, X q-continuous. σpx, X q is the smallest topology that makes the functionals ι φ pθq : Θpφq continuous as a function of Θ P X, where φ P X. In particular, these are the only σpx, X qcontinuous functionals by Theorem 5.13 (ii) in the notes. Therefore, there exists φ P X s.t. fpθq ι φ pθq Θpφq for all Θ P X. It then follows that }φ} X but then a contradiction. sup φpxq sup rιpxqspφq sup fpιpxqq sup fpθq ď 1; xpb X xpb X xpb X ΘPK 1 ă fpξq Ξpφq ď }Ξ} X }φ} X ď }φ} X, Marco Vitturi, Room 4606, James Clerk Maxwell Building, University of Edinburgh, Peter Guthrie Tait Road, Edinburgh, EH9 3FD. address: m.vitturi@sms.ed.ac.uk 9 the closure is obviously taken in the topology σpx, X q.