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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.

JohnK
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1 Answers1

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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.

user764828
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  • Thanks so much. Could I also ask whether the claim still holds if I replace $>$ with a weak inequality in $\inf_{||f||=D} L(f) > L(0)$? – JohnK Mar 29 '20 at 14:01
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    The argument doesn't work, since the lower level sets may not be bounded, which stops them from being weakly compact. If there was a point $x$ (possibly other than $0$) inside the open unit ball such that $L(x) < \inf_{|f| = D} L(f)$, then the argument proceeds as before. Otherwise, this infimum is equal to the infimum over the whole closed disk. If there is a point $y$ such that $L(y)$ is less than this infimum, it lies outside the unit disk. A standard convexity argument shows that $L$ must take a strictly smaller value on some point on the sphere, a contradiction. – user764828 Mar 29 '20 at 14:09
  • Thanks again, your answer provides the background that I was looking for. – JohnK Mar 29 '20 at 14:11
  • I think you meant to say in the third paragraph that the lower level sets are closed and convex, not compact. Also how does the condition imply that the level sets are bounded? – JohnK Mar 29 '20 at 20:07
  • @JohnK Yes, thank you. – user764828 Mar 29 '20 at 23:40
  • Could you please explain why the level sets are bounded? – JohnK Mar 30 '20 at 07:57
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    @JohnK Well, all we really need is one lower level set $L^{-1}[\inf L, \inf L + \varepsilon]$ to be bounded, since the lower level sets are nested. Since $L(0) < \inf_{|f| = D} L(f)$, this means that $\varepsilon = \inf_{|f| = D} L(f) - \inf L > 0$, and for such $\varepsilon$, $L^{-1}[\inf L, \inf L + \varepsilon]$ must be contained in the closed ball radius $D$, as the convexity of the lower level sets would provide a contradiction otherwise. – user764828 Mar 30 '20 at 09:16
  • Is the infimum taken over the unit ball or the whole space? Sorry for being dense. – JohnK Mar 30 '20 at 09:22
  • @JohnK $\inf L$ refers to the infimum over the whole space. – user764828 Mar 30 '20 at 10:28