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Let $\ell^1$ be the space of absolutely summable real or complex sequences. Let us say that a sequence $(x_1, x_2, \ldots)$ of vectors in $\ell^1$ converges weakly to $x \in \ell^1$ if for every bounded linear functional $\varphi \in (\ell^1)^*$, $\varphi(x_n) \rightarrow \varphi(x)$ as $n \to \infty$. How may I show that weak convergence, in this sense, is the same as the usual convergence-in-norm? It's clear the weak convergence implies pointwise convergence, but that's not good enough to conclude strong convergence...

By linearity, it suffices to prove that if $\varphi(x_n) \longrightarrow 0$ for every $\varphi \in (\ell^1)^*$, then $\| x_n \| \longrightarrow 0$. Let $x_n(k)$ be the $k$-th component of the vector $x_n$. Then, $x_n(k) \longrightarrow 0$ for every $k$, so $\sup_n |x_n(k)| < \infty$ for each $k$, and this implies $$\lim_{N \to \infty} \lim_{n \to \infty} \sum_{k=1}^{N} |x_n(k)| = 0$$ This is almost what I want, but the limits are the wrong way around. The obvious next thing to try is to construct some clever functional, or even a family of clever functionals, but I can't think of anything useful here. I can see that pointwise convergence alone is not good enough — if $x_n$ is the standard basis vector, then $x_n \longrightarrow 0$ pointwise, but $\| x_n \| = 1$ for all $n$. The fact that it doesn't converge strongly can be detected by the linear functional $\varphi(x_n) = \sum_k x_n(k)$, but I'm at a loss as to how to generalise this.

Davide Giraudo
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Zhen Lin
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    This is a nontrivial theorem. One place to read a proof is J. B. Conway's *A Course in Functional Analysis*, where it appears as Theorem V.5.2. – Nate Eldredge Jun 01 '11 at 18:22
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    As @Nate said, this needs some tricks. A nice indirect proof is outlined in exercise E 2.4.7 of Pedersen's *Analysis Now*. Unfortunately, I have to run now, but I might add some ideas later on. – t.b. Jun 01 '11 at 18:29
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    @Theo: Pedersen's (sketched) proof is much nicer than Conway's. Thanks for the reference. – Nate Eldredge Jun 01 '11 at 18:50
  • @Nate: I didn't expect it to be difficult — it was part of an old exam question. The syllabus has probably changed since then though. – Zhen Lin Jun 01 '11 at 19:19
  • Pedersen's argument is in fact more or less the same as the argument by girdav below. I haven't checked *very* thoroughly but it looks correct. – t.b. Jun 02 '11 at 01:16
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    If it was on an old exam, then probably it had been done in that course... – GEdgar Feb 14 '13 at 14:07

1 Answers1

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Let $\left\{y^{(n)}\right\}\subset \ell^1$ a sequence which converges weakly to $0$ in $\ell^1$. We assume this sequence doesn't converge in norm, there exists $\varepsilon >0$ such that $\lVert y^{(n)}\rVert\geqslant 3\varepsilon$ (if it's not the case, we will take a subsequence). We will show that there exists $x\in \ell^{\infty}$ and a subsequence $\left\{y^{(k_j)}\right\}$ such that $\langle x,y^{(k_j)}\rangle >\varepsilon$. Let $n_0$ such that $\sum_{n\geqslant n_0+1}|y^{(0)}_n|<\varepsilon$. For $0\leqslant k\leqslant n_0$, we take $x_k = \operatorname{sgn}y^{(0)}_k$. For all $x\in \ell^{\infty}$ whose $n_0$ first coordinates are $x_k$ we have $\langle x,y^{(0)}\rangle>\varepsilon$. From the weak convergence, we can find $k_1$ such that for $k\geqslant k_1$ we have $\sum_{n=0}^{n_0} x_ny_n^{(k)}<\varepsilon$. We can find $n_1>n_0$ and $x_{n_0+1},\cdots,x_{n_1}$ with $|x_j|\leqslant 1$ such that if $x\in \ell^{\infty}$ with the $n_1$-first coordinates are $x_j$ we have $\langle x,y^{(1)}\rangle>\varepsilon$. By this way, we will get a subsequence $\left\{ y^{(k_j)}\right\}$ and a $x\in \ell^{\infty}$ such that $\langle x,y^{(k_j)}\rangle>\varepsilon$. This contradicts the weak convergence to $0$.

