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Math Help - Weak Convergence

  1. #1
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    Weak Convergence

    Hi,

    I have the following problem. Prove that if x_n \to x weakly AND  \|x_n\| \to \|x\|, then  x_n \to x strongly. (i.e.  \|x_n-x\| \to 0).

    My proof so far.

    Since x_n \to x weakly  \Rightarrow f(x_n) \to f(x)  \forall f \in X^*. I also know that f(x_n) \to f(x)  \forall f \in X^*  \Rightarrow  \hat{x}_n(f) \to \hat{x}(f). I think it is useful to use this hat notation because I know that \|\hat{x}\|=\|x\|. So I'll start with

    \|x_n-x\| \le something.

    Details missing. Then,

    |\|\hat{x}_n\|-\|\hat{x}\||= |\|x_n\|-\|x\|| \to 0.

    Therefore, \|x_n-x\| \to 0.

    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  x_n \to x weakly does not imply \|x_n\| \to \|x\|.

    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.
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  2. #2
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    I'm assuming that this is meant to take place in the context of a Banach space.

    For the counterexample, take X to be the space c_0 of sequences converging to zero, with the sup norm. (The dual space can be identified with \ell^1, the space of all sequences with absolutely convergent sum.) Let x_n be the sequence with a 1 in the n'th coordinate and zeros everywhere else. Then x_n\to0 weakly but not strongly.

    However, a similar example shows that the result claimed in the original problem is false. In fact, take x_n to be the sequence with a 1 in the first and n'th coordinates and zeros everywhere else: x_n(k) = 1 if k=1 or n, and x_n(k) = 0 for all other values of k. Then x_n\to x_1 weakly but not strongly, and \|x_n\| = \|x_1\| = 1.

    Edit. The result is true in the case where X is a Hilbert space. Is that what is intended here?
    Last edited by Opalg; August 13th 2010 at 11:30 AM. Reason: afterthought
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  3. #3
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    Hi again,

    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.
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  4. #4
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    Quote Originally Posted by Benmath View Post
    Prove that if x_n \to x weakly AND  \|x_n\| \to \|x\|, then  x_n \to x strongly. (i.e.  \|x_n-x\| \to 0).
    So suppose that we are working in a Hilbert space, with inner product \langle.,.\rangle. It follows from the weak convergence that \langle x_n,x\rangle\to\langle x,x\rangle = \|x\|^2. Then

    \|x_n-x\|^2 = \langle x_n-x,x_n-x\rangle = \langle x_n,x_n\rangle - 2\text{re}\langle x_n,x\rangle + \langle x,x\rangle \to \|x\|^2-2\|x\|^2 + \|x\|^2=0.

    Quote Originally Posted by Benmath View Post
    The next part of the question says find a counterexample to show that  x_n \to x weakly does not imply \|x_n\| \to \|x\|.
    For the counterexample, you can take \{x_n\} to be the set of vectors in an orthonormal sequence. Then x_n\to0 weakly but not in norm.
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