One of the most known theorem of existence of a minimizer of a functional is the following.
${\bf Theorem:}$ Let $H$ be an Hilbert space and $A\subset H$ be a nonempty, bounded, closed and convex subset. If $E:H\to\mathbb R$ is a lower semicontinuous and convex functional, hence $E$ achieves its minimum in $A$, which means that exists $a\in A$ such that $E(a) =\displaystyle\inf_{x\in A} E(x)$.
An interesting variant of that Theorem assures that if $A$ is not bounded, the results remains true if we ask for $E$ to be coercive.
My question is: the above Theorem holds in a Hilbert space (in general, of infinite dimension). What happens if $H$ is finite dimensional, e.g. if $H=\mathbb R$? The result still holds? If yes, how to prove that?
Thank you in advance!
In $\mathbb{R}^n$, the demonstration can take advantage of the fact that the bounded closed subsets are the compact subsets. The hypothesis "$A$ is compact" is sufficient: convexity is not needed.
Proof: let $M$ be the minimum of $E$ on $A$ (possibly -$\infty$); we can find points $x_n \in A$ where $E(x_n)$ approaches $M$ as close as we want (in the case of $M=-\infty$, this means $\forall N, \exists x_n \in A, E(x_n)<N$).
As $A$ is compact, we can extract a subsequence $x_{i_n}$ from $x_n$ such that $x_{i_n}$ converges to a point $y$ in $A$.
$E$ being lower semicontinuous, $E(y) \le \lim_{n \rightarrow +\infty} E(x_{i_n}) = M$.
So $E(y) \le M$ (which implies $M \ne -\infty$).
As $y \in A$, $E(y) \ge M$. So $E(y) = M$: the minimum is attained by $E$ in $A$.
In the variant where $A$ is not bounded and $E$ is coercive, we use the fact that "$E$ is coercive" is equivalent to $\forall N, \{x \in A, E(x) \le N\}$ is empty or bounded.
Choosing a $N$ such that $B=\{x \in A, E(x) \le N\}$ is non-empty, searching the minimum of $E$ in $A$ is equivalent to searching the minimum of $E$ in $B$.
We take $C$ the closure of $B$. The closure of a bounded subset in a metric space is bounded, so $C$ is bounded.
$C$ being the intersection of all closed subsets that contain $B$, and $A$ being closed and containing $B$, $C$ is included in $A$.
So $C$ is compact, we can apply the first result which shows that $E$ attains its minimum in $C$, and this point is in $A$.