Expected utility function is continuous over discrete probability distributions

Question

I'm having trouble proving that the follwing function is continuous:
Let $A$ be a non-empty set (not necessarily finite) and $$ X = \left\{ x : A \to [0, 1] \ \middle| \ \text{supp}(x) \ \text{is finite and} \sum_{a \in \text{supp}(x)} x(a) = 1 \right\}. $$ Here $\text{supp}(x) = \{a \in A \ \mid \ x(a) \neq 0\}$ and of course $X$ is convex and is toplogical subspace with the topology induced by the sup norm $\|x\| = \sup\{|x(a)| \ \mid \ a \in A\}$.
Supppose that there is $f : X \to \mathbb{R}$ such that for all $t \in (0, 1)$ and $x, y \in X$, $$ f(t x + (1 - t) y) = t f(x) + (1 - t) f(y). $$ Then $f$ is continuous (the topology in $\mathbb{R}$ is the usual one).
I know that in finite dimension this is easy to prove and for the special case of a convex function $f : \mathbb{R}^n \to \mathbb{R}$ this is well-known, but in the infinite case this isn't always true (here's a discussion). However, $X$ is the convex hull of the set of indicator functions of $\{a\} \subset A$ and $f$ is convex.
In this particular case I think continuity is true.
Any hints are welcome!

Answer 1

Not true. Let $A=\mathbb N$. Then the set $X$ resembles the space of sequences with finite support $c_{00}$. In this case, one can explicitly construct an discontinuous linear function.

Define $$ f(x) = \sum_j j x_j. $$ Then $f$ is linear but discontinuous on $X$:

Let $x^k$ be the $k$-th unit vector in $X$, i.e., the characteristic function of $\{k\}$: $x^k(k)=1$, $x^k(i)=0$ for all $i\ne k$.

Then $$\frac1{\sqrt k}(x^k-x^1)+x^1 \to x^1$$ in the sup-norm, but $$f(\frac1{\sqrt k}(x^k-x^1)+x^1 ) = \sqrt k -\frac1{\sqrt k} +1\to\infty.$$