Weak Convergence

**Benmath** · August 12th 2010, 05:29 PM

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.

**Opalg** · August 13th 2010, 08:47 AM

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?

**Benmath** · August 15th 2010, 06:29 PM

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.

**Opalg** · August 15th 2010, 11:55 PM

Originally Posted by Benmath

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

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

Originally Posted by Benmath

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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