2

Possible Duplicate:
Weak-to-weak continuous operator which is not norm-continuous

I saw in some text that if an operator $T:Y^\star \rightarrow X^\star$ (where $X$ and $Y$ are separable Banach spaces) is $w^\star - w^\star$ - continuous then $T$ is the adjoint of some operator $S:X\rightarrow Y$. Can anyone write a short proof for this claim or give some reference?

Community
  • 1
user25640
  • 1,574
  • 11
  • 21
  • 2
    See part b) of the answer to http://math.stackexchange.com/q/55679 –  Nov 06 '12 at 13:11
  • I don't think it's a duplicate. It is true that the answer is in that question, but will anybody who is looking for this find the answer? If I search the site using the title of this question as my search string, the answers mentioned in the comments do not appear, at least not in the first page. – Martin Argerami Nov 06 '12 at 19:05
  • @Martin: Anybody searching for this title will find this question, which has a perfectly good link to the previous one. – Chris Eagle Nov 07 '12 at 10:57

0 Answers0