||y||ϕ(y) = ϕ(||y||y ) where ||||y||y || = 1 so ||y||1 u ∈ ϕ[S]. Contradiction. Hence U ⊂ ϕ[B]. Thus ϕ[B] is open and ϕ−1 is continuous.
(*) Proof of Proposition 15.
34. Let M be a finite dimensional subspace of a Hausdorff topological vector space X. If x /∈ M , let N be the finite dimensional subspace spanned by x and M . Then N has the usual topology so x is not a point of closure of M . Hence M is closed.
35. Let A be a linear mapping of a finite dimensional Hausdorff vector space X into a topological vector space Y . The range of A is a finite dimensional subspace of Y so it has the usual topology. If xn→ x in X where xn =P a(n)i ei and x =P aiei, then a(n)i → ai for each i. Since the range of A has the usual topology, Axn=P a(n)i Aei→P aiAei = Ax. Hence A is continuous.
*36. Let A be a linear mapping from a topological vector space X to a finite dimensional topological space Y . If A is continuous, then its kernel M is closed by Q14. Conversely, suppose M is closed. There is a unique linear mapping B : X/M → Y such that A = B ◦ ϕ where ϕ : X → X/M is the natural homomorphism, which is a continuous open map by Q32. Then ker ϕ is closed and ϕ is an open map so X/M is a Hausdorff topological vector space. Furthermore, B is one-to-one and Y is finite dimensional so X/M is finite dimensional. By Q35, B is continuous. Hence A is continuous.
*37. Let X be a locally compact Hausdorff vector space. Let V be a neighbourhood of θ with ¯V compact and αV ⊂ V for each α with |α| < 1. The set {x +13V : x ∈ ¯V } is an open cover of ¯V so ¯V can be covered by a finite number of translates x1+13V, . . . , xn+13V . To show that x1, . . . , xn span X, it suffices to show that they span V . Let x ∈ V . Then x = xk1+13u1 for some k1∈ {1, . . . , n} and u1∈ U . Now 13U is covered by 13(x1+13V ), . . . ,13(xn+13V ) so x = xk1+ 13xk2 +19u2. Continuing in this way, we have x ∈ xk1+13xk2+ · · · +3r−21 xkr+31rU for each r. Let yr= xk1+13xk2+ · · · +3r−21 xkr. Then yris in the span of S, which is finite dimensional and thus closed by Q34.
Now for each y ∈ ¯V , there exists εy> 0 and an open neighbourhood Uyof y such that δUy⊂ V whenever
|δ| < εysince scalar multiplication is continuous at h0, yi. The open sets Uycover ¯V so ¯V ⊂ Uy1∪· · ·∪Uyn. Then δ ¯V ⊂ V whenever |δ| < min1≤k≤nεyk.
Choose N such that δ0V ⊂ V whenever |δ¯ 0| < 3−N. Let δ > 0 be given. Choose M such that 3−Mδ−1 < 3−N. When m ≥ M , we have 3−mδ−1 < 3−N so 3−mδ−1V ⊂ V . i.e. 3¯ −mV ⊂ δV . Thus¯ x − ym∈ δV for m ≥ M . Thus x − ym→ θ and ym→ x. Hence x is in the span of S. It follows that x1, . . . , xn span X so X is finite dimensional.
10.6 Weak topologies
38a. Suppose xn → x weakly. Then f (xn) → f (x) for each f ∈ X∗. Thus |f (xn)| ≤ Cf for each f ∈ X∗ and each n. Now let ϕ : X → X∗∗ be the natural homomorphism so that ϕ(xn)(f ) = f (xn). Then
|ϕ(xn)(f )| ≤ Cf for each f ∈ X∗ and each n. Since X∗ is a Banach space, h||ϕ(xn)||i is bounded but
||ϕ(xn)|| = ||xn|| so h||xn||i is bounded.
*38b. Let hxni be a sequence in `p, 1 < p < ∞, and let xn = hξm,ni∞m=1. Suppose hxni converges weakly to x = hξmi. By part (a), h||xn||i is bounded. Each bounded linear functional F on `pis given by F (xn) =P
mξm,nηm for some hηmi ∈ `q. Conversely, taking the m-th term to be 1 and the remaining terms to be 0 gives a sequence hηmi ∈ `qso that F (xn) =P ξm,nηm= ξm,nis a bounded linear functional on `p. Thus hF (xn)i converges to F (x). i.e. hξm,ni∞m=1 converges to ξm.
