TITLE: Proof that $\lim_{n\to\infty}\|x_n-x\|=0$ with weak convergence
QUESTION [5 upvotes]: I want to prove that in a normed linear space $H$, that if $x_n$ is weak convergent to $x$, and $\lim_{n\to\infty} \|x_n\| = \|x\|$ then:
$$\lim_{n\to\infty}\|x_n-x\|=0$$

Can I please have a hint? Also does $\langle x,x\rangle^{\frac12}=\|x\|$ or something?

The impression I have is, essentially $\lim_{n\to\infty}\|x_n\|-\|x\|=\lim_{n\to\infty}\|x_n - x\|$ due to the weak convergence, otherwise in the general case:
$$\|x\|-\|y\|=\|x - y\|$$ could be true, but only if they are in opposite directions.

My definition of weak convergence is $\lim_{n\to\infty} \langle x_n,y\rangle =\langle x,y\rangle$, so I suspect that this is a hilbert space.

REPLY [8 votes]: This property is called "(H)" property and the space is called a "Radon Riesz" space.
As example of this space we can say every Hilbert space as the proof given by @bartgol and $L^p$ for $1<p<\infty$, and schur space (in particular $\ell^1$).