Schur's Theorem: In $\ell^1$ weak convergence of $x_n$ is the same as convergence in the norm

Solution 1:

The crux of the matter is the idea of the "gliding hump". Intuitively, if our sequence of $\ell^1$ sequences doesn't converge in norm to $0$, we can construct a functional on $\ell^1$ such that the $\ell^1$ mass of our sequence is always carried down the line, like a hump gliding to infinity. Put another way, if we have $x_n \rightharpoonup 0$, but $\|x_n\| \not\to 0$, there will be a subsequence with $\|x_m\|> \epsilon$ for all $m$, and some $\epsilon$. Then we will find a functional which preserves most of this "mass" (like, say, 98% of the $\epsilon$), and then we will contradict weak convergence. So here we go.

Suppose we have a sequence $\{x^n\}_{n \in \mathbb{N}}$ such that $x^n \rightharpoonup 0$ but $\|x^n\| \not\to 0$. Then there exists an $\epsilon >0$ such that $\|x^m\| > \epsilon$ for infinitely many $m$. Pass then to the subsequence $\{x^m\}_{m \in \mathbb{N}}$. Of course it suffices to consider $x^n \rightharpoonup 0$ otherwise if the weak limit is $x$ we pass to $x^n - x$. Now, choose $N_1, M_1, N_2$ and $M_2$ in the following way:

\begin{align} &N_1 \Rightarrow \sum_{k=N_1}^\infty |x_k^1| < \frac{\epsilon}{100}&\\ &M_1 \Rightarrow \sum_{k=1}^{N_1} |x_k^{M_1}| < \frac{\epsilon}{100}&\\ &N_2 \Rightarrow \sum_{k=N_2}^\infty |x_k^{M_1}| < \frac{\epsilon}{100}&\\ &M_2 \Rightarrow \sum_{k=1}^{N_2}|x_k^{M_2}|< \frac{\epsilon}{100} \end{align} where choices of $N_i$ follow from the fact that the tail ends of $x^{M_{i-1}}$ can be made small, and choices of $M_i$ can be made by the termwise convergence (note these are finite sums) given by weak convergence. In general, take \begin{align} &N_i \Rightarrow \sum_{k=N_i}^\infty |x_k^{M_{i-1}}| < \frac{\epsilon}{100}&\\ &M_i \Rightarrow \sum_{k=1}^{N_i} |x_k^{M_i}| < \frac{\epsilon}{100}& \end{align}

where it goes without saying (but we say it anyway) that $\Rightarrow$ fulfills the role of "such that". Now, we see that in this construction the bulk of the $\ell^1$ mass of $x^{M_j}$ lies in the $x_k^{M_j}$ with $N_j + 1 \le k \le N_{j+1}$. More precisely, 98 percent of the mass lies in the $x_k^{M_j}$ in that interval.

Now define the function $f \in (\ell^1)^*$ corresponding to the bounded sequence $y \in \ell^\infty$ given by $y_k =$ sgn$(x_k^{M_j})$ for $N_{j-1} \le k < N_j$. Then $\displaystyle f(x^{M_J}) = \sum_{k=1}^\infty y_kx^{M_j}_k = \sum_{k=0}^{N_{j-1}}y_kx_k^{M_j} + \sum_{k=N_{j-1}+1}^{N_j}y_kx_k^{M_j} + \sum_{k=N_j +1}^{\infty}y_kx_k^{M_j}$
so then $\displaystyle \left|\sum_{k=1}^\infty y_kx^{M_j}_k \right| \ge \left| \sum_{k=N_{j-1}+1}^{N_j}y_kx_k^{M_j} - \left|\left|\sum_{k=0}^{N_{j-1}}y_kx_k^{M_j}\right| - \left|\sum_{k=N_j +1}^{\infty}y_kx_k^{M_j}\right|\right|\right| \ge | \frac{98}{100}\epsilon - \frac{1}{100}\epsilon - \frac{1}{100}\epsilon |= \frac{24}{25}\epsilon > 0$

for all $j$. But then this contradicts weak convergence to 0 along the subsequence $\{x^{M_j}\}_{j \in \mathbb{}}$. So therefore weak convergence implies strong convergence.

I'll leave proving strong convergence implies weak convergence to you.