Conversely, suppose h||xn||i is bounded and for each m we have ξm,n → ξm. Then ||xn|| ≤ C and
||x|| ≤ C for some C. Let F ∈ (`p)∗ = `q. Let en be the sequence having 1 as the n-th term and 0 elsewhere. Then span{en} is dense in `q so there exists hFni with Fn ∈ span{en} and Fn → F . Given ε > 0, there exists N such that ||FN − F || < ε/3C. Since FN ∈ span{en} and ξm,n → ξm
for each n, there is an M such that |FN(xn) − FN(x)| < ε/3 for n ≥ M . Thus for n ≥ M , we have
|F (xn) − F (x)| ≤ |F (xn) − FN(xn)| + |FN(xn) − FN(x)| + |FN(x) − F (x)| < ε. Hence hxni converges weakly to x.
38c. Let hxni be a sequence in Lp[0, 1], 1 < p < ∞. Suppose h||xn||i is bounded and hxni converges to x in measure. By Corollary 4.19, every subsequence of hxni has in turn a subsequence that converges a.e. to x. By Q6.17, every subsequence of hxni has in turn a subsequence hxnkji such that for each y ∈ Lq[0, 1] we have R xnkjy →R xy. Now for each bounded linear functional F on Lp[0, 1], we have F (x) = R xy for some y ∈ Lq[0, 1]. Thus every subsequence of hxni has in turn a subsequence hxnkji such that F (xnkj) → F (x). By Q2.12, F (xn) → F (x). Hence hxni converges weakly to x.
38d. Let xn = nχ[0,1/n] for each n. Then xn → 0 and ||xn||1 = 1 for all n. In particular, hxni is a sequence in L1[0, 1] converging to 0 in measure. Let y = χ[0,1] ∈ L∞[0, 1]. Then F (x) =R xy is a bounded linear functional on L1[0, 1] and F (0) = 0 while F (xn) = ||xn||1= 1 for each n so F (xn) does not converge to F (0). Hence hxni does not converge weakly to 0.
(*) See Q6.17
38e. In `p, 1 < p < ∞, let xn be the sequence whose n-th term is one and whose remaining terms are zero. For any bounded linear functional F on `p, there exists hyni ∈ `q such that F (xn) = yn. Then F (xn) = yn → 0 since hyni ∈ `q. Thus xn → 0 in the weak topology. If hxni converges in the strong topology, then it must converge to 0 but ||xn||p= 1 for all n so it does not converge to 0 and thus does not converge in the strong topology.
38f. Let xn be as in part (e), and define yn,m = xn + nxm. Let F = {yn,m : m > n}. Note that the distance between any two points in F is at least 1 so there are no nonconstant sequences in F that converge in the strong topology. Any sequence in F that converges must be a constant sequence so its limit is in F . Hence F is strongly closed.
38g. Let F be as in part (f). The sets {x : |fi(x)| < ε, i = 1, . . . , n} where ε > 0 and f1, . . . , fn ∈ (`p)∗ form a base at θ for the weak topology. Given ε > 0 and f1, . . . , fn ∈ (`p)∗, fi(ym,n) is of the form ξin+ nξmi where hξnii ∈ `q. Choose n such that |ξni| < ε/2 for all i. Then choose m > n such that
|ξim| < ε/2n for all i. Then |fi(ym,n)| ≤ |ξni| + n|ξmi | < ε. Thus F ∩ {x : |fi(x)| < ε, i = 1, . . . , n} 6= ∅ and θ is a weak closure point of F .
