# Simple example on uniformly convex spaces

In the lectures we showed the following result:

Theorem: Let $$(E,\|\cdot\|_E)$$ be a uniformly convex space. Consider a sequence $$\{x_n \}\rvert_{n\in\mathbb{N}} \subset E$$ and $$x \in E$$ such that it converges weakly to $$x\in E$$ $$x_n\rightharpoonup x ,$$ and the sequence of the norms converges to the norm of $$x\in E$$, i.e. $$\|x_n\|_E \longrightarrow \|x\|_E.$$ Then the sequence $$\{x_n \}\rvert_{n\in\mathbb{N}} \subset E$$ is strongly convergent $$x_n \longrightarrow x.$$

This means that weak convergence, together with the convergence of the norms imply strong converge in uniformly convex spaces.

Question: Could you please provide a counterexample on a non uniformly convex space (maybe sequence space of bounded sequences $$\ell^\infty$$?) where this result does not hold?

Concretely: A sequence on a non uniformly convex space such that it is weak convergent, and the sequence of the norms converges, but the sequence itself is not strongly convergent.

I would be grateful to read any possible counterexample. Thanks!

Consider the sequence $$x_n := (e_1+e_n) \in \ell^\infty$$.
Then it can be shown that $$e_n\rightharpoonup 0$$ (see here) which implies $$x_n\rightharpoonup e_1$$.
It can also be calculated that $$x_n\not\to e_1$$ in the norm convergence and that $$\|x_n\|\to \|e_1\|=1$$.
• Dear Sir. Thanks for your answer. To clarify, you can pick $\varepsilon = \frac{1}{2}$ such that $\| x_n - e_1 \|_{\infty} = \| e_n \|_{\infty} > \varepsilon$ for all $n\in \mathbb{N}$. This would show that the convergence does not follow strongly. Correct?