Is the minimizer of a strictly convex functional on a reflexive space strongly exposed?

Question

Bonjour,

I would like to understand when the minimizer $x^*$ of a strictly convex functional $F:X\rightarrow \mathbb{R}$ is strongly exposed, i.e. $$ F(x_n)\rightarrow F(x^*)=\min_{x\in X}F(x) \Longrightarrow \Vert x_n-x^*\Vert\rightarrow0 $$

In particular $X$ is a Banach reflexive space ( $W_0^{1,p}(\Omega)$ for instance) and the functional is lower semi continuous, strictly convex and coercive.

Under these hypothesis we have existence and uniqueness of minimizers, but what I would like to prove is that the minimizer is strongly exposed.

I know that if I could prove that $x^*$ is a denting point for $F$ I'd have the result but, truth is that to show the denting property is very similar to prove the statement. Is there any general result or I should rely on the specific form of my functional?

My functional is in the form $F(\phi)=\int_{\Omega} G(x,\phi(x))dx$ with $G:\Omega\times\mathbb R\rightarrow \mathbb R$ smooth, strictly convex and $$ \frac{\vert a\vert^p}{C}-C\leq G(x,a)\leq C{\vert a\vert^p}+C $$

Thank you very much!

Answer 1

I do not have a full answer, but some hints which might be helpful.

Let me sketch a possible proof for the exposedness of $x^*$.

Let $\{x_n\}$ be a sequence with $F(x_n) \to F(x^*)$. At first, you try to show the boundedness of $\{x_n\}$ in $X$ (depending on your definition of coercivity, this might follow already from coercivity). Since $X$ is a Banach space, there is a weakly convergent subsequence (without relabeling - we care about this later) with $x_n \rightharpoonup x^+$. Since $F$ is weakly lower semicontinuous, we infer $F(x^+) \le F(x^*)$, hence $x^+ = x^*$.

It remains to prove the strong convergence $x_n \to x^*$. For this, we assume that $X$ is a strongly convex Banach space (e.g., a Hilbert space, or $W^{1,p}(\Omega)$, $L^p(\Omega)$ with $p \in (1,\infty)$), hence $x_n \rightharpoonup x^*$ together with $\|x_n\| \to \|x^*\|$ implies $x_n \to x^*$.

For this, you have to use the special structure of $F$. As an example, I will discuss the case $F(x) = \|x\|^p + G(x)$ for some $p \in (1,\infty)$ and $G : X \to \mathbb{R}$ is assumed to be weakly continuous. We have $$ F(x^*) = \|x^*\|^p + G(x^*) \le \liminf_{n \to \infty} \|x_n\|^p + \lim_{n \to \infty} G(x_n) \le \lim_{n \to \infty} F(x_n) = F(x^*). $$ Hence, all inequalities can be replaced by equalities and, in particular, we have $\|x_n\| \to \|x^*\|$. Together with the strong convexity of $F$, we infer $x_n \to x^*$.

In your case, you might be able to employ the growth condition on $G$ in order to write down a similar argument.

Finally, we mention that the convergence of the entire sequence follows from a standard subsequence-subsequence argument.