I have the following problem. Prove that if weakly AND , then strongly. (i.e. .
My proof so far.
Since weakly . I also know that . I think it is useful to use this hat notation because I know that . So I'll start with
Details missing. Then,
I just need help filling in the middle details. I was thinking of adding and subtracting something, but I just can't figure it out.
The next part of the question says find a counterexample to show that weakly does not imply .
I am bad with counterexamples. I know I need to come up with a sequence that converges weakly, which means I need a function such that this sequence converges, but I cannot think of one. Thank you for any ideas.
I'm assuming that this is meant to take place in the context of a Banach space.
For the counterexample, take to be the space of sequences converging to zero, with the sup norm. (The dual space can be identified with , the space of all sequences with absolutely convergent sum.) Let be the sequence with a 1 in the n'th coordinate and zeros everywhere else. Then weakly but not strongly.
However, a similar example shows that the result claimed in the original problem is false. In fact, take to be the sequence with a 1 in the first and n'th coordinates and zeros everywhere else: if k=1 or n, and for all other values of k. Then weakly but not strongly, and .
Edit. The result is true in the case where is a Hilbert space. Is that what is intended here?
I had to go back and look at the problem again. Yes, it is a Hilbert space. I'm still having trouble filling in the middle details of the proof. Thanks.