Uniqueness of weak limit in Hilbert space

Question

Let $H$ be a Hilbert space and $\{x_n\}$ a sequence in $H$ and $y,z\in H$.

Suppose that $x_n\rightarrow y$ and $x_n \rightarrow z$ weakly.

Now let $\varepsilon > 0$, and $w\in H$. Then there exists $N>0$ such that for all $n > N$, $$|\langle x_n, w\rangle - \langle y, w\rangle|<\frac{\varepsilon}{2}$$ and $$|\langle x_n, w\rangle - \langle z, w\rangle|<\frac{\varepsilon}{2}.$$

So take $n > N$. Then we have $$|\langle y-z, w\rangle| = |\langle y - x_n + x_n - z, w\rangle| \leq |\langle y - x_n,w\rangle| + |\langle x_n - z, w\rangle| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$

So since $\varepsilon$ was arbitrary we may conclude that $\langle y-z, w\rangle = 0$. But then since $w$ was arbitrary, we may conclude that $y=z$. Therefore weak limits are unique.

Does this proof work?

Answer 1

Here is a much simpler version: $$\begin{split} \|x-z\|^2 &= \langle x-z, x-z\rangle =\langle x,x-z\rangle - \langle z,x-z\rangle\\&= \lim_{n\to\infty}\langle x_n, x-z\rangle- \lim_{n\to\infty}\langle x_n, x-z\rangle = \lim_{n\to\infty}\langle x_n-x_n,x-z\rangle =0. \end{split} $$ (the main idea is to use the test element $w:=x-z$)