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?

  • $\begingroup$ 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

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.
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.