The unit vectors \(e_1,e_2,e_3,\dots\) of \(\ell^2\) all have norm \(1\) and are pairwise at distance \(\sqrt2\) , so no subsequence converges. Yet in every way that a single measurement can detect, they tend to zero. Each coordinate of \(e_n\) is eventually \(0\) . For any fixed \(y\in\ell^2\) the pairing \(\sum_k(e_n)_ky_k=y_n\) tends to \(0\) , because the terms of a convergent series do. The sequence runs off to infinity through ever new dimensions, and no bounded functional can follow it.
The functions \(\sin(n\pi x)\) on \([0,1]\) do the same thing by oscillating. Their size, measured by \(\int_0^1\sin^2(n\pi x)\,dx=\tfrac12\) , does not shrink, but their integral against any fixed function tends to zero as the oscillations cancel. In both cases there is a perfectly good notion of limit, namely the zero vector, provided that convergence is tested through functionals instead of through the norm. This chapter develops that notion. Its reward is the recovery of what Theorem 1.10 took away: with respect to the weaker convergence, bounded sets do have convergent subsequences, in every reflexive space and in every dual of a separable space. Its cost is that the norm, and every other nonlinear quantity, is no longer continuous.
6.1 Convergence tested by functionals
Definition 6.1 (weak convergence). A sequence \((x_n)\) in a normed space \(X\) converges weakly to \(x\in X\) , written \(x_n\rightharpoonup x\) , if \(f(x_n)\to f(x)\) for every \(f\in X^*\) .
Convergence in norm is also called strong convergence in this context. By Theorem 3.1, weak convergence in \(\ell^p\) for \(1\le p<\infty\) means \(\sum_kx^{(n)}_ky_k\to\sum_kx_ky_k\) for every \(y\in\ell^q\) ; and by the duality of \(L^p\) and \(L^q\) quoted in Section 3.1, weak convergence in \(L^p\) for \(1\le p<\infty\) means \(\int f_ng\to\int fg\) for every \(g\in L^q\) .
Proposition 6.2 (basic properties of weak convergence). Let \(x_n\rightharpoonup x\) in a normed space \(X\) . Then: (a) the weak limit is unique; (b) \(\sup_n\lVert x_n\rVert<\infty\) ; (c) \(\lVert x\rVert\le\liminf_n\lVert x_n\rVert\) ; (d) if \(T\in\mathcal B(X,Y)\) , then \(Tx_n\rightharpoonup Tx\) in \(Y\) . Moreover, strong convergence implies weak convergence, and in a finite-dimensional space the two coincide.
Proof. (a) If also \(x_n\rightharpoonup x'\) , then \(f(x)=f(x')\) for all \(f\in X^*\) , so \(x=x'\) by Corollary 3.6. (b) The set \(\{x_n\}\) is bounded under every functional, since convergent scalar sequences are bounded; apply Corollary 4.7. (c) If \(x=0\) there is nothing to prove. Otherwise let \(f\) be a norming functional for \(x\) (Corollary 3.6). Then \(\lVert x\rVert=f(x)=\lim_nf(x_n)\) and \(\lvert f(x_n)\rvert\le\lVert x_n\rVert\) , so \(\lVert x\rVert\le\liminf_n\lVert x_n\rVert\) . (d) For \(g\in Y^*\) the composite \(g\circ T\) belongs to \(X^*\) , so \(g(Tx_n)\to g(Tx)\) . Strong convergence implies weak because \(\lvert f(x_n)-f(x)\rvert\le\lVert f\rVert\lVert x_n-x\rVert\) . In \(\mathbb K^d\) the coordinate functionals are bounded, so weak convergence gives coordinatewise convergence, which is convergence in norm.
Property (b) used the uniform boundedness principle and property (a) the Hahn–Banach theorem; weak convergence is a notion that only works because of the two preceding chapters. Property (c) says the norm is weakly lower semicontinuous: mass can be lost in a weak limit, never gained.
Checking \(f(x_n)\to f(x)\) for every functional would be laborious. With a bound on the norms, a spanning set suffices.
Proposition 6.3 (testing on a spanning set). Let \((x_n)\) be a sequence in a normed space \(X\) and \(x\in X\) . Let \(\Gamma\subseteq X^*\) be a set whose linear span is dense in \(X^*\) . Then \(x_n\rightharpoonup x\) if and only if \(\sup_n\lVert x_n\rVert<\infty\) and \(f(x_n)\to f(x)\) for every \(f\in\Gamma\) .
Proof. Necessity is Proposition 6.2(b). For sufficiency let \(C\) bound \(\lVert x_n\rVert\) and \(\lVert x\rVert\) . Convergence on \(\Gamma\) extends to the span of \(\Gamma\) by linearity. Given \(f\in X^*\) and \(\varepsilon>0\) choose \(g\) in the span with \(\lVert f-g\rVert\le\varepsilon\) . Then
and the last term tends to \(0\) .
For \(1<p<\infty\) the span of the coordinate functionals is \(c_{00}\) , which is dense in \(\ell^q=(\ell^p)^*\) . Hence in \(\ell^p\) with \(1<p<\infty\) , and likewise in \(c_0\) , a sequence converges weakly if and only if it is bounded and converges coordinatewise. For \(p=1\) the dual is \(\ell^\infty\) , in which \(c_{00}\) is not dense, and the criterion fails.
Example 6.1: Unit vectors converge weakly in some spaces and not in others
In \(\ell^p\) with \(1<p<\infty\) and in \(c_0\) , the sequence \((e_n)\) is bounded and each coordinate \((e_n)_k\) is \(0\) once \(n>k\) . By Proposition 6.3, \(e_n\rightharpoonup 0\) . Against \(y=(1/k)\) , for instance, the pairing is \(1/n\) . The norms do not follow: \(\lVert e_n\rVert=1\) for all \(n\) while the weak limit has norm \(0\) , so the inequality of Proposition 6.2(c) reads \(0\le 1\) and is strict.
