Functional analysis-Hilbert spaces Let $ X$ be an inner product space.
Show that $ X$ is a Hilbert space if and only if for each continuous linear functional $ L$ on $ X$,there exists $ z\in X$ such that $ L(x)=\langle x,z\rangle $ .
Here,one part is exactly the Riesz Representation Theorem.
How can I prove the converse result?That is, If for each continuous linear functional $ L$ on $ X$,there exists $ z\