How to show that this functional is convex

Question

Consider the following question (from a text in portuguese on Critical Point Theory I am reading):

Let $E$ be a Hilbert space, $a: E \times E \longrightarrow \Bbb{R}$ a continuous coercive bilinear form and $f \in E^*$. Given a closed, convex subset $C$ of $E$, show that there exists an unique $u_0 \in C$ such that $$ \psi(u_0) = \inf_C \psi(u), $$ where $$ \psi(u) = \frac12a(u, u) - f(u). $$

I can see that the functional $\psi$ is continuous (hence lower semicontinuous) and coercive, and that if we take the intersection of $C$ with any ball the resulting set is weakly compact (as a closed, bounded, convex set os a reflexive space), but this is not helping since lower semicontinuity does not imply weak lower semicontinuity.

Any hints or comments will be the most appreciated.

Thanks in advance and kind regards.

EDIT

I have found in Kavian's Introduction à la Théorie des Points Critiques that a convex lower semicontinuous functional on a convex closed set is weakly lower semicontinuous. This solves the problem, once I show that $\psi$ is convex. I would now appreciate help on this issue, with which I am having trouble.

Answer 1

I would show it more straight forward: Let $v_n \rightharpoonup v$ be a weakly convergent sequence. Then: $$ a(v_n,v_n)-a(v,v)=a(v_n-v,v_n-v)+a(v,v_n-v)+a(v_n-v,v) \geq $$ $$ a(v,v_n-v)+a(v_n-v,v) $$ The inequality comes from the the coervity of $a(u,u)$ i.e. $a(v_n-v,v_n-v) \geq c||v_n-v||_H^2\geq0$.
Now $a(v,v_n-v)$ defines a linear, continuous function if we fix the first component i.e. we consider $g(w)=a(v,w) \in H^*$ and note that $g(v_n-v)=a(v,v_n-v) \to 0 $ by the defintion of weak convergence. Similar, we argue that $a(v_n-v,v) \to 0$. Overall, we see that $$ \liminf_{n \to \infty}a(v_n,v_n) \geq a(v,v) $$ and you have weak, lower-semicontinuity. I hope this answers your question despite me not showing convexity of the associated bilinear form (I was thinking about using differential calculus in Hilbert spaces to do so). If you have any question, just leave a comment.

Answer 2

I think I got this. Lemme 15.4 in the first chapter of Kavian's book reads (translated):

Let $K$ be a convex set and $J: K \longrightarrow \Bbb{R}$ a Gateaux - differentiable function. Then the following are equivalent:

• $J$ is strictly convex

• For all $x, y \in K$, $x \neq y$, we have $J(y) > J(x) + J'(x)[y - x]$

• $J'$ is strictly monotone.

Now, of course $\psi$ is Gateaux - differentiable, since it is Fréchet - differentiable: $$ \psi'(u)[v] = \frac12 a(u, v) + \frac12a(v, u) - f(v) $$

We have that $\frac12 a(u-v, u-v) > 0$, by coercivity. Now, we note that $$ 0 < \frac12 a(u-v, u-v) = \frac12 a(u, u) - f(u) + f(u) - \frac12 a(u, v) - \frac12 a(v, u) + \frac12 a(v, v). $$ Hence, $$ \psi(u) = \frac12 a(u, u) - f(u) \\ > \frac12 a(u, v) + \frac12 a(v, u) - \frac12 a(v, v) - f(u) \\ = \frac12 (v, v) - f(v) + \frac12 a(v, u) - \frac12 a(v, v) + \frac12 a(u, v) - \frac12 a(v, v) - f(u) + f(v) \\ = \psi(v) - \psi'(v)[u-v] $$ and therefore $\psi$ is strictly convex. From this we can deduce that the point where the infimum $c$ is attained is unique: if there were two points where the infimum is attained, say $u$ and $v$, we would have $$ c \leq \psi\left(\frac{u + v}{2} \right) < \frac12 (\psi(u) + \psi(v)) = c, $$ absurd.