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

Question

I'm having a really hard time with nearly every part of this proof, any help would be appreciated.

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

Definition: For $x_n \in \ell^1$ convergence in the norm: $$ x_n \to x \iff \|\ x_n - x \|_{\ell^1} \to 0.$$

Definition: For $x_n \in \ell^1$, weak convergence: $$x_n \rightharpoonup x \iff \phi x_n \to \phi x \hspace{1cm} \forall x_n \in \ell^1 , \space\ \space\ \forall \phi \in \ell^{1*}.$$

This problem is from Muscat's "Functional Analysis" text. It breaks Schur's Theorem up into the following parts:

1) If the statement were false there would be unit $x_n = (a_{ni}) \in \ell^1$ such that $x_n \rightharpoonup 0$

Proof: Taking a unit $(a_{ni}) \in \ell^1$ we have that $\sum_{i=1}^{\infty} |a_{ni}| = 1$. Then, for some $\phi \in \ell^{1*}$ we (somehow?) obtain that $\phi x_n \to \phi x = 0$.

2) For each $n$ there is an $N_n$, such that $\sum_{i=1}^{N_n} |a_{ni}| > \frac{4}{5}$.

Proof: Since $(a_{ni})$ is convergent in $\ell^1$ as a unit, we have that $$\sum_{i=1}^{\infty} |a_{ni}| = 1 \hspace{1cm} \forall n$$ as a requirement for convergent series is that their tail sequence goes to $0$, so that for $\epsilon > 0$ there exists some $N \in \mathbb{N}$ where $$\sum_{i=N}^{\infty} |a_{ni}| < \epsilon.$$ Hence, it follows that we may find an $N_n$ where $$\sum_{i=1}^{N_n} |a_{ni}| > \frac{4}{5} \hspace{1cm} \forall n$$

3) Each coefficient converges to $0$ as $n \to \infty$, so $\forall k , \exists M, n \geq M \Rightarrow \sum_{i < k} |a_{ni} | < \frac{1}{5}$.

Proof: Mirroring what we did above in part (2) and choosing $\epsilon = \frac{1}{5}$.

4) A subsequence of $\left\{ x_n \right\}$ exists with $$\sum_{i < N_n - 1} |a_{ni}| < \dfrac{1}{5} , \sum_{i=N_{n-1}}^{N_n} |a_{ni} | > \dfrac{3}{5}, \sum_{i > N_n} |a_{ni} | < \dfrac{1}{5} .$$

Proof:

5) Let $y := (|a_{ni}| / a_{ni}) \in \ell^{\infty}$ where for each $i$, $n$ is such that $N_{n-1} \leq i < N_n$. Show $|y \cdot x_n | \geq \frac{1}{5}$ to obtain a contradiction.

Proof:

Answer 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 $\displaystyle \left|\sum_{k=1}^\infty y_kx^{M_j}_k \right| \ge \left| \sum_{k=N_{j-1}+1}^{N_j}y_kx_k^{M_j} - \left|\sum_{k=0}^{N_{j-1}}y_kx_k^{M_j}\right| - \left|\sum_{k=N_j +1}^{\infty}y_kx_k^{M_j}\right|\right| \ge | \frac{98}{100}\epsilon - \frac{1}{100}\epsilon - \frac{1}{100}\epsilon |= \frac{96}{100}\epsilon > 0$

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.