Davide Giraudo
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  • Why is it ok to take a subsequence? – Johan Nov 23 '12 at 19:28
  • Because if a subsequence doesn't converge weakly to $0$, it will also be the case for the whole sequence. – Davide Giraudo Nov 23 '12 at 20:43
  • Why does it hold $\langle x, y^{(0)} \rangle>\varepsilon$? I can't get it... Thanks. – Romeo Jan 12 '13 at 23:09
  • We have, as $\lVert y^{(n)}\rVert\geqslant 3\varepsilon$ and $\sum_{n\geqslant n_0+1}|y_n^(0)|<\varepsilon$ that $\sum_{n=0}^{n_0}|y_n^{(0)}|\geqslant 2\varepsilon$. – Davide Giraudo Jan 13 '13 at 11:09
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    I am not sure that you can really arrange $\langle x, y^{k_1}\rangle < \varepsilon$ with the construction as stated. I think you also have to choose $k_1$ large enough that $\sum_{n=0}^{n_0} |y_n^{(k)}| < \varepsilon$ for $k \geq k_1$, so that "enough of the $\ell^1$-norm is contained in the indices after $n_0$". – PhoemueX May 26 '14 at 19:04
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    @Davide: OK, so you have $\langle x, y^{(0)} \rangle \geq 2\epsilon + \sum_{n=n_0+1}^\infty x_ny_n^{(0)}$. I don't see $\langle x, y^{(0)}\rangle > \epsilon$ following from that. Do you mean to have $x$ be the sequence whose first $n_0$ coordinates are $x_k$ and which has zeroes everywhere else? – Zach Blumenstein Nov 13 '15 at 19:27
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    @ZachBlumenstein, I share your concern. It seems to me that the conclusion of the following sentence ("From the weak convergence...") also only makes sense if $x_n = 0$ for all $n > n_0$. – Garrett Jun 13 '16 at 14:16
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    For those with concerns about missing details in this answer, please take a look at https://math.stackexchange.com/questions/2208898/schurs-theorem-in-ell1-weak-convergence-of-x-n-is-the-same-as-convergenc –  May 29 '17 at 19:11
  • @DavideGiraudo I think that $\langle x,y^{(1)}\rangle>\varepsilon$ should be $\langle x,y^{(k_1)}\rangle>\varepsilon$ in the line before the last one. – RBH Apr 16 '20 at 12:11
  • @DavideGiraudo Could you please tell me if we can keep making the same choice for $x_k$ as in the first time with $y^{(0)}$? – RBH Apr 16 '20 at 12:50
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    @DavideGiraudo: I don't now if this is an overkill, but the Schur property for $\ell_1$ can be proved using the Vitali-Hanh-Saks Theorem. Here is one solution along that line https://math.stackexchange.com/a/3739389/121671 – Oliver Díaz Jun 29 '20 at 22:31
  • Even if one solves the problem mentioned by @PhoemueX, which can be solved, one still has the problem that one has no statement what $\sum_{n = 0}^{n_{j-1}} x_n y^{(j)}_n$ is. This could aswell be $- \sum_{n = 0}^{ n_{j-1}} |y^{(j)}_n|$. Same for $\sum_{n = n_{j} +1}^{\infty} x_ny^{(j)}_n$. To also fix this one could chose the $n_j's$ so that: $\sum_{n = n_{j-1} +1}^{n_j} |y^{j}| - \sum_{n = 0}^{n_{j-1}} |y^{j}| - \sum_{n = n_j+1}^{\infty} |y^{j}| \ge \epsilon$ – crush3dice Jan 18 '21 at 11:57