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

  • 1
    $\begingroup$ 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? $\endgroup$
    – rarc
    Feb 26 '19 at 19:59
  • $\begingroup$ yes, this is correct. $\endgroup$
    – supinf
    Feb 26 '19 at 20:17

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