Equivalence of weak convergences Let $X$ be a normed space. Let $(x_n)$ be a sequence in $X$. We say that $x_n\to x\in X$ weakly as $n\to \infty$ if $\ell(x_n)\to \ell(x)$ as $n\to \infty$ for all $\ell\in X^*$. 
I found a note, where it says in a remark (with no explanation) that, if $X$ were Hilbert space, then $x_n\to x$ weakly as $n\to \infty$ if and only if $\langle x_n,z\rangle\to \langle x,z\rangle$ as $n\to \infty$ for all $z\in X$. I'd like to know why this holds.
Assume $X$ is Hilbert space. If $\ell \in X^*$, then there exists a unique $z\in X$ such that $\ell(x)=\langle x,z\rangle$ for all $x\in H$, by the Riesz Representation Theorem. To prove the "if" direction, my thought goes like this: suppose $\langle x_n,z\rangle\to \langle x,z\rangle$ as $n\to \infty$ for all $z\in X$. Then it also holds for some $z\in X$, which implies $\ell(x_n)\to \ell(x)$ as $n\to \infty$. So the "if" direction makes sense. Why does "only if" direction hold? I thought about defining a map $\phi:X\to X^*$ defined by $(\phi(z))(x)=\langle x,z\rangle$ for all $x,z\in X$, but I am not sure about it.

Yes, it remains to show that $\phi(z)$ is actually a bounded map: $|\phi(z)(x)|=|\left<x,z\right>|\leq\|x\|\|z\|$, so $\|\phi(z)\|\leq\|z\|<\infty$, so $\phi(z)\in X^{\ast}$.
