# Strong and weak convergence in $\ell^1$

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.

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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 1 '11 at 18:22
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 1 '11 at 18:29
@Theo: Pedersen's (sketched) proof is much nicer than Conway's. Thanks for the reference. –  Nate Eldredge Jun 1 '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 1 '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 2 '11 at 1:16

## 1 Answer

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 $0$.

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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
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