1
$\begingroup$

I was asking myself the following question: let $X$ be a Banach space, dual of a separable Banach space, if $I: X \to R$ is a convex weak lower semicontinous functional, does it follow that $I$ is weak-* lower semicontinous?

If not, does the claim is true if we consider sequential lower semicontinuity?

Because in Dacorogna's book on direct methods of calculus of variations there is a proof of weak-* sequential lower semicontinuity in $W^{1,\infty}$ of a particular functional which goes like: the functional is strong l.s.c. thus weak-* seq. L.s.c, applying the general theorem on Banach spaces which states that strong l.s.c. implies WEAK l.s.c.

Now I don't see how to say it easy for general spaces, essentially because the proof of strong l.s.c. implies weak l.s.c. is based on the fact that in every locally convex t.v.s. strongly closed convex sets are weakly closed. Any ideas?

$\endgroup$
  • $\begingroup$ If this was true, wouldn't it imply that every (real) bounded functional on $X $ is given by an element of the "predual"? At least this should follow if we consider weak-$\ast $-lower semicontinuity instead of weak-$\ast $-sequential lower semicontinuity. $\endgroup$ – PhoemueX May 2 '17 at 20:53
1
$\begingroup$

The proof of the "general theorem" uses the following arguments:

  • Since $I$ is convex, its epigraph is convex.
  • Since $I$ is l.s.c., the epigraph is closed.
  • Closed and convex sets are weakly closed.
  • Hence, the epigraph is weakly closed and $I$ is weakly l.s.c.

The third step does not work with the weak-* topology, there are closed and convex sets which are not weak-* closed.

$\endgroup$
  • $\begingroup$ That was exactly what i meant $\endgroup$ – jJjjJ May 3 '17 at 6:45

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.