Infinite convex combination is in the closure of the convex hull.

convex-hullsfunctional-analysistopological-vector-spaces

Let $V$ be a topological vector space and $\{x_i\}_{i \in I}$ be a net in $V$. Further, let $\{\lambda_i\}_{i\in I}$ be a net in $[0,1]$ such that
$$\sum_{i \in I}\lambda_i =1.$$ Assume $x= \sum_{i \in I} \lambda_i x_i$ exists in $V$. That is, $$x= \lim_F \sum_{i \in F} \lambda_i x_i$$
where $F$ ranges over all finite subsets of $I$.

Is it true that $x$ is in the closure of the convex hull of $\{x_i:i \in I\}$?

Intuitively, this seems clear, but I would need to show that $x$ is the limit of (finite) convex combinations of $\{x_i: i \in I\}$, but I can't see why this is true.

If necessary, I can assume that $V$ is a normed space although I think the result holds more generally.

Best Answer

Yes, it holds in any topological vector space: $x=\lim (\sum_{i \in F} \lambda_i) \frac {\sum_{i \in F} \lambda_ix_i} {\sum_{i \in F} \lambda_i}$ and $\frac {\sum_{i \in F} \lambda_i x_i} {\sum_{i \in F} \lambda_i}$ is in the convex hull of $\{x_i: i \in I\}$.

As pointed out by QuantumSpace we must take limit along subnet to avoid division by $0$.

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