Show that in $\ell^1(\mathbb{N})$ weak convergence is equivalent to strong convergence

I know that weak convergence is equivalent to strong convergence when the space has finite dimension, but $$\ell^1(\mathbb{N})$$ is not, so I have no idea where to begin.

• This is a basic theorem due to Schur. See Rudin's FA for a proof. – Kavi Rama Murthy Aug 2 '20 at 23:50
• [Maybe a stupid question] Doesn't weak convergence requires an inner product? How do you define an inner product in $\ell^1(\mathbb N)$? – Dmitry Aug 2 '20 at 23:56
• @Dmity Weak convergence replaces inner product with duality product. The real sequence $\langle y , x_n \rangle_{L^1,L^\infty}$ converges to the real number $\langle y , x \rangle_{L^1,L^\infty}$ for every $y \in L^\infty$ means that $x_n$ converges weakly to $x$. In the case of an inner product, it is the same definition, but the dual space is isomorphic to the space itself. – rubikscube09 Aug 2 '20 at 23:58
• Note that this is true for sequences, but fails quite badly for nets (in other words, the strong and weak topology do not coincide). – MaoWao Aug 3 '20 at 1:11

Here is my attempt at a hint, but I haven't actually seen the proof, so maybe this won't work. Suppose there is a sequence that converges weakly to $$0$$, but not strongly to $$0$$. Then w.l.o.g. there is a sequence $$x_n \rightharpoonup 0$$ and $$\|x_n\| \ge 1$$. Now choose a subsequence as follows. Given $$x_{n_1}, x_{n_2}, \dots, x_{n_k}$$, we know that there exists $$N_k > N_{k-1}$$ such that the sums used to calculate the norms of these vectors is almost all concentrated on $$[1,N_k]$$. Next, the individual components of $$x_n$$ converge to $$0$$. So we can choose $$x_{n_{k+1}}$$ so that the sum used to calculate the norm is almost all concentrated on $$[N_k+1,\infty)$$.
Now choose $$y \in \ell^\infty$$ so that the components of $$y$$ on $$[N_{k-1}+1,N_k]$$ are the signs of the components of $$x_{n_k}$$. Then see that $$\langle x_{n_k}, y\rangle \not\to 0$$.