Exercise 33, § 1

Let E be a uniformly convex Banach space (V, p. 67, exerc. 31). Then there exists a number $\theta$ such that $\frac{3}{4} \leq \theta < 1$ and such that the relation $\|x - y\| \geq \frac{1}{2} \sup (\|x\|, \|y\|)$ in E implies $\| \frac{1}{2}(x + y) \| \leq \theta \sup (\|x\|, \|y\|)$. a) Let $(x_n)$ be a sequence of points of E such that $\|x_n\| \leq M$ and such that the sequence tends to 0 for $\sigma(E, E')$. Show that, if for an index $p,\ \|x_p\| \geq \frac{1}{2} M$, then there exists $q > p$ such that $\|x_p - x_q\| > \frac{1}{2} M$, and consequently $\| \frac{1}{2}(x_p + x_q) \| \leq \theta M$ (argue by reductio ad absurdum, by observing that for all $x' \in E'$ such that $\|x'\| = 1$, we have $\langle x_p, x' \rangle = \lim_{n \to \infty} \langle x_p - x_n, x' \rangle$).

Deduce that there exists a strictly increasing mapping $\ell$ from $\mathbf{N}$ into itself such that $\| \frac{1}{2}(x_{\ell(2n)} + x_{\ell(2n+1)}) \| \leq M \theta$, and such that if $x_n^{(1)} = \frac{1}{2}(x_{\ell(2n)} + x_{\ell(2n+1)})$, the sequence $(x_n^{(1)})$ tends to 0 for $\sigma(E, E')$ and that $\|x_n^{(1)}\| \leq M \theta$ for all $n$. b) Show that there exists a sequence $(x_{n_k})$ extracted from $(x_n)$ such that, if we put $y(k) = x_{n_k}$, we have the following property : for every integer $p > 1$, every integer $q < p$ and every integer $i$ such that $1 \leq i \leq 2^{p-q}$,

(Iterate the procedure of a) by constructing a sequence $(x_n^{(k+1)})$ from the sequence $(x_n^{(k)})$ in the same way as $(x_n^{(1)})$ is constructed from $(x_n)$; then use a suitable « method of diagonalization ».) c) Let $r$ and $q$ be two integers $> 1$. Deduce from b) that if $r2^q \leq k \leq (r+1)2^q$, then (decompose the sum on the left into several subsets, by varying $h$ from 1 to $2^q$, then from $(j-1)2^q + 1$ to $j2^q$ for $2 \leq j \leq r$, then from $r2^q + 1$ to $k$). d) Show that for every bounded sequence $(x_n)$ in E, there exists an extracted sequence $(x_{n_k})$ such that the sequence of the averages $(x_{n_1} + \cdots + x_{n_k})/k$ converges for the initial topology of E (the Banach-Saks-Kakutani theorem). (Using the fact that E is reflexive, reduce to the case when the sequence $(x_n)$ converges to 0 for $\sigma(E, E')$, and use c) for $q$ and $r$ large enough.)