closed strong vs. closed in weak convergence

Question

Studying a chapter about weak topologies and weak convergence I though the following which I have no idea how to prove or disprove it. So here it is:

Question: Does there exist Banach space $X$ and a closed subset $F \subset X$ such that $F$ is not closed with respect to weak convergence?

Any ideas?

Norbert · Accepted Answer · 2014-01-14T13:43:17.213

4

Yes, if $X$ is infinite dimensional. For example $$ \operatorname{cl}_{\sigma(X,X^*)}(S_X)=B_X $$ See exercisre 3.46 in Banach space theory. The basis for linear and non-linear analysis. M. Fabian, P. Habala, P. Hajek, V. Montesinos, V. Zizler.

If $X$ is finite dimensional, then weak and strong toplogy coincide. See proposition 3.13 in the same book. Thus closedness in both topologies is the same thing.

edited Jan 14 '14 at 13:43

answered Jan 14 '14 at 13:26

Norbert

58,398

Thank you very much for your reply! THank you also for the reference of the book! – passenger Jan 14 '14 at 13:45
You are wellcome! – Norbert Jan 14 '14 at 13:45
1

@passenger You can find a proof of the first fact here at MSE. – David Mitra Jan 14 '14 at 14:30
@DavidMitra: Yes I found it there! Thank you for your comment-reply! – passenger Jan 14 '14 at 14:43

closed strong vs. closed in weak convergence

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