Simple example on uniformly convex spaces

Question

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!

Answer 1

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 $\|x_n\|\to \|e_1\|=1$.