Question about the proof of the fact that minimum is attained for a l.s.c. convex function over a convex compact set. I quote here the proof of a result given in Haim Brezis Functional Analysis, Sobolev Spaces and partial differential Equations:
I haven't been able to conclude where exactly is this hypothesis used:

A is closed and convex.


As far as I understand, one concludes the existence of minimum in the convex, closed and bounded set $\tilde{A}$, but this is done without using the fact that $A$ is closed and convex, 
The fact that $\tilde{A}$ is closed  and convex simply follows from the lower semicontinuity and convexity of the function $\varphi$ and we obtain the existence of minimum in the set $\tilde{A}$ Then, as $\tilde{A} \subset A$ we obtain the minimum in the whole $A$. 
 A: No the assumption for $A$ to be closed and convex are important, because if we put 
$$
B=\{x\in A \; , \; \varphi(x)\leq \varphi(a) \}
$$
so using the fact that $\varphi$ is lsc prove that $B$ is closed in the trace topology of $A$, since $A$ is closed $B$ is closed in $E$, and the fact that $A$ is convex and $\varphi$ is convex prove that $B$ is convex, and finally $(5)$ prove that $B$ is bounded  so in particular $B$ is a compact in $\sigma(E,E^*)$.
If $A$ is not closed the theorem will fall in fact :
$E=\mathbb{R}$ and $A=]0,+\infty[$ and $\varphi(t)=t$ it's clear that $\inf_{t\in A}\varphi(t)=0$ who is not achieved in any point of $A$.
If $A$ is closed but not convex the question is more complicated because every sci in a compact set achieve the minimum, so we need an infinite dimensional example,but the theorem fall also in this case in fact :
$E=l^2(\mathbb{N})$ and $A=\{(x_n)_n\; , \; \sum_{k\geq0}|x_k|^2=1\}$ and $\varphi((x_n)_n)=x_1$ it's clear that $\inf_{x\in A}\varphi(x)=-\infty$ who is not achieved in any point of $A$.
