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Thread: Weak Convergence

  1. #1
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    Weak Convergence

    Hi,

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

    My proof so far.

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

    $\displaystyle \|x_n-x\| \le $ something.

    Details missing. Then,

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

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

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

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

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

    Edit. The result is true in the case where $\displaystyle X$ is a Hilbert space. Is that what is intended here?
    Last edited by Opalg; Aug 13th 2010 at 10:30 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 $\displaystyle x_n \to x$ weakly AND $\displaystyle \|x_n\| \to \|x\|$, then $\displaystyle x_n \to x$ strongly. (i.e. $\displaystyle \|x_n-x\| \to 0)$.
    So suppose that we are working in a Hilbert space, with inner product $\displaystyle \langle.,.\rangle$. It follows from the weak convergence that $\displaystyle \langle x_n,x\rangle\to\langle x,x\rangle = \|x\|^2$. Then

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

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