Does an uncountable convex combination of elements of a set lie in the convex hull of the set in finite dimension?

Question

Suppose that $\mathcal{F}$ is a finite-dimensional vector space and that $C\subseteq\mathcal{F}$ is a convex subset of $\mathcal{F}$.

Is it true that an uncountable convex-combination of elements of $C$, $\int f(x)C(x)dx$ is in $C$,

where $\int f(x)dx=1$, $f(x)\geq0$ and $C(x)\in C$? What are the conditions for this to happen?

What if $\mathcal{F}$ is infinite-dimensional? (Here I suppose that the answer is no)

To be more precise, I'm interested in the case where $\mathcal{F}$ is a finite-dimensional vector space of functions $\mathbb{R}^n\to\mathbb{R}^m$ and $C$ is a convex subset. If $X$ is a random variable on $C$, is it true that the function $\mathbb{E}_X[g_X]$ (defined as the function $x\mapsto\mathbb{E}_X[g_X(x)]$ ) is in $C$ if the functions $g_X$ are in $C$?

I'm asking this question because I could only see this problem treated for $\mathbb{R}^n$. My understanding is that for $\mathbb{R}^n$ this is true.

Answer 1

$\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}\newcommand\si{\sigma}\newcommand\Si{\Sigma}$

1. You should have defined the meaning of the integrals $\int f(x)\,dx$ and $\int f(x)C(x)\,dx$. These should be properly understood as integrals with respect to a probability measure, say $\mu$, defined on a $\si$-algebra over $\F$ such that $C\in\F$ and $\mu(C)=1$. Also, it is then not a good idea to denote elements of $C$ by $C(x)$. So, the question can then be properly restated as follows:

Suppose that $\Si$ and $\mu$ are as above. Is it then always true that $\int c\,\mu(dc)\in C$?

2. In the case when $\F=\R^d$ we have the easy "yes" answer (and you said you know how to treat this case); here we need to assume the natural condition that all affine subspaces of $\F$ are in $\Si$; in particular, it would be enough for $\Si$ to contain the Borel $\si$-algebra over $\R^d$. The case of any finite-dimensional $\F$ then immediately follows, because then $\F$ is linearly isomorphic to $\R^d$ for some $d$.

3. For infinite-dimensional $\F$, the answer is "no" in general. E.g., let $\F=C_0[0,\infty)$, the topological vector space of all continuous functions on $[0,\infty)$ vanishing at $\infty$, with the topology defined by the $\sup$ norm. Let $\Si$ be the Borel $\si$-algebra over $\F$. Let $C$ be the convex hull (or the span) of the set $C_*:=\{c_x\colon x\in(0,\infty)\}$, where $c_x$ is the (decreasing exponential) function in $\F$ defined by the formula $c_x(t):=e^{-xt}$ for $t\in[0,\infty)$. Let $\mu$ be the probability measure on $\Si$ defined by the formula $$\mu(B):=\int_{(0,\infty)}dx\,e^{-x}\,1(c_x\in B)$$ for all $B\in\Si$. Then $\mu(C)=\mu(C_*)=1$. Let $$b:=\int c\,\mu(dc),$$ so that $$b(t)=\int_{(0,\infty)}dx\,e^{-x}\,e^{-tx}=\frac1{t+1}$$ for all $t\in[0,\infty)$. Then $b\notin C$ -- because, in contrast with $b$, all functions in $C$ decrease to $0$ exponentially fast near $\infty$.

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Here is another, possibly simpler, "infinite-dimensional" counterexample. Let $\F=\R^\R$, the set of all real-valued functions on $\R$. Let $\Si$ be the powerset of $\F$. Let $C$ be the set of all polynomial functions in $\F$. Let the probability measure on $\Si$ be defined by the condition $$\mu(\{p_n\})=\frac1{en!}$$ for $n=0,1,\dots$, where $p_n(t):=t^n$ for $t\in\R$. Then for $$g:=\int c\,\mu(dc)$$ and $t\in\R$ we have $$g(t)=\sum_{n=0}^\infty p_n(t)\mu(\{p_n\})=e^{t-1}.$$ So, $g\notin C$, because $g'=g\ne0$, whereas $c'\ne c$ for any nonzero $c\in C$.

The latter counterexample shows that even a countable convex combination of points of a convex subset $C$ of an infinite-dimensional linear space does not have to be in $C$.