# Definition of convex set with countable sum

The definition of a convex set I'm used to involves the sum of term: $C$ is convex iff $\forall x_1, x_2 \in C, \forall \theta \in [0, 1], \theta \,x + (1 - \theta)\, y \in C$.

I am aware of the definition with $n \in \mathbb{N}$ terms which can be derived by induction and used equivalently.

However I have read that it can even be generalized to series : if $(\theta_i)_{i \in \mathbb{N}} \in (\mathbb{R}_+)^\mathbb{N}$ and $(x_i)_{i \in \mathbb{N}} \in C^\mathbb{N}$ are such that $\sum_0^\infty \theta_i = 1$, then $\sum_0^\infty \theta_i x_i \in C$.

I cannot manage to prove this result, mostly because a convex is not necessarily closed. I only have that $\sum_0^\infty \theta_i x_i \in \bar{C}$ as the limit of $\sum_0^n \theta_i x_i + (\sum_{n+1}^\infty \theta_i) x_0$, which is in $C$ for every $n$ by the definition which uses finite sums.

Any suggestion on this matter?

## 1 Answer

The result is false in infinite dimension, see below, but true in finite dimension.

Let $V$ be a real finite dimensional normed vectorspace and $C\subset V$ convex. Let $x_i\in C$, $i\in {\Bbb N}$ and let $\theta$ be a probability on ${\Bbb N}$. We assume that $\sum_{n\geq 1} \theta_n |x_n|<+\infty$ but $m=\sum_i \theta_i x_i\notin C$.

Since $m\in {\rm Cl} \; C$ we must have $m\in \partial C$. Now using Hahn-Banach (in finite dimension) we may find a bounded linear functional $\ell: V\rightarrow {\Bbb R}$ so that $$x\in {\rm Cl}\; C \Rightarrow \ell(x)\geq \ell(m)$$ Then we must have $\ell(x_i)=\ell(m)$ for every $i$. The subspace $V_1=\{x\in V: \ell(x)=\ell(m)\}$ has dimension one less than $V$ and $C_1=C\cap V_1$ is still convex non-empty (contains all the $x_i$) and $m \in ({\rm Cl}\; C_1) \setminus C_1$. So we may restrict the problem to $V_1$ and apply the same argument to $C_1$. Proceeding by induction we end up with $m$ on a line and all $x_i$ being on one side. But then they must all equal $m$ and $m$ was in $C$ in the first place.

A counter example in infinite dimension: Let $V=\ell^2({\Bbb N})$ and let $F\subset V$ be the set of sequences with only finitely many non-zero elements. $F$ is clearly convex. Let $e_n=(0,0,...,0,1,0,...)$ be the canonical basis vector with 1 at the n'th place. If $\theta_n>0$ for all $n$ then $\sum p_n e_n \notin F$.

• $V_1$ is an affine subset, not a subspace, isn't it? Don't you need an argument for $C \ne C_1$? Where have you needed that the space is finite dimensional? – user251257 Sep 5 '16 at 19:51
• Yes, affine subspace. It doesn't matter if $C=C_1$. The importance in the proof is that the dimension is 1 below. Finite dimension because I want the induction to end (with a line, or rather a point). I thought about infinite dimension but didn't manage to make a proof going (it is probably still true) – H. H. Rugh Sep 5 '16 at 21:20
• ah that right with induction. But there is nothing in your argument that says you may select a different functional $\ell$ each time. So the dimension needs not decrease. – user251257 Sep 5 '16 at 22:06
• You probably need some argument with non empty relative interior, which is in fact only true in finite dimensional case. – user251257 Sep 5 '16 at 22:15
• In fact it does imply a different functional. Convexity of $C_1$ in $V_1$ means existence of a non-trivial linear functional of $V_1$, so automatically independent of the first (and by induction it follows for the rest) – H. H. Rugh Sep 5 '16 at 22:15