# “Passing Through Convex Combinations” - Constructing a Norm Convergent Net from a Weakly Convergent Net

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 $$\operatorname{conv}\{ x_\alpha: \alpha \in I \} \subseteq B(0,R)$$.
• 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! – LMW Apr 22 '19 at 23:19