Continuity of support functionals Let $X$ be a normed space(not necessarily Banach) and let $G$ be a $w^*-$compact subset of $X^*,$ the dual space of $X.$ Consider the function $f:X\to \Bbb R$ defined by
$$f(x)= \max_{g^*\in G}\{\langle g^*,x\rangle\}.$$ Is $f$ continuous? If not, what about lower semicontinuous? Could you present a example of a discontinuous $f$? 
I can prove that if $X$ is Banach the function is Lipschitz(by using the Uniform Boundedness Theorem). However, in the general case I am still curious. 
 A: 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.
A: It is continuous. 
The $ epi f$ is intersection of all closed half space in the form $\{(x , \alpha) \in X \times R : ~~ \langle g^* , x \rangle  \leq \alpha \}$ where $g^* \in G$. 
This shows the \epi f is closed and convex and Hence $f$ is convex and lower semicontinuous  which implies $f$ is continuous. 
