Is a bounded convex function $g$ that is non-negative on this particular convex set also non-negative on the its closure?

Question

Setup :

Let $S$ be the simplex on $\mathbb N$, i.e. the set of probability distribution on the natural numbers. Suppose we have $p\in S$ such that, for all $n\geq 1$, $p_n > 0$. For any $\emptyset\neq A\subseteq \mathbb N$, let $p^A\in S$ be defined as $p^A_n=\frac{p_n\cdot\mathbf 1(n\in A)}{\sum_{k\in A} p_n}$. Let $T$ be the convex hull of $\{ p^A : A\subseteq \mathbb N \}$, then for any reasonable topology on $S$, the closure of $T$ is $S$. We can for instance assume that $S$ is a subset of the Hilbert space on sequences with inner product

\begin{align*} \langle q, r\rangle &= \sum_{n\geq 1} \frac{q_n r_n}{p_n} \end{align*}

Question :

If $g:S\to\mathbb R$ is convex and bounded on $S$ as well as non-negative on $T$, then is $g$ non-negative on $S$ ?

Attempts :

In order to disprove this, I tried centering my space a $p$, then we can have a basis $\{ q_i \}_{i\in\mathcal I}$ for the span of $T$ and we can extend this basis to the span of $S$ as $\{ q_i \}_{i\in \mathcal J}$ with $\mathcal I\subsetneq \mathcal J$. Then I set $g=0$ on $S$ and I tried setting $g(q_i)=-1$ for one of the $i\in\mathcal J\setminus \mathcal I$, but I think that this is bound to fail since $g$ is linear and bounded on $T$ and I believe this should imply, since it is also bounded on $S$, that it is linear and continuous on $S$, which means that $g=0$ on $S$, but then $g(q_i)=-1$ is impossible.

Answer 1

Choose a nonprincipal ultrafilter $\omega$. Given a point $x\in S$ with barycentric coordinates $(m_1,m_2,\dots)$, let $$f(x)=\lim_{n\to \omega}(m_n\cdot n!).$$ It is an affine function on $S$ with values in $[0,\infty]$. Note that $$g=-\min\{f,1\};$$ is convex, bounded, and it is negative at some points.

Choose a point $p=(m_1,m_2,\dots)$ so that the sequence $m_n$ converges to zero very fast, say $m_n>n!\cdot m_{n+1}$ for all $n$. Note that $f(p^A)=0$ for any $A$. Therefore, $h|_T\equiv0$.