Let $\mathcal{H}$ denote a Hilbert space of functions and let $L: \mathcal{H} \to \mathbb{R}$ denote a continuous (hence bounded) convex functional. I have been able to show that for some $D>0$
$$\inf_{||f||=D} L(f) > L(0), $$
and I have been wondering whether from this it follows that there exists a (possibly nonunique) minimizer of $L$ in the ball $\{f \in \mathcal{H}: ||f|| \leq D \}$. I know that this holds in nice Euclidean spaces but I also know that infinite-dimensional spaces sometimes defy intuition. Thank you for your help.
Best Answer
Yes, there is a (possibly non-unique) minimiser inside the unit ball.
Hilbert spaces belong to a class of spaces known as reflexive spaces, a condition equivalent to their closed unit ball being compact in the weak topology, the weakest topology that makes the continuous linear functionals in the dual of the space still continuous.
The Hahn-Banach separation theorem also shows us that bounded (i.e. are subsets of some closed ball) convex subsets of a Banach space are weakly closed. Since closed subsets of a compact topological space are compact, this implies that closed, bounded convex subsets of a Hilbert space are weakly compact.
Now, the lower level sets of a continuous convex functional are definitely closed and convex. The condition that $L(0)$ is less than the function values on the unit sphere implies that the lower level sets are bounded too, and hence they are all compact.
We can express the set of minimisers of $L$ as an intersection of nested weakly compact sets: $$\operatorname{argmin} L = \bigcap_{\varepsilon > 0} L^{-1}\left[\inf L, \inf L + \varepsilon\right],$$ which must be non-empty by Cantor's Intersection Theorem (compact version). It also must be a weakly compact, convex set too.
Hope that helps.