# Convexe and serie

1. Feb 19, 2016

### Calabi

Hello every one, let be V an open convex a a normed vectorial space E.
Let be $(a_{n}) \in \mathbb{R}^{n}$ with $\sum_{i \in \mathbb{N}} a_{i} = 1$.
Let be $(v_{n}) \in V^{\mathbb{N}}$ as $\sum_{i \in \mathbb{N}} a_{i}v_{i}$ exists.
Does necessarly $\sum_{i \in \mathbb{N}} a_{i}v_{i} \in V$ please?

Thank you in advance and have a nice afternoon.

2. Feb 19, 2016

### Calabi

I suppose a positive.
In fact for my need I can suppose $a_{n} = \frac{1}{2^{n+1}}$.

3. Feb 19, 2016

### mathman

Your original statement needs proofreading. I assume you meant V an open convex set (but I'm not sure). Also $V^N$? In any case, the sum is not necessarily in V, if it exists, but in the closure of V. The sum existence depends on whether on not E is complete.

4. Feb 20, 2016

### Samy_A

Here is a theorem about open convex sets that might help (based on theorem 3.4 in Rudin's Functional Analysis).

Suppose $X$ is a topological vector space over $\mathbb R$, $A$ and $B$ disjoint, nonempty, convex subsets of $X$ with $A$ open, then there exists a continuous linear function $f: X \to \mathbb R$ and a real number $\gamma$ such that $f(a) \lt \gamma \leq f(b)$ for all $a \in A,\ b\in B$.

This should help answer the question.

For reference, the theorem from the book:

Last edited: Feb 20, 2016
5. Feb 20, 2016

### Calabi

No I suppose my sum exist . And I clearly wroght V is an open convex.