Suppose hzki = hymk,nki = hxnk+ nkxmki is a sequence from F that converges weakly to zero. Given ε > 0 and hξni ∈ `q, there exists N such that |ξnk+ nkξmk| < ε for k ≥ N . Suppose {mk} is bounded above. Then some m is repeated infinitely many times. Let ξm= 1 and ξn = 0 otherwise. For each N there exists k ≥ N such that mk = m so |ξnk+ nkξmk| = |nk| ≥ 1. Thus {mk} is not bounded above and we may assume the sequence hmki is strictly increasing. Now suppose {nk} is bounded above. Then some n is repeated infinitely many times. Let ξn = 1 and ξm = 0 otherwise. For each N there exists k ≥ N such that nk = n so |ξnk+ nkξmk| = 1. Thus {nk} is not bounded above and we may assume the sequence hnki is strictly increasing. Now let ξmk = 1/nk for each k and ξm= 0 if m 6= mk for any k. Then hξni ∈ `q and |ξnk+ nkξmk| ≥ 1 for all k. Contradiction. Hence there is no sequence hzki from F that converges weakly to zero.
38h. The weak topology of `1 is the weakest topology such that all functionals in (`1)∗ = `∞ are continuous. A base at θ is given by the sets {x ∈ `1 : |fi(x)| < ε, i = 1, . . . , n} where ε > 0 and f1, . . . , fn ∈ `∞. A net h(x(α)n )i in `1 converges weakly to (xn) ∈ `1 if and only ifP
nx(α)n yn converges toP
nxnyn for each (yn) ∈ `∞.
If a net h(x(α)n )i in `1converges weakly to (xn) ∈ `1, then for each n, taking (yn) ∈ `∞where yn= 1 and ym= 0 for m 6= n, we have x(α)n converging to xn for each n.
If the net h(x(α)n )i in `1 is bounded, say by M , and x(α)n converges to xn for each n, then P
n|xn| ≤ P
n|x(α)n −xn|+P
n|x(α)n | ≤ M so (xn) ∈ `1. If (yn) ∈ `∞, then |P
n(x(α)n −xn)yn| ≤ ||(yn)||∞P
n|x(α)n − xn| → 0 soP
nx(α)n yn converges toP
nxnyn and h(x(α)n )i converges weakly to (xn).
For k ∈ N, let (x(k)n ) ∈ `1 where x(k)k = k and x(k)n = 0 if n 6= k. Then the sequence h(x(k)n )i is not bounded and x(k)n converges to 0 for each n. However, taking (yn) ∈ `∞where yn= 1 for all n, we have P
nx(k)n yn= k, which does not converge to 0 so h(x(k)n )i does not converge weakly to θ.
The weak* topology on `1as the dual of c0is the weakest topology such that all functionals in ϕ[c0] ⊂ `∞ are continuous. A base at θ is given by the sets {f ∈ `1 : |f (xi)| < ε, i = 1, . . . , n} where ε > 0 and x1, . . . , xn∈ c0. A net h(x(α)n )i in `1 is weak* convergent to (xn) ∈ `1 if and only ifP
nx(α)n yn converges
toP
nxnyn for all (yn) ∈ c0.
Using the same arguments as above and replacing `∞ by c0, we see that if a net h(x(α)n )i in `1 is weak*
convergent to (xn) ∈ `1, then x(α)n converges to xn for each n. We also see that if the net is bounded and x(α)n converges to xn, then the net is weak* convergent to (xn).
For k ∈ N, let (x(k)n ) ∈ `1 where x(k)k = k and x(k)n = 0 if n 6= k. Then the sequence h(x(k)n )i is not bounded and x(k)n converges to 0 for each n. However, taking (yn) ∈ c0 where yn = 1/n for all n, we haveP
nx(k)n yn = 1, which does not converge to 0 so h(x(k)n )i is not weak* convergent to θ.
39a. Let X = c0, F = `1 and F0 the set of sequences with finitely many nonzero terms, which is dense in `1. Consider the sequence h(x(k)n )i in c0 where x(k)k = k2 and x(k)n = 0 if n 6= k. For any sequence (yn) ∈ F0, we have P
nx(k)n yn = k2yk → 0. Thus the sequence h(x(k)n )i converges to zero in the weak topology generated by F0. Now if the sequence h(x(k)n )i converges in the weak topology generated by F , the weak limit must then be zero. Let (zn) be the sequence in F where zn = 1/n2 for each n. Then P
nx(k)n zn = 1. Thus the sequence h(x(k)n )i does not converge in the weak topology generated by F . Hence F and F0 generate different weak topologies for X.
