Strong and weak convergence in $\ell^1$

Question

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.

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 $0$.