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.


1 Answer 1


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, $$\|s_m\|_\infty = \max_{i \in \Bbb{N}} \max_{k \in (K_{i-1}, K_i]} |s_m(k)| \le \frac{2M}{m}.$$


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