Continuity of support functionals

Is $f(x)$ well-defined?

Yes, if $G$ is not empty. Let $x\in X$ be given. The function $\psi_x:X^*\to\mathbb R$, $x^*\mapsto \langle x^*, x\rangle$ is continuous from the $w^*$-topology of $X^*$ to $\mathbb R$. Therefore, the image of $G$ under $\psi_x$ is compact. Because the maximum of each non-empty compact subset of $\mathbb R$ exists and is real-valued, $f(x)$ is well-defined.

Is $f$ continuous?

I think I found a counterexample.

Let $X=\ell^2_0$, which is the subspace of $\ell^2$ which consists of sequences with only finitely many nonzero entries. Clearly, $X$ is not a Banach space and $\ell^2$ is its dual space. We set $$ G := \{\sqrt{n}e_n : n\in\mathbb N\} \cup \{0\}\subset X^*. $$ Then it can be shown that $G$ is a $w*$-compact subset of $X^*$ (I can fill in the details if requested).

Let us show that $f$ is not continuous. Let $x_n:=n^{-1/2}e_n$ be a sequence in $\ell^2_0$. Clearly, $x_n\to 0$. However, we have $f(x_n)=\langle \sqrt{n} e_n,n^{-1/2}e_n\rangle = 1$ and $f(0)$, but not $f(x_n)\to f(0)$. Therefore $f$ is not continuous.

This seems to contradict the answer of @Red shoes.

Is $f$ lower semicontinuous?

Yes, $f$ should be lower semicontinuous. This is because of the arguments in the answer of @Red shoes.

Under what conditions on $G$ is $f$ continuous?

It can be shown that $f$ is continuous if and only if $G$ is bounded (and if and only if $f$ is continuous at $0$).

First, suppose that $G$ is bounded. Then it can be seen that $f$ is locally bounded. Since $f$ is also lower semicontinuous it follows that $f$ is continuous.

Now, suppose that $f$ is continuous at $0$. As we can see in the question Unbounded subdifferential of a convex functional, this implies that $\partial f(0)$ is bounded. Now we claim that $G\subset \partial f(0)$ holds. Indeed, for each $g^*\in G$ we have $ f(x)-f(0) = f(x) \geq \langle g^*, x \rangle $ and therefore $g^*\in\partial f(0)$. Since $\partial f(0)$ is bounded this implies that $G$ has to be bounded, too.