In \(\ell^1\) the same sequence does not converge weakly. The constant sequence \(y=(1,1,1,\dots)\) belongs to \(\ell^\infty=(\ell^1)^*\) and \(\varphi_y(e_n)=1\) for every \(n\) , whereas a weak limit would have to be \(0\) by the coordinate functionals. In fact \(\ell^1\) has the remarkable property, due to Schur, that every weakly convergent sequence converges in norm; we shall not prove it, but the unit vectors show the mechanism, since \(\ell^\infty\) contains enough functionals to see them all at once.
Boundedness cannot be dropped from Proposition 6.3. In \(\ell^2\) the sequence \(\sqrt n\,e_n\) converges to \(0\) coordinatewise, but it is unbounded, so by Proposition 6.2(b) it does not converge weakly; Example 4.3 exhibited a \(y\in\ell^2\) against which the pairings are unbounded.
Example 6.2: Oscillation
In \(L^2[0,1]\) let \(f_n(x)=\sin(n\pi x)\) . Then \(\lVert f_n\rVert_2^2=\int_0^1\sin^2(n\pi x)\,dx=\tfrac12\) , so \(\lVert f_n\rVert_2=1/\sqrt2\approx 0.7071\) for every \(n\) . For the indicator function of an interval \([a,b]\subseteq[0,1]\) ,
So \(\int f_ng\to 0\) for every step function \(g\) . Step functions are dense in \(L^2[0,1]\) , a standard fact of Measure Theory, and the sequence is bounded; by Proposition 6.3, \(f_n\rightharpoonup 0\) in \(L^2[0,1]\) . Against the smooth function \(g(x)=x\) the pairing can be computed exactly by parts:
which is about \(0.0796\) in absolute value for \(n=4\) and \(0.0099\) for \(n=32\) .
The convergence is not strong: for \(n\ne m\) the product \(\sin(n\pi x)\sin(m\pi x)\) integrates to \(0\) over \([0,1]\) , so \(\lVert f_n-f_m\rVert_2^2=\tfrac12+\tfrac12=1\) . And weak convergence does not survive squaring. Since \(f_n^2=\tfrac12-\tfrac12\cos(2n\pi x)\) and the cosine term tends weakly to \(0\) by the same argument, \(f_n^2\rightharpoonup\tfrac12\) , which is not the square of the weak limit of \(f_n\) . The weak limit records the local average of the functions, and the average of a square is not the square of the average.
6.2 The weak and weak-star topologies
On a dual space there are two natural ways to test convergence, depending on whether one uses all functionals on \(X^*\) or only those that come from \(X\) .
Definition 6.4 (weak-star convergence). A sequence \((f_n)\) in \(X^*\) converges weak-star to \(f\in X^*\) , written \(f_n\overset{*}{\rightharpoonup}f\) , if \(f_n(x)\to f(x)\) for every \(x\in X\) .
Weak-star convergence is pointwise convergence of functionals. Weak convergence in \(X^*\) requires \(\Phi(f_n)\to\Phi(f)\) for all \(\Phi\in X^{\ast\ast}\) , and since \(X^{\ast\ast}\) contains the evaluations \(Jx\) , weak convergence implies weak-star convergence. The two coincide when \(X\) is reflexive. If \(X\) is a Banach space, a weak-star convergent sequence is bounded and its limit satisfies \(\lVert f\rVert\le\liminf_n\lVert f_n\rVert\) , by Corollary 4.5.
Behind the sequential definitions stand two topologies. A basic weak neighbourhood of a point \(x_0\in X\) is a set
with finitely many \(f_i\in X^*\) ; a set is weakly open if it contains a basic weak neighbourhood of each of its points. The weak-star topology on \(X^*\) is defined in the same way with finitely many points \(x_i\in X\) in place of the \(f_i\) . A sequence converges for one of these topologies exactly when it converges weakly, respectively weak-star. Every weakly open set is open for the norm, so the weak topology has fewer open sets, fewer closed sets, and more compact sets.
Proposition 6.5 (weak neighbourhoods are large). In an infinite-dimensional normed space \(X\) , every basic weak neighbourhood of a point contains a line through that point. Consequently no nonempty weakly open set is bounded, and every point of the open unit ball lies in the weak closure of the unit sphere \(\{x:\lVert x\rVert=1\}\) .
Proof. The map \(x\mapsto(f_1(x),\dots,f_m(x))\) from \(X\) to \(\mathbb K^m\) is linear, and since \(X\) is infinite-dimensional it has a nonzero vector \(z\) in its kernel. Then \(x_0+tz\in U(x_0;f_1,\dots,f_m;\varepsilon)\) for every scalar \(t\) . For the last statement let \(\lVert x_0\rVert<1\) and let \(U\) be a basic weak neighbourhood of \(x_0\) with \(z\) as above. The function \(t\mapsto\lVert x_0+tz\rVert\) is continuous on \([0,\infty)\) , is less than \(1\) at \(t=0\) , and tends to infinity, so it equals \(1\) for some \(t\) ; that point lies in \(U\) and on the sphere.
So the weak topology is strictly coarser than the norm topology whenever the dimension is infinite, and the norm is not weakly continuous: the sphere, where the norm equals \(1\) , is weakly dense in the ball. Chapters 7 to 10 use weak convergence only through sequences, and the reader may think sequentially throughout, with one caution: in general the weak closure of a set is not obtained by taking limits of sequences (Exercise 16).
Example 6.3: Weak-star convergence that is not weak
Regard \(\ell^1\) as the dual of \(c_0\) (Theorem 3.1). For \(x\in c_0\) the pairing with \(e_n\in\ell^1\) is \(x_n\) , which tends to \(0\) . So \(e_n\overset{*}{\rightharpoonup}0\) in \(\ell^1\) . But Example 6.1 showed that \((e_n)\) does not converge weakly in \(\ell^1\) . The functional \((1,1,1,\dots)\in\ell^\infty\) that detects the difference belongs to the second dual of \(c_0\) and not to \(c_0\) .
