Convex combination of countably many points dense in the closure of a convex set

Question

Fix any positive integer $n$ and suppose that $C$ is a non-empty, bounded convex subset of $\mathbb R^n$.

Let $\overline C$ denote the closure of $C$ (with respect to the Euclidean topology on $\mathbb R^n$).

Suppose that $\{y^m\}_{m=1}^{\infty}$ is a countable, dense subset of $\overline C$ and that $(\lambda_m)_{m=1}^{\infty}$ is a sequence of strictly positive numbers summing up to $1$.

My conjecture is that $$y\equiv\sum_{m=1}^{\infty}\lambda_my^m\in C.$$

Clearly, this infinite convex combination must be contained in $\overline C$, but I think the positivity of the weights and the denseness of $\{y^m\}_{m=1}^{\infty}$ in $\overline C$ actually force $y$ to be in $C$.

Can anyone help me come up with a rigorous argument supporting this hunch?

Answer 1

Let $x$ be a point in the relative interior of $C$, $y$ be any point in the closure of $C$ and $\lambda$ a number strictly between $0$ and $1$. Then $\lambda x+(1-\lambda)y$ is in the relative interior of $C$. Indeed, if $O\subseteq C$ is a relatively open neighborhood of $x$, then $\lambda O+(1-\lambda)y$ is a neighborhood of $\lambda x+(1-\lambda)y$ included in the closure of $C$. There must be a smaller neighborhood fully included in $C$.

The sequence must contain a point in the relative interior since it is dense, and you can write the infinite sum as a proper convex combination of this point and a point in the closure of $C$; so the result follows from the argument above.

Answer 2

The result is true for an arbitrary probability measure on $C$.

In this case, we have the measure $p = \sum_m \lambda_m 1_{ \{ y^m \} }$.

Here is a slightly tedious approach involving induction on dimension, some details omitted.

Let $\bar{y}= \int_C y dp(y)$, where $p$ is a probability measure. We claim that $\bar{y} \in C$.

If $\phi$ is a linear functional, we see that $\phi(\bar{y}) = \int_C \phi(y) dp(y)$ and hence if $\phi(c) \le \alpha$ for all $c \in C$ we see that $\phi(y) \le \alpha$ and hence $y \in \overline{C}$.

Suppose $n = 1$ and $\bar{y} \in \overline{C} \setminus C$. Then either $\bar{y} > y$ for all $y \in C$ or $\bar{y} < y$ for all $y \in C$ (since $\bar{y}$ can only be one of two points). So, if we assume the former we see that if $\int_C \phi(y) dp(y) = \bar{y}$ then $y = \bar{y}$ for ae. $[p]$ $y \in C$ which is a contradiction, and similarly for the 'other' side. Hence we see that $y \in C$.

Now suppose the result is true for dimensions $1,...,n-1$.

Again suppose $\bar{y} \in \overline{C} \setminus C$. There is a linear functional $ \phi$ such that $\phi(y) \le \phi(\bar{y})$ for all $y \in C$. As above, it follows that $\phi(y) = \phi(\bar{y})$ for ae. $[p]$ $y \in C$. Let $H=\phi^{-1}(\phi(\bar{y}))$, and let $C_0 = C \cap H$. Note that $p(C \setminus C_0) = 0$ and so $\bar{y} = \int_{C_0} y dp(y)$. Since $H$ has dimension $n-1$, we see that $\bar{y} \in C_0 \subset C$.