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.

  • 9
    $\begingroup$ 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. $\endgroup$ Jun 1, 2011 at 18:22
  • 5
    $\begingroup$ 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. $\endgroup$
    – t.b.
    Jun 1, 2011 at 18:29
  • 4
    $\begingroup$ @Theo: Pedersen's (sketched) proof is much nicer than Conway's. Thanks for the reference. $\endgroup$ Jun 1, 2011 at 18:50
  • $\begingroup$ @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. $\endgroup$
    – Zhen Lin
    Jun 1, 2011 at 19:19
  • 2
    $\begingroup$ If it was on an old exam, then probably it had been done in that course... $\endgroup$
    – GEdgar
    Feb 14, 2013 at 14:07

1 Answer 1


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 $0$.

  • 10
    $\begingroup$ 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 $n_0$". $\endgroup$
    – PhoemueX
    May 26, 2014 at 19:04
  • 1
    $\begingroup$ @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? $\endgroup$ Nov 13, 2015 at 19:27
  • 3
    $\begingroup$ @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 $n > n_0$. $\endgroup$
    – Garrett
    Jun 13, 2016 at 14:16
  • 3
    $\begingroup$ For those with concerns about missing details in this answer, please take a look at math.stackexchange.com/questions/2208898/… $\endgroup$
    – user138530
    May 29, 2017 at 19:11
  • 1
    $\begingroup$ @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 math.stackexchange.com/a/3739389/121671 $\endgroup$ Jun 29, 2020 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.