Let $ \mathfrak{A} $ be a Banach algebra (or Banach space, I don't think it really matters) and let $ (x_{\alpha}) \subset \mathfrak{A} $ and $ x \in \mathfrak{A} $ be such that:

$$ \mathrm{wk} - \lim_{\alpha} x_{\alpha} = x $$

How can I construct a new net $ (x_{\beta}) $ depending on $ (x_{\alpha}) $ such that:

$$ \mathrm{st} - \lim_{\beta} x_{\beta} = x $$

and what sort of properties of the net $ (x_{\alpha}) $ can $ (x_{\beta}) $ inherit? E.g., if $ (x_{\alpha}) $ is bounded, can $ (x_{\beta}) $ be made to also be bounded by the same bound?

I have seen this technique of "passing through convex combinations" mentioned a few times, but never really understood it completely. I am quite rusty when it comes to convexity though. I imagine that Mazur's Theorem and looking at the convex hull of some set is used.

Any help would be appreciated. Thanks!

Edit: I just found Mazur's Lemma, which states: If $ X $ is a Banach space and if $ (x_n) $ is a weakly convergent sequence that weakly converges to $ x $ then there exists a sequence $ (y_n) $ of convex combinations of elements from $ \{ x_n : n \in \mathbb{N} \} $ such that $ (y_n) $ strongly converges to $ x $. Is there a Mazur's Lemma for weakly convergent nets?


The proof of Mazur's Lemma adapts immediately to give you existence of a net $(x_\beta)$ with the desired properties.

By assumption, $x$ lies in the weak closure of $\operatorname{conv} \{ x_\alpha: \alpha \in I \}$, using the net characterisation of the closure. For convex sets, the weak closure and strong closure coincide and so $x$ lies in the strong closure of $\operatorname{conv} \{ x_\alpha: \alpha \in I \}$. This means that there is a net $(x_\beta)$ (or even a sequence) with values in $\operatorname{conv} \{ x_\alpha: \alpha \in I \}$ converging to $x$ in the norm topology.

It is of course obvious that if $(x_\alpha)$ is bounded, say $x_\alpha \in B(0,R)$ for all $\alpha$, then $(x_\beta)$ above is also bounded with the same bound since then $\operatorname{conv}\{ x_\alpha: \alpha \in I \} \subseteq B(0,R)$.

  • $\begingroup$ Ah, I think I see now. But shouldn't a net $ (x_{\beta}) $ be just in $ \mathrm{conv} \{ x_{\alpha} : \alpha \in I \} $ and not necessarily in the strong closure of $ \mathrm{conv} \{ x_{\alpha} : \alpha \in I \} $? Thanks! $\endgroup$ – LMW Apr 22 '19 at 23:19
  • $\begingroup$ Sure, that was a slip up on my part. $\endgroup$ – Rhys Steele Apr 22 '19 at 23:40

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