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My main question is (I give details about my reflexion below):

If $E$ is a Banach space which is the dual space of a separable space, and $f:E\to\mathbb{R}$ is convex, lower semi-continuous and coercive (for its usual norm), is this true that there exists a minimizer to $f$ over $E$?

Now I explain what I understand about the problem. I know two results:

1/ If $E$ is a Banach space and $f$ is $\alpha$-convex (meaning $f-\alpha\|\cdot\|^2$ is convex) and lower-semi-continuous, then it has a minimum

2/ If $E$ is a reflexive Banach space and $f$ is convex (or simply quasi-convex) and lower-semi-continuous, then it has a minimum

Proof of 1/ only uses Cauchy criterion in the manner of the proof of projection theorem in Hilbert spaces (though there is an extra hard part here which is to prove that $f$ is above a parabola, which uses Hahn-Banach theorem as a first step; but in many cases (including projection) this part is trivial)

Proof of 2/ (which matters more to me here) uses weak convergence in $E$ (testing convergence with every $l\in E^*$), weak sequential compactness of bounded sets in reflexive space, and that convex closed sets are weakly closed (and therefore convex lower-semi-continuous functions are weakly lower-semicontinuous)

For my initial problem, I shall use weak* convergence instead of weak convergence, as $E$ is the dual of a space (meaning testing convergence with every $x\in F$ where $E=F^*$). I can handle easily weak* sequential compactness (sequential version of Banach Alaoglu Theorem, thanks to separability of $E$).

But it remains to question (sequential) weak* lower-semi-continuity of convex functions. I read (for example in a the book of Brézis [Functional Analysis and PDE]) that closed convex sets aren't necessarily weakly* closed. Which seems to indicate that the answer to my question should be "no". But in the book they only refer to topological closedness, which isn't necessarily the same as sequentially closed; and more importantly, I'd rather have a convincing counter-example that just a theoretical reason. I'd like to get a feeling of what wrong can happen here regarding existence/non-existence of a minimizer.

I am happy if anyone could give a good reference on the topic.

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Here is an example showing that the answer is no in general.

Let us consider $X = \ell^\infty$ and let $\varphi : X \to \mathbb R$ be a Banach limit. In particular, $\varphi$ is continuous, but not weak-$\star$ continuous.

Now, we define the nonlinear functional $f : X \to \mathbb R$ via $$f(x) := \sum_{n=1}^\infty \frac{x_n}{n^2} + \max(\varphi(-x),0) + 4 \, \max(\|x\|-1,0).$$ It is easy to check that this functional meets your requirements. Moreover, for any $x \in \ell^\infty$ we have \begin{align*} f(x) &\ge -\frac{\pi}{6}\,\|x\| + 0 + 4 \, \max(\|x\|-1,0)\\ &\ge - \frac{\pi}{6} \end{align*} and the last inequality holds by equality iff $\|x\|=1$. This shows that the infimum of $f$ is at least $-\pi^2/6$.

By considering sequences of the form $x = (1,\ldots,1,0,\ldots)$, we can check that the infimum is really $-\pi^2/6$.

However, for $x \in X$ with $f(x) = -\pi^2/6$ we have $\|x\| = 1$ and, consequently, $$\sum_{n=1}^\infty \frac{x_n}{n^2} = -\pi^2/6$$ which implies $x_n \equiv -1$. But then, $$\varphi(-x) = 1$$ and this yields the contradiction $f(x) = 1-\pi^2/6$.

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  • $\begingroup$ Thanks gerw. This is perfect. I didn't know Banach limits were classical counterexamples for non weak* continuous functionals (which should have been the first thing to look at). From there it seems I understand how you built your counterexample. Also, it is clear that there is no issue of compactness (as the minimizing sequence converges) but a default of (ls)continuity. Well, thanks; I looked in many books of optimization, but none of them were referring precisely to that question (though as I mentionned, they gave hints that about the answer). $\endgroup$ – John Steinbeck Oct 2 '17 at 12:25

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