I would like a quick proof verification for my attempt at a proof of the following theorem:

Let $H$ be a hilbert space, $x_n \to x$ $\iff$ ($\|x_n\| \to \|x\|$ and $x_n \rightharpoonup x$)

$(\Leftarrow)$ $\|x_n-x\|^2=\langle x_n -x, x_n - x \rangle = \|x_n\| + \|x\| - 2\langle x,x_n\rangle$


$$\lim_{n\to\infty} \|x_n-x\|^2 = \lim_{n\to\infty} (\|x_n\| + \|x\| - 2\langle x,x_n\rangle) =2\|x\| - 2\|x\| = 0$$

Hence $x_n \to x$.

$(\Rightarrow)$ By Riesz lemma we have a unique $f_y(x) = \langle y,x\rangle, \forall x\in H$.

$|f(x_n)-f(x)|=|f(x_n-x)|=|\langle y, x_n-x\rangle|\leq\|y\|\|x_n-x\|\to 0$

Which implies $\langle y ,x_n\rangle \to \langle y,x\rangle$ which is weak convergence $x_n\rightharpoonup x$.

Now $|\|x_n\| - \|x\| | \leq^{(*)} \|x_n-x\|\to 0$ (IS $(*)$ TRUE?)


Yes, $(*)$ is true, since (by the triangle inequality) $$\|x_n\| = \|x_n - x + x\| \le \|x_n - x\| + \|x\|$$ and, similarly, $$\|x\| \le \|x_n - x\| + \|x_n\|.$$ Together, you get $$|\|x_n\|-\|x\|| \le \|x_n - x\|.$$

The remained of your proof is fine.


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