A second instance lives in \(\ell^\infty=(\ell^1)^*\) . Let \(t_n=(0,\dots,0,1,1,1,\dots)\) with \(n\) zeros. For \(x\in\ell^1\) the pairing is the tail \(\sum_{k>n}x_k\) , which tends to \(0\) ; for \(x=(2^{-k})\) it equals \(2^{-n}\) . So \(t_n\overset{*}{\rightharpoonup}0\) while \(\lVert t_n\rVert_\infty=1\) . The convergence is not weak: a Banach limit \(\Lambda\) (Exercise 6 of Chapter 3Chapter 3) is a bounded functional on \(\ell^\infty\) with \(\Lambda(t_n)=1\) for all \(n\) , because each \(t_n\) converges to \(1\) as a sequence.
6.3 Convex sets do not see the difference
A norm-closed set need not be weakly closed; the unit sphere is an example. For convex sets the two notions agree, and this single fact is what makes weak convergence usable in the calculus of variations.
Theorem 6.6 (Mazur). Let \(K\) be a convex subset of a normed space \(X\) that is closed for the norm. If \(x_n\in K\) and \(x_n\rightharpoonup x\) , then \(x\in K\) . Consequently, whenever \(x_n\rightharpoonup x\) there are finite convex combinations of the \(x_n\) that converge to \(x\) in norm.
Proof. Suppose \(x\notin K\) . By Corollary 3.11, applied to the real part of a functional in the complex case, there are \(F\in X^*\) and a real \(\gamma\) with \(\operatorname{Re}F(k)\le\gamma<\operatorname{Re}F(x)\) for all \(k\in K\) . But \(F(x_n)\to F(x)\) and \(\operatorname{Re}F(x_n)\le\gamma\) , a contradiction. For the second statement let \(K\) be the norm closure of the set of finite convex combinations of the \(x_n\) . It is convex and closed and contains every \(x_n\) , so it contains \(x\) .
Corollary 6.7 (weak lower semicontinuity). Let \(\psi:X\to\mathbb R\) be convex and continuous for the norm. If \(x_n\rightharpoonup x\) , then \(\psi(x)\le\liminf_n\psi(x_n)\) .
Proof. Let \(c>\liminf_n\psi(x_n)\) . Infinitely many \(x_n\) lie in the set \(\{\psi\le c\}\) , which is convex because \(\psi\) is convex and closed because \(\psi\) is continuous. That subsequence converges weakly to \(x\) , so by Theorem 6.6, \(\psi(x)\le c\) . Let \(c\) decrease to the limit inferior.
Proposition 6.2(c) is the special case \(\psi(x)=\lVert x\rVert\) .
Example 6.4: Averages of a weakly convergent sequence converge in norm
For \(e_n\rightharpoonup 0\) in \(\ell^2\) , Mazur’s theorem promises convex combinations tending to \(0\) in norm, and the plain averages already work:
For the oscillating functions of Example 6.2, orthogonality gives the same law: \(\lVert\tfrac1N(f_1+\dots+f_N)\rVert_2=\tfrac1{\sqrt2}\cdot\tfrac1{\sqrt N}\) . With \(N=100\) the two norms are \(0.1\) and about \(0.0707\) . Averaging spreads the mass over many orthogonal directions and the norm, which adds in squares, decays.
In \(\ell^1\) the same averages have \(\lVert a_N\rVert_1=N\cdot\tfrac1N=1\) for every \(N\) . They do not tend to \(0\) , and neither does any other convex combination of unit vectors, since the \(\ell^1\) norm of a convex combination of distinct \(e_n\) is the sum of the coefficients. This is consistent with Example 6.1: if \((e_n)\) converged weakly to \(0\) in \(\ell^1\) , Mazur’s theorem would be contradicted.
6.4 Compactness regained
Theorem 6.8 (sequential Banach–Alaoglu). Let \(X\) be a separable normed space. Then every bounded sequence in \(X^*\) has a weak-star convergent subsequence.
Proof. Let \(\lVert f_n\rVert\le C\) and let \(\{z_1,z_2,\dots\}\) be dense in \(X\) . The scalars \(f_n(z_1)\) are bounded by \(C\lVert z_1\rVert\) , so some subsequence of \((f_n)\) converges at \(z_1\) . A further subsequence converges at \(z_2\) , and so on. Let \((g_j)\) be the diagonal sequence, whose \(j\) -th term is the \(j\) -th term of the \(j\) -th subsequence. From the \(i\) -th term on it is a subsequence of the \(i\) -th subsequence, so \(g_j(z_i)\) converges for every \(i\) .
Now let \(x\in X\) and \(\varepsilon>0\) , and choose \(z_i\) with \(\lVert x-z_i\rVert\le\varepsilon\) . Then
so \((g_j(x))\) is a Cauchy sequence of scalars and converges. Define \(f(x)=\lim_jg_j(x)\) . Then \(f\) is linear and \(\lvert f(x)\rvert\le C\lVert x\rVert\) , so \(f\in X^*\) , and \(g_j\overset{*}{\rightharpoonup}f\) .
The proof is the diagonal argument that also proves the Arzelà–Ascoli theorem. Separability supplied a countable set on which to diagonalise, and the uniform bound carried convergence from that set to the whole space. Without separability sequences are not enough, but the topological statement survives in full generality.
Theorem 6.9 (Banach–Alaoglu). For every normed space \(X\) , the closed unit ball \(B_{X^*}\) of the dual is compact in the weak-star topology.
Proof. For each \(x\in X\) let \(D_x=\{\lambda\in\mathbb K:\lvert\lambda\rvert\le\lVert x\rVert\}\) , a compact set, and let \(P=\prod_{x\in X}D_x\) with the product topology. By Tychonoff’s theorem from general topology, a product of compact spaces is compact, so \(P\) is compact. An element of \(P\) is a function \(p\) on \(X\) with \(\lvert p(x)\rvert\le\lVert x\rVert\) , and a functional \(f\in B_{X^*}\) is exactly such a function that is also linear. Under this identification the weak-star topology on \(B_{X^*}\) is the topology inherited from \(P\) , since both are defined by finitely many evaluations \(p\mapsto p(x_i)\) . It remains to see that the linear elements form a closed subset of \(P\) . For fixed \(x,y\in X\) and scalars \(\alpha,\beta\) , the map \(p\mapsto p(\alpha x+\beta y)-\alpha p(x)-\beta p(y)\) is continuous on \(P\) , being built from three coordinate evaluations, so its zero set is closed; the linear elements are the intersection of these closed sets over all \(x,y,\alpha,\beta\) . A closed subset of a compact space is compact.
