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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?

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

  • 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 $J(x) = j$.
  • Conclude.
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  • $\begingroup$ Thanks! It is the classical approach: construct a minimizing sequence. So it is a generalized version of the extreme value theorem. $\endgroup$ – dknew Jan 22 '16 at 9:28
  • $\begingroup$ 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? $\endgroup$ – dknew Jan 25 '16 at 15:09
  • $\begingroup$ 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 $j \in \mathbb{R}$. $\endgroup$ – gerw Jan 26 '16 at 8:14
  • $\begingroup$ Thanks. OT: Do you know why the website is not parsing the latex in my comment? $\endgroup$ – dknew Jan 26 '16 at 11:49
  • $\begingroup$ I think the problem is the double index: you have to write x_{n_k} instead of x_n_k. $\endgroup$ – gerw Jan 26 '16 at 12:30

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