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?


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 $\sum p_n e_n \notin F$.

  • $\begingroup$ $V_1$ 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? $\endgroup$ – user251257 Sep 5 '16 at 19:51
  • $\begingroup$ 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) $\endgroup$ – H. H. Rugh Sep 5 '16 at 21:20
  • $\begingroup$ 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. $\endgroup$ – user251257 Sep 5 '16 at 22:06
  • $\begingroup$ You probably need some argument with non empty relative interior, which is in fact only true in finite dimensional case. $\endgroup$ – user251257 Sep 5 '16 at 22:15
  • $\begingroup$ 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) $\endgroup$ – H. H. Rugh Sep 5 '16 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.