Now suppose S is a bounded subset of X. We may assume that S contains θ. Let F be a set of functionals in X∗ and let F0 be a dense subset of F (in the norm topology on X∗). Note that in general, the weak topology generated by F0 is weaker than the weak topology generated by F . A base at θ for the weak topology on S generated by F is given by the sets {x ∈ S : |fi(x)| < ε, i = 1, . . . , n} where ε > 0 and f1, . . . , fn∈ F . A set in a base at θ for the weak topology on S generated by F0 is also in a base at θ for the weak topology generated by F . Suppose x ∈ S so that ||x|| ≤ M and |fi(x)| < ε for some ε > 0 and f1, . . . , fn ∈ F . For each i, there exists gi ∈ F0 such that ||fi− gi|| < ε/2M . If |gi(x)| < ε/2 for i = 1, . . . , n, then |fi(x)| ≤ |fi(x) − gi(x)| + |gi(x)| ≤ ||fi− gi|| ||x|| + |gi(x)| < ε for i = 1, . . . , n. Thus any set in a base at θ for the weak topology on S generated by F contains a set in a base at θ for the weak topology generated by F0. Hence the two weak topologies are the same on S.
*39b. Let S∗be the unit sphere in the dual X∗of a separable Banach space X. Let {xn} be a countable dense subset of X. Then {ϕ(xn)} is a countable dense subset of ϕ[X]. By part (a), {ϕ(xn)} generates the same weak topology on S∗ as ϕ[X]. i.e. {ϕ(xn)} generates the weak* topology on S∗. Now define ρ(f, g) = P 2−n1+|f (x|f (xn)−g(xn)|
n)−g(xn)|. Then ρ is a metric on S∗. Furthermore ρ(fn, f ) → 0 if and only if
|fn(xk) − f (xk)| → 0 for each k (see Q7.24a) if and only if |ϕ(xk)(fn) − ϕ(xk)(f )| → 0 for each k if and only if fn→ f in the weak* topology. Hence S∗is metrizable.
40. Suppose X is a weakly compact set. Every x∗ ∈ X∗ is continuous so x∗[X] is compact in R, and thus bounded, for each x∗ ∈ X∗. For each x ∈ X and x∗ ∈ X∗, there is a constant Mx∗ such that
|ϕ(x)(x∗)| = |x∗(x)| ≤ Mx∗. Thus {||ϕ(x)|| : x ∈ X} is bounded. Since ||ϕ(x)|| = ||x|| for each x, we have {||x|| : x ∈ X} is bounded.
41a. Let S be the linear subspace of C[0, 1] given in Q28 (S is closed as a subspace of L2[0, 1]. Suppose hfni is a sequence in S such that fn → f weakly in L2. By Q28c, for each y ∈ [0, 1], there exists ky ∈ L2 such that for all f ∈ S we have f (y) =R kyf . Now R fnky →R f ky for each y ∈ [0, 1] since ky∈ L2= (L2)∗. Thus fn(y) → f (y) for each y ∈ [0, 1].
41b. Suppose hfni is a sequence in S such that fn → f weakly in L2. By Q38a, h||fn||2i is bounded.
By Q28b, there exists M such that ||f ||∞≤ M ||f ||2 for all f ∈ S. In particular, ||fn||∞≤ M ||fn||2 for all n. Hence h||fn||∞i is bounded. Now hfn2i is a sequence of measurable functions with |fn|2≤ M0 on [0, 1] and fn2(y) → f2(y) for each y ∈ [0, 1] as a consequence of part (a). By the Lebesgue Convergence Theorem, ||fn||22→ ||f ||22 and so ||fn||2→ ||f ||2. By Q6.16, fn→ f strongly in L2.
*41c. Since L2is reflexive and S is a closed linear subspace, S is a reflexive Banach space by Q23d. By Alaoglu’s Theorem, the unit ball of S∗∗ is weak* compact. Then the unit ball of S is weakly compact since the weak* topology on S∗∗ induces the weak topology on S when S is regarded as a subspace of S∗∗. By part (b), the unit ball of S is compact. Thus S is locally compact Hausdorff so by Q37, S is finite dimensional.