# Weakly lower semicontinuous functional on a bounded closed and convex set

Let $J$ be a sequentially weakly lower semicontinuous functional on $C$ with values on the real line. Moreover let $C$ be a bounded, closed and convex subset of a Hilbert space $H$.

Is it true that the functional attains his minimum?

• On which space does $J$ act? What is $C$? But I suspect the answer is yes. – gerw Jan 21 '16 at 19:03
• Thanks. I didn't notice I have made mess with a copy and paste. – dknew Jan 22 '16 at 9:04
• $C$ is weakly compact. Forgetting everything else, we have a lower semicontinuous function on a compact set. – Daniel Fischer Jan 22 '16 at 9:07
• @DanielFischer: $J$ is only sequentially weakly lower semicontinuous, but this doesn't matter, since $C$ is also sequentially weakly compact. – gerw Jan 22 '16 at 9:14
• @gerw I overlooked that word, thanks. – Daniel Fischer Jan 22 '16 at 9:17

• Define $j = \inf_{x \in C}J(x)$
• Take a sequence $\{x_n\} \in C$ with $J(x_n) \to j$.
• Find a weakly convergent subsequence of $\{x_n\}$ with limit $x \in C$.
• Proof $J(x) = j$.
• The proof is something like: $$\inf J(x) \leq J(x*) \leq \lim \inf_k J(x_n_k) = \lim_n J(x_n) = \inf J(x)$$ But if the infimum is infinity the sequence $J(x_n_k)$ is not convergent and the last equality is not true. Am I wrong? – dknew Jan 25 '16 at 15:09
• In case $j = +\infty$, you have $C = \emptyset$, which is not interesting. In case $j = -\infty$, you get a contradiction with $J(x) \le \liminf_{k} J(x_{n_k}) = j$. Hence $j \in \mathbb{R}$. – gerw Jan 26 '16 at 8:14
• I think the problem is the double index: you have to write x_{n_k} instead of x_n_k. – gerw Jan 26 '16 at 12:30