Compactness in Theorem 6.9 does not by itself give convergent subsequences. In \((\ell^\infty)^*\) the coordinate functionals \(\delta_n(x)=x_n\) lie in the unit ball, and no subsequence \((\delta_{n_j})\) converges weak-star: the vector \(x\in\ell^\infty\) with \(x_{n_j}=(-1)^j\) and all other coordinates \(0\) gives \(\delta_{n_j}(x)=(-1)^j\) . Here \(X=\ell^\infty\) is not separable, so Theorem 6.8 does not apply.
For weak convergence in the space itself, as opposed to weak-star convergence in a dual, reflexivity is the hypothesis that transfers the result.
Theorem 6.10 (weak sequential compactness in reflexive spaces). Let \(X\) be a reflexive Banach space. Then every bounded sequence in \(X\) has a weakly convergent subsequence.
Proof. Suppose first that \(X^*\) is separable. A bounded sequence \((x_n)\) in \(X\) gives a bounded sequence \((Jx_n)\) in \(X^{\ast\ast}=(X^*)^*\) , and Theorem 6.8 applied to the separable space \(X^*\) yields a subsequence with \(Jx_{n_j}\overset{*}{\rightharpoonup}\Phi\) for some \(\Phi\in X^{\ast\ast}\) . By reflexivity \(\Phi=Jx\) for some \(x\in X\) . Unwinding the definitions, \(f(x_{n_j})=(Jx_{n_j})(f)\to(Jx)(f)=f(x)\) for all \(f\in X^*\) , that is, \(x_{n_j}\rightharpoonup x\) .
In general, let \(Y\) be the closed linear span of the \(x_n\) . It is separable, and it is reflexive because a closed subspace of a reflexive space is reflexive (Exercise 12). Then \(Y^{\ast\ast}\) , being isometric to \(Y\) , is separable, and therefore \(Y^*\) is separable (Exercise 8 of Chapter 3Chapter 3, applied to the space \(Y^*\) ). By the first part a subsequence converges weakly in \(Y\) to some \(x\in Y\) ; since the restriction to \(Y\) of any \(f\in X^*\) belongs to \(Y^*\) , the subsequence converges weakly to \(x\) in \(X\) as well.
The theorem applies to \(\ell^p\) and \(L^p\) for \(1<p<\infty\) , and to every Hilbert space once Chapter 8 has shown that Hilbert spaces are reflexive. It fails in \(c_0\) , \(\ell^1\) , \(L^1\) , and \(C[0,1]\) , none of which is reflexive.
Example 6.5: Bounded sequences with no weakly convergent subsequence
In \(c_0\) let \(s_n=e_1+\dots+e_n=(1,\dots,1,0,0,\dots)\) , so \(\lVert s_n\rVert_\infty=1\) . If a subsequence converged weakly to some \(s\in c_0\) , then applying the coordinate functionals would give \(s_k=\lim_j(s_{n_j})_k=1\) for every \(k\) . But \((1,1,1,\dots)\notin c_0\) . So \((s_n)\) has no weakly convergent subsequence, and by Theorem 6.10 this proves again that \(c_0\) is not reflexive. Regarded in \(\ell^\infty=(\ell^1)^*\) , the same sequence does converge weak-star, to \((1,1,1,\dots)\) : the limit exists in the larger dual space and has simply left \(c_0\) .
In \(L^1[0,1]\) let \(g_n=n\,\mathbf 1_{[0,1/n]}\) , a spike of height \(n\) and width \(1/n\) , so \(\lVert g_n\rVert_1=1\) . For a continuous function \(\phi\) , the integral \(\int_0^1g_n\phi=n\int_0^{1/n}\phi\) is the average of \(\phi\) over \([0,1/n]\) and tends to \(\phi(0)\) ; for \(\phi=\cos\) it equals \(n\sin(1/n)\) , which is about \(0.99833\) at \(n=10\) . Suppose a subsequence converged weakly in \(L^1\) to some \(g\) . Testing against continuous \(\phi\) vanishing on \([0,\delta]\) shows \(\int g\phi=0\) for all such \(\phi\) , whence \(g=0\) almost everywhere on \([\delta,1]\) for every \(\delta>0\) , so \(g=0\) ; but testing against \(\phi=1\) gives \(\int g=1\) . Hence no subsequence converges weakly. The mass of \(g_n\) concentrates at a point, and the natural limit is the evaluation functional \(\phi\mapsto\phi(0)\) , which is not represented by any integrable function.
6.5 Existence of minimizers
The standard use of weak compactness is to prove that a minimum is attained. In \(\mathbb R^n\) one argues with a minimizing sequence, compactness, and continuity. In a reflexive space the same three steps go through with weak convergence, provided that the set and the function are convex.
Theorem 6.11 (direct method). Let \(X\) be a reflexive Banach space, \(K\subseteq X\) a nonempty closed convex set, and \(\psi:X\to\mathbb R\) a convex function that is continuous for the norm. Assume either that \(K\) is bounded, or that \(\psi(x)\to\infty\) as \(\lVert x\rVert\to\infty\) with \(x\in K\) . Then \(\psi\) attains its infimum over \(K\) .
Proof. Let \((x_n)\) be a sequence in \(K\) with \(\psi(x_n)\to\inf_K\psi\) . Under either assumption \((x_n)\) is bounded: in the second case, an unbounded minimizing sequence would have a subsequence along which \(\psi\to\infty\) . By Theorem 6.10 a subsequence converges weakly to some \(x\) ; by Theorem 6.6, \(x\in K\) ; by Corollary 6.7, \(\psi(x)\le\liminf_j\psi(x_{n_j})=\inf_K\psi\) . So the infimum is finite and is attained at \(x\) .
Taking \(\psi(x)=\lVert x-x_0\rVert\) shows that in a reflexive space every nonempty closed convex set contains a point nearest to any given \(x_0\) . The nearest point need not be unique in general; Chapter 7 shows that in a Hilbert space it is, and gives a proof there that avoids weak convergence altogether.
Example 6.6: A closed convex set with no element of smallest norm
In the real space \(C[0,1]\) let \(\varphi(f)=\int_0^{1/2}f-\int_{1/2}^1f\) and \(K=\{f:\varphi(f)=1\}\) , a closed convex set. By Exercise 4 of Chapter 3Chapter 3, \(\lVert\varphi\rVert=1\) and the norm is not attained. For \(f\in K\) we have \(1=\varphi(f)\le\lVert f\rVert_\infty\) , and equality would mean that \(f\) attains the norm of \(\varphi\) ; so every element of \(K\) has norm strictly greater than \(1\) . On the other hand the functions \(f_n\) of that exercise have \(\varphi(f_n)=1-\tfrac1n\) , so \(f_n/\varphi(f_n)\in K\) has norm \(n/(n-1)\) , which is \(1.25\) for \(n=5\) and tends to \(1\) . The infimum of the norm over \(K\) is \(1\) and it is not attained. The hypothesis of Theorem 6.11 that fails is reflexivity.
Compare the same problem in the reflexive space \(\ell^2\) : minimize \(\lVert x\rVert_2\) over \(K=\{x:\sum_kx_k/2^k=1\}\) . With \(y=(2^{-k})\) and \(\lVert y\rVert_2^2=\tfrac13\) , every \(x\in K\) satisfies \(1=\sum_kx_ky_k\le\lVert x\rVert_2\lVert y\rVert_2\) , so \(\lVert x\rVert_2\ge\sqrt3\) , and \(x=3y\) belongs to \(K\) with \(\lVert 3y\rVert_2=\sqrt3\approx 1.732\) . The minimum is attained.
Common mistake. Passing to the limit in a nonlinear expression along a weakly convergent sequence. If \(x_n\rightharpoonup x\) one may conclude \(f(x_n)\to f(x)\) for a fixed bounded functional and \(Tx_n\rightharpoonup Tx\) for a fixed bounded operator, and nothing else without further argument. The norm may drop in the limit; squares and products do not converge to the square or product of the limits (Example 6.2); and a pairing \(f_n(x_n)\) in which both factors converge only weakly may fail to converge to \(f(x)\) , as \(\sum_k(e_n)_k(e_n)_k=1\) shows. One strongly convergent factor is enough to repair the last statement (Exercise 5).
Weak convergence trades the continuity of the norm for compactness. In a Hilbert space the trade is especially favourable, because the norm comes from a pairing and weak limits can be computed by taking inner products. The next two chapters develop that special geometry.
Exercises
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Using Theorem 3.1 directly, show that \(e_n\rightharpoonup 0\) in \(c_0\) , and that no subsequence of \((e_n)\) converges weakly in \(\ell^1\) .
Hint
A functional on \(c_0\) is \(\varphi_y\) with \(y\in\ell^1\) , and \(\varphi_y(e_n)=y_n\to 0\) because the terms of a convergent series tend to \(0\) . In \(\ell^1\) , a weak limit \(x\) of a subsequence would satisfy \(x_k=\lim_j(e_{n_j})_k=0\) for each \(k\) , so \(x=0\) , while the functional \(y\in\ell^\infty\) with \(y_{n_j}=(-1)^j\) and \(y_k=0\) elsewhere gives \(\varphi_y(e_{n_j})=(-1)^j\) , which does not converge.
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In \(\ell^2\) let \(x^{(n)}=e_1+e_n\) for \(n\ge 2\) . Find the weak limit \(x\) , compute \(\lVert x^{(n)}\rVert_2\) and \(\lVert x\rVert_2\) , and say what Proposition 6.2(c) asserts in this case.
Answer
\(x=e_1\) : the sequence is bounded and converges coordinatewise to \(e_1\) . \(\lVert x^{(n)}\rVert_2=\sqrt2\) for all \(n\) and \(\lVert x\rVert_2=1\) . Proposition 6.2(c) asserts \(1\le\sqrt2\) ; the inequality is strict because the part \(e_n\) of the vector escapes weakly to \(0\) .
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Compute \(\int_0^1x^2\sin(n\pi x)\,dx\) in closed form and verify that it tends to \(0\) at the rate \(1/n\) .
Solution
Write \(a=n\pi\) . Integrating by parts twice, an antiderivative of \(x^2\sin(ax)\) is
\[ -\frac{x^2\cos(ax)}{a}+\frac{2x\sin(ax)}{a^2}+\frac{2\cos(ax)}{a^3} . \]At \(x=1\) , \(\sin(a)=0\) and \(\cos(a)=(-1)^n\) , giving \(-(-1)^n/a+2(-1)^n/a^3\) ; at \(x=0\) the value is \(2/a^3\) . Hence
\[ \int_0^1x^2\sin(n\pi x)\,dx=\frac{(-1)^{n+1}}{n\pi}+\frac{2\bigl((-1)^n-1\bigr)}{n^3\pi^3} . \]The second term is \(0\) for even \(n\) and \(-4/(n^3\pi^3)\) for odd \(n\) ; in either case it is of order \(n^{-3}\) . So the integral equals \((-1)^{n+1}/(n\pi)\) up to an error of order \(n^{-3}\) , and tends to \(0\) at the rate \(1/n\) . For \(n=1\) the value is \(\tfrac1\pi-\tfrac4{\pi^3}\approx 0.1893\) . The leading term comes entirely from the boundary point \(x=1\) , where \(x^2\) does not vanish; a test function vanishing at both endpoints would give faster decay. This is the general pattern for oscillatory integrals: smoothness of the test function is converted into decay, one power of \(n\) for each integration by parts.
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Show that \(\cos^2(n\pi x)\rightharpoonup\tfrac12\) in \(L^2[0,1]\) , using the identity \(\cos^2\theta=\tfrac12+\tfrac12\cos 2\theta\) and the method of Example 6.2; conclude that \(u_n\rightharpoonup u\) does not imply \(u_n^2\rightharpoonup u^2\) .
Solution
By the identity, \(\cos^2(n\pi x)=\tfrac12+\tfrac12\cos(2n\pi x)\) , so it suffices to prove \(c_n\rightharpoonup 0\) in \(L^2[0,1]\) for \(c_n(x)=\cos(2n\pi x)\) . The sequence is bounded: \(\lVert c_n\rVert_2^2=\int_0^1\cos^2(2n\pi x)\,dx=\tfrac12\) . For the indicator of an interval \([a,b]\subseteq[0,1]\) ,
\[ \int_a^b\cos(2n\pi x)\,dx=\frac{\sin(2n\pi b)-\sin(2n\pi a)}{2n\pi},\qquad\text{of absolute value at most }\frac1{n\pi}. \]So the pairing of \(c_n\) with every step function tends to \(0\) . Step functions are dense in \(L^2[0,1]=(L^2[0,1])^*\) , and Proposition 6.3 gives \(c_n\rightharpoonup 0\) . Therefore \(\cos^2(n\pi x)\rightharpoonup\tfrac12\) .
Now let \(u_n(x)=\cos(n\pi x)\) . The same computation, with \(n\pi\) in place of \(2n\pi\) , shows \(u_n\rightharpoonup 0\) . So \(u=0\) and \(u^2=0\) , whereas \(u_n^2\rightharpoonup\tfrac12\ne 0\) . Squaring is not weakly continuous. The number \(\tfrac12\) is the mean value of \(\cos^2\) over a period: the weak limit of a rapidly oscillating function is its local average, and the average of the square exceeds the square of the average unless the function is constant.
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Let \(x_n\rightharpoonup x\) in a normed space \(X\) and \(f_n\to f\) in the norm of \(X^*\) . Prove that \(f_n(x_n)\to f(x)\) , and give an example in \(\ell^2\) in which \(f_n\overset{*}{\rightharpoonup}f\) and \(x_n\rightharpoonup x\) but \(f_n(x_n)\) does not converge to \(f(x)\) .
Solution
By Proposition 6.2(b) there is \(C\) with \(\lVert x_n\rVert\le C\) for all \(n\) . Then
\[ \lvert f_n(x_n)-f(x)\rvert\le\lvert(f_n-f)(x_n)\rvert+\lvert f(x_n)-f(x)\rvert\le C\lVert f_n-f\rVert+\lvert f(x_n)-f(x)\rvert . \]The first term tends to \(0\) because \(f_n\to f\) in norm, and the second because \(x_n\rightharpoonup x\) and \(f\) is a fixed functional. So \(f_n(x_n)\to f(x)\) .
For the counterexample take \(X=\ell^2\) , \(x_n=e_n\) , and \(f_n=\varphi_{e_n}\) , the \(n\) -th coordinate functional. Then \(x_n\rightharpoonup 0\) by Example 6.1, and \(f_n(x)=x_n\to 0\) for every \(x\in\ell^2\) , so \(f_n\overset{*}{\rightharpoonup}0\) (indeed weakly, since \(\ell^2\) is reflexive). But \(f_n(x_n)=1\) for every \(n\) , which does not converge to \(f(x)=0\) . In the estimate above, the term \((f_n-f)(x_n)\) is the one that cannot be controlled: each of its two factors tends to zero only when tested against a fixed partner, and here the partner moves with \(n\) .
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Let \(X\) and \(Y\) be Banach spaces and \(T:X\to Y\) a linear map such that \(x_n\rightharpoonup x\) implies \(Tx_n\rightharpoonup Tx\) . Prove that \(T\) is bounded, using Exercise 15 of Chapter 4Chapter 4.
Hint
For \(g\in Y^*\) , the linear functional \(g\circ T\) maps weakly convergent sequences to convergent scalar sequences; in particular it maps norm-null sequences to null sequences, so it is continuous at \(0\) and hence bounded. Now Exercise 15 of Chapter 4Chapter 4 applies.
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Let \(x^{(n)}\rightharpoonup x\) in \(\ell^2\) and suppose also that \(\lVert x^{(n)}\rVert_2\to\lVert x\rVert_2\) . Prove that \(x^{(n)}\to x\) in norm.
Solution
Write \(\langle u,v\rangle=\sum_ku_k\overline{v_k}\) , so that \(\lVert u\rVert_2^2=\langle u,u\rangle\) . Expanding,
\[ \lVert x^{(n)}-x\rVert_2^2=\lVert x^{(n)}\rVert_2^2-2\operatorname{Re}\langle x^{(n)},x\rangle+\lVert x\rVert_2^2 . \]The map \(u\mapsto\langle u,x\rangle\) is the bounded functional \(\varphi_{\bar x}\) on \(\ell^2\) , so weak convergence gives \(\langle x^{(n)},x\rangle\to\langle x,x\rangle=\lVert x\rVert_2^2\) . By hypothesis \(\lVert x^{(n)}\rVert_2^2\to\lVert x\rVert_2^2\) . Therefore
\[ \lVert x^{(n)}-x\rVert_2^2\to\lVert x\rVert_2^2-2\lVert x\rVert_2^2+\lVert x\rVert_2^2=0 . \]So weak convergence together with convergence of the norms is strong convergence. The unit vectors show that neither hypothesis can be dropped: \(e_n\rightharpoonup 0\) while \(\lVert e_n\rVert_2=1\) does not tend to \(\lVert 0\rVert_2\) . The proof used only that the square of the norm is given by the pairing, so it holds in every Hilbert space; in \(\ell^1\) , by contrast, the conclusion holds for a different and deeper reason, Schur’s theorem.
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Find the error: “The sequence \((e_n)\) is bounded in \(\ell^1=c_0^*\) , so by Theorem 6.8 a subsequence converges weak-star, necessarily to \(0\) . Weak-star convergence is weak convergence, so \(e_{n_j}\rightharpoonup 0\) in \(\ell^1\) , contradicting Example 6.1.”
Answer
The first sentence is correct: \(e_n\overset{*}{\rightharpoonup}0\) in \(\ell^1=c_0^*\) . The error is the claim that weak-star convergence is weak convergence. That holds when the predual is reflexive, and \(c_0\) is not. Weak convergence in \(\ell^1\) tests against all of \(\ell^\infty\) , weak-star convergence only against \(c_0\) .
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Show that the coordinate functionals \(\delta_n(x)=x_n\) form a sequence in the unit ball of \((\ell^\infty)^*\) with no weak-star convergent subsequence, and explain why this contradicts neither Theorem 6.8 nor Theorem 6.9.
Solution
Each \(\delta_n\) is linear with \(\lvert\delta_n(x)\rvert\le\lVert x\rVert_\infty\) , so \(\lVert\delta_n\rVert\le 1\) . Let \((\delta_{n_j})\) be any subsequence, with \(n_1<n_2<\cdots\) . Define \(x\in\ell^\infty\) by \(x_{n_j}=(-1)^j\) and \(x_k=0\) for all other \(k\) . Then \(\delta_{n_j}(x)=(-1)^j\) , which has no limit as \(j\to\infty\) . Since weak-star convergence would require \(\delta_{n_j}(x)\) to converge for every \(x\in\ell^\infty\) , the subsequence does not converge weak-star. The subsequence was arbitrary.
Theorem 6.8 assumes that the space \(X\) whose dual is considered is separable. Here \(X=\ell^\infty\) , which is not separable (Section 1.6), so the theorem is silent. Theorem 6.9 asserts that the unit ball of \((\ell^\infty)^*\) is compact for the weak-star topology, and that is true. Compactness guarantees that the sequence \((\delta_n)\) has a cluster point, a functional every weak-star neighbourhood of which contains \(\delta_n\) for infinitely many \(n\) . It does not guarantee a convergent subsequence, because the weak-star topology on this ball is not given by a metric. The cluster points are functionals that assign to each bounded sequence a generalised limit along some part of the index set, close relatives of the Banach limits of Chapter 3.
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Compute \(\lVert\tfrac1N(e_1+\dots+e_N)\rVert_p\) in \(\ell^p\) for \(1\le p\le\infty\) , and relate the cases \(p=1\) and \(p>1\) to Theorem 6.6.
Answer
For \(1\le p<\infty\) , \(\lVert a_N\rVert_p=(N\cdot N^{-p})^{1/p}=N^{1/p-1}\) , and for \(p=\infty\) it is \(1/N\) . For \(p>1\) this tends to \(0\) , in agreement with \(e_n\rightharpoonup 0\) and Theorem 6.6. For \(p=1\) it equals \(1\) for all \(N\) , consistent with the fact that \((e_n)\) does not converge weakly in \(\ell^1\) .
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Let \(M\) be a closed subspace of a normed space and \(x_n\in M\) with \(x_n\rightharpoonup x\) . Prove that \(x\in M\) , and show that the unit sphere of \(\ell^2\) does not have the corresponding property.
Answer
A closed subspace is a norm-closed convex set, so Theorem 6.6 applies. For the sphere: \(e_n\rightharpoonup 0\) with \(\lVert e_n\rVert_2=1\) and \(\lVert 0\rVert_2=0\) ; the sphere is closed and not convex.
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Prove that a closed subspace \(Y\) of a reflexive Banach space \(X\) is reflexive.
Solution
Let \(J_X\) and \(J_Y\) be the canonical embeddings. Given \(\Psi\in Y^{\ast\ast}\) we must find \(y\in Y\) with \(J_Yy=\Psi\) .
The restriction map \(R:X^*\to Y^*\) , \(Rf=f|_Y\) , is linear with \(\lVert Rf\rVert\le\lVert f\rVert\) . Define \(\Phi=\Psi\circ R\) , so that \(\Phi(f)=\Psi(f|_Y)\) for \(f\in X^*\) . Then \(\Phi\in X^{\ast\ast}\) . Since \(X\) is reflexive there is \(x\in X\) with \(J_Xx=\Phi\) , that is,
\[ f(x)=\Psi(f|_Y)\qquad\text{for all }f\in X^* . \]The point \(x\) lies in \(Y\) . If not, then since \(Y\) is closed, \(\operatorname{dist}(x,Y)>0\) , and Corollary 3.7 gives \(f\in X^*\) with \(f|_Y=0\) and \(f(x)\ne 0\) . But then \(f(x)=\Psi(0)=0\) , a contradiction.
\(J_Yx=\Psi\) . Let \(g\in Y^*\) . By Theorem 3.5, \(g\) extends to some \(f\in X^*\) , so \(g=f|_Y\) . Then \(\Psi(g)=\Psi(f|_Y)=f(x)=g(x)=(J_Yx)(g)\) .
Hence \(J_Y\) is onto and \(Y\) is reflexive. The Hahn–Banach theorem was used twice, once to keep \(x\) inside \(Y\) and once to lift functionals from \(Y\) to \(X\) ; closedness of \(Y\) entered through the positive distance.
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Let \(X\) be a reflexive Banach space and \(f\in X^*\) . Prove that there is \(x\in X\) with \(\lVert x\rVert\le 1\) and \(f(x)=\lVert f\rVert\) , and deduce from Example 3.1 and Exercise 4 of Chapter 3Chapter 3 that \(c_0\) and \(C[0,1]\) are not reflexive.
Solution
If \(f=0\) any \(x\) works, so let \(f\ne 0\) . Apply Corollary 3.6 to the normed space \(X^*\) and its nonzero element \(f\) : there is \(\Phi\in X^{\ast\ast}\) with \(\lVert\Phi\rVert=1\) and \(\Phi(f)=\lVert f\rVert\) . Because \(X\) is reflexive, \(\Phi=Jx\) for some \(x\in X\) , and since \(J\) is an isometry, \(\lVert x\rVert=\lVert\Phi\rVert=1\) . Then
\[ f(x)=(Jx)(f)=\Phi(f)=\lVert f\rVert . \]So every bounded functional on a reflexive space attains its norm on the closed unit ball.
Contrapositively, a space carrying a bounded functional that does not attain its norm is not reflexive. On \(c_0\) , the functional \(\varphi(x)=\sum_kx_k/2^k\) has norm \(1\) and \(\lvert\varphi(x)\rvert<1\) whenever \(\lVert x\rVert_\infty\le 1\) (Example 3.1), so \(c_0\) is not reflexive. On \(C[0,1]\) , the functional \(\varphi(f)=\int_0^{1/2}f-\int_{1/2}^1f\) has norm \(1\) , not attained (Exercise 4 of Chapter 3Chapter 3), so \(C[0,1]\) is not reflexive. A theorem of James, which is much harder, states the converse: a Banach space on which every bounded functional attains its norm is reflexive.
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Find the element of smallest norm in \(K=\{x\in\ell^2:\sum_kx_k/3^k=1\}\) and its norm, and prove that it is the unique minimizer.
Solution
Let \(y=(3^{-k})_{k\ge 1}\) , so that \(K=\{x:\langle x,y\rangle=1\}\) with \(\langle x,y\rangle=\sum_kx_ky_k\) , and
\[ \lVert y\rVert_2^2=\sum_{k\ge 1}9^{-k}=\frac{1/9}{1-1/9}=\frac18 . \]For \(x\in K\) the Cauchy–Schwarz inequality gives \(1=\lvert\langle x,y\rangle\rvert\le\lVert x\rVert_2\lVert y\rVert_2\) , hence
\[ \lVert x\rVert_2\ge\frac1{\lVert y\rVert_2}=\sqrt8=2\sqrt2\approx 2.828 . \]The vector \(x^\star=y/\lVert y\rVert_2^2=8y\) lies in \(K\) , since \(\langle 8y,y\rangle=8\cdot\tfrac18=1\) , and \(\lVert 8y\rVert_2=8/\sqrt8=2\sqrt2\) . So the minimum is \(2\sqrt2\) , attained at \(x^\star=8y=(\tfrac83,\tfrac89,\tfrac8{27},\dots)\) .
Uniqueness: if \(x\in K\) has \(\lVert x\rVert_2=2\sqrt2\) , then equality holds in the Cauchy–Schwarz inequality, which forces \(x\) to be a scalar multiple of \(y\) ; the constraint \(\langle x,y\rangle=1\) then fixes the multiple as \(8\) . Existence could also have been obtained from Theorem 6.11, with \(\psi\) the norm and \(K\) closed and convex in the reflexive space \(\ell^2\) ; the direct computation gives the minimizer explicitly, which the theorem does not.
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In \(\ell^2\) describe the basic weak neighbourhood \(U(0;\varphi_{e_1},\varphi_{e_2};1)\) in coordinates, show that it contains the line \(\{te_3:t\in\mathbb R\}\) , and deduce that the open unit ball of \(\ell^2\) is not weakly open.
Answer
\(U=\{x\in\ell^2:\lvert x_1\rvert<1,\ \lvert x_2\rvert<1\}\) , a cylinder that constrains two coordinates and leaves the rest free. It contains \(te_3\) for every \(t\) , so it is unbounded. Every weak neighbourhood of \(0\) contains such a set after shrinking, hence is unbounded and cannot be contained in the unit ball; so the open unit ball contains no weak neighbourhood of \(0\) .
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Show that \(0\) belongs to the weak closure of the set \(A=\{\sqrt n\,e_n:n\ge 1\}\) in \(\ell^2\) , by proving that every basic weak neighbourhood of \(0\) meets \(A\) , and that nevertheless no sequence of elements of \(A\) converges weakly to \(0\) .
Hint
A basic weak neighbourhood of \(0\) is given by \(y^{(1)},\dots,y^{(m)}\in\ell^2\) and \(\varepsilon>0\) ; it contains \(\sqrt n\,e_n\) when \(n\sum_i\lvert y^{(i)}_n\rvert^2<\varepsilon^2\) . If that failed for all \(n\) , then \(\sum_n\sum_i\lvert y^{(i)}_n\rvert^2\ge\varepsilon^2\sum_n1/n=\infty\) , contradicting square-summability. A sequence in \(A\) converging weakly would be bounded (Proposition 6.2(b)), so it would take only finitely many values, and no single \(\sqrt n\,e_n\) equals \(0\) .
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For \(g_n=n\,\mathbf 1_{[0,1/n]}\) in \(L^1[0,1]\) compute \(\int_0^1g_n\phi\) for \(\phi(x)=x^2\) , for \(\phi(x)=e^x\) , and for \(\phi=\mathbf 1_{[1/2,1]}\) , and identify the limit in each case.
Answer
\(\int g_nx^2=n\cdot\frac1{3n^3}=\frac1{3n^2}\to 0\) . \(\int g_ne^x=n(e^{1/n}-1)\to 1\) . \(\int g_n\mathbf 1_{[1/2,1]}=0\) for \(n\ge 2\) . Each limit is the value of \(\phi\) at \(0\) .
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Minimize \(\psi(x)=\tfrac12\lVert x\rVert_2^2-\sum_kx_k/2^k\) over \(\ell^2\) : show that the hypotheses of Theorem 6.11 hold, find the minimizer by completing the square, and give the minimum value.
Answer
With \(y=(2^{-k})\) , \(\psi(x)=\tfrac12\lVert x-y\rVert_2^2-\tfrac12\lVert y\rVert_2^2\) . It is convex, continuous, and tends to infinity with \(\lVert x\rVert_2\) , and \(\ell^2\) is reflexive. The minimizer is \(x=y\) and the minimum is \(-\tfrac12\lVert y\rVert_2^2=-\tfrac16\) .