I am trying to understand if the following statement holds true:
Let $X$ be a finite dimensional Hilbert space (e.g. $X=\overline{\mathbb{R}}$). Let $I:X\to\mathbb{R}$ be a convex, coercive and lower semicontinuous functional. Let $A\subset X$ be a non empty and closed subset. Then $I$ achieves its minimum, i.e. $\bar{x}\in A$ exists such that $$I(\bar{x}) = \inf_{x\in A} I(x).$$
I know that in a finite dimensional Hilbert space, the continuity of the functional and the compactness of the set $A$ guarantee the existence of a minimizer for $I$.
In the case of the statement above I have lower semicontinuity and coerciveness which imply the boundedness from below of the functional (which clearly does not guarantee the existence of a minimizer).
I was wondering if coercivity + convexity of $I$ together with the only closedness of the set imply that the functional implies the existence of a minimizer. If yes, how to prove it?
In general, do you find that the previous statement contains some unnecessary conditions on $I$ and/or on $A$?
Finally, what about the assumption of convexity for the set $A$ which is required in the case of an infinite dimensional Hilbert space? It seems to disappear in the finite dimensional case (see e.g. Existence of minimizers in a finite dimensional Hilbert space).
I hope someone could help me in clarifying these doubts.
Thank you in advance.
Best Answer
If $X$ is a finite-dimensional Hilbert space, $I$ is lower semi-continuous, $A$ is closed, and $A$ is bounded or $I$ is coercive then you get the existence of a solution. Using the indicator function of $A$ defined by $$ \delta_A(x):= \begin{cases}0 & \text{ if } x\in A, \\ +\infty & \text{ if } x\not \in A \end{cases}. $$ Then $\delta_A$ is lower semicontinuous iff $A$ is closed, $\delta_A$ is coercive iff $A$ is bounded.
And the function $f: X\to \mathbb R\cup \{+\infty\}$ has a global minimum, if $f$ is coercive, lower semicontinuous, and proper (there is $x$ such that $f(x)<+\infty$).
Proof: Let $x_n$ be such that $\lim f(x_n) = \inf_{x\in X} f$. Since $f$ is proper, the infimum is $<+\infty$, since $f$ is coercive, $(x_n)$ is bounded and has a converging subsequence with limit $x$. Due to lower semicontinuity, the limit $x$ is the minimum of $f$.
In infinite-dimensional Hilbert spaces, boundedness does only imply the existence of a weakly converging subsequence. Hence, we need $f$ to be weakly lower semicontinuous. Usually this is achieved by assuming $f$ is lower semicontinuous and convex.