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.