# $c_0$ satisfies the weak Banach-Saks property

We say that a Banach space $$X$$ satisfies the Banach-Saks property if any bounded sequence $$(x_n)$$ has a subsequence $$(x_{n_k})$$ such that $$\frac{x_{n_1}+...+x_{n_k}}{k}$$ converges in norm. If the same property holds for all weakly convergent sequences, then we say that $$X$$ has the weak Banach-Saks property.

Let $$c_0$$ be the space of sequences convergent to $$0$$ with supremum norm. How can one prove that it satisfies the weak Banach-Saks property?

I found this as an exercise in Diestel's Sequences and series in Banach spaces with no hint at all, searching other sources didn't lead to anything. The only hint I got is to use the sliding hump argument, but I'm not entirely sure what that means.

First, some notation clarification: as every point $$x$$ in $$c_0$$ or $$\ell^1$$ is a sequence, I will denote the the $$k$$th entry in $$x$$ by $$x(k)$$. I also make no guarantee that this is the most efficient way of doing this.

Suppose $$x_n$$ is weakly convergent. Then $$x_n$$ must be bounded; let $$M > 0$$ be an upper bound on $$\|x_n\|_\infty$$. Assume without loss of generality that $$x_n \rightharpoonup 0$$.

The idea here is very much a sliding hump argument. The "hump" here is the "large" values of $$x_n$$. Since $$x_n(k) \to 0$$ as $$k \to \infty$$, we know that the values are eventually close to $$0$$, but also since $$x_n \rightharpoonup 0$$, we have $$x_n(k) \to 0$$ as $$n \to \infty$$. We just need to sample sequence points with essentially separate humps, so that the humps don't build on each other in summation. We obviously can't stop the terms $$x_n$$ having "large" entries (otherwise $$x_n \to 0$$ strongly), but we just want to separate these large entries out.

## Defining the subsequence

Let's choose $$n_1 = 1$$. Since $$x_{n_1}(k) \to 0$$ as $$k \to \infty$$, there exists some $$K_1 \in \Bbb{N}$$ such that $$k > K_1 \implies |x_{n_1}(k)| < \frac{M}{2}.$$ We are going to choose $$n_2 > \max\{n_1, K_1\}$$ such that $$|x_{n_2}(k)| < M/2$$ for all $$k = 1, \ldots, K_1$$, which must exist because of weak convergence.

Next, since $$x_{n_1}(k), x_{n_2}(k) \to 0$$ as $$k \to \infty$$, there exists some $$K_2 \in \Bbb{N}$$, strictly greater than $$K_1$$, such that $$k > K_2 \implies \max\{|x_{n_1}(k)|, |x_{n_2}(k)|\} < \frac{M}{2^2}.$$ Choose $$n_3 > \max\{n_2, K_2\}$$ such that $$|x_{n_3}(k)| < M/2^2$$ for $$k = 1 \ldots, K_2$$.

Continue defining this subsequence inductively in this same manner. Assume that, for some $$m \ge 1$$, we have already defined $$x_{n_1}, \ldots, x_{n_m}$$, along with $$K_1, \ldots, K_m$$ such that $$(n_i)_{i=1}^m$$ and $$(K_i)_{i=1}^m$$ are both strictly increasing sequences of natural numbers, that $$k > K_m \implies \max_{1 \le i \le m} |x_{n_i}(k)| < \frac{M}{2^m},$$ and that $$(\forall i = 1, \ldots, m)(\forall k = 1, \ldots K_i) \; |x_{n_i}(k)| < \frac{M}{2^m}.$$ Then choose $$n_{m+1} > \max\{n_m, K_m\}$$ such that $$|x_{n_{m+1}}(k)| < \frac{M}{2^{m+1}}$$ for all $$k = 1, \ldots, m+1$$. Also, since $$x_{n_1}(k), \ldots, x_{n_{m+1}}(k) \to 0$$ as $$k \to \infty$$, choose a natural number $$K_{m+1} > K_m$$ such that $$k > K_{m+1} \implies \max_{1 \le i \le m+1} |x_{n_i}(k)| < \frac{M}{2^{m+1}}.$$ And so we have inductively defined our subsequence!

## Proving the subsequence works

Our definition of $$(K_m)_{m \in \Bbb{N}}$$ makes it a strictly increasing sequence, which must therefore diverge to $$\infty$$. It also means that we can use these sequence points to partition $$\Bbb{N}$$ into intervals: begin with $$[1, K_1]$$, then $$(K_1, K_2]$$, $$(K_2, K_3]$$, etc (note: these are intervals in $$\Bbb{N}$$, not $$\Bbb{R}$$). Define, for the sake of convenience, $$K_0 = 0$$, and all of our intervals take the form $$(K_{i-1}, K_i]$$.

For $$m \in \Bbb{N}$$, define $$s_m = (x_{n_1} + \ldots + x_{n_m})/m$$. For $$i \in \Bbb{N}$$, let us consider $$\max_{k \in (K_{i-1}, K_i]} |s_m(k)|$$ as a function of $$m$$ and $$i$$. In particular, we wish to bound it above.

By our construction, only $$x_{n_i}$$ in our subsequence takes values above $$2^{-1}M$$ at $$k \in (K_{i-1}, K_i]$$. We also know that only $$x_{n_i}$$ and $$x_{n_1}$$ take values above $$2^{-2}M$$, and only $$x_{n_i}, x_{n_1}, x_{n_2}$$ take values above $$2^{-3}M$$, etc. So, if $$1 \le i \le m$$ and $$k \in (K_{i-1}, K_i]$$, then $$|s_m(k)| \le \frac{\sum_{\substack{1 \le j \le m \\ i \neq j}} 2^{-j}M + M}{m} \le \frac{\sum^\infty_{j=1} 2^{-j}M + M}{m} = \frac{2M}{m},$$ since $$M$$ is an upper bound on $$\|x_n\|_\infty$$ for all $$n$$. Otherwise, if $$i > M$$, then we have $$|s_m(k)| \le \frac{\sum_{j=1}^m 2^{-j}M}{m} \le \frac{\sum^\infty_{j=1} 2^{-j}M}{m} = \frac{M}{m} \le \frac{2M}{m}.$$ Thus, $$\|s_m\|_\infty = \max_{i \in \Bbb{N}} \max_{k \in (K_{i-1}, K_i]} |s_m(k)| \le \frac{2M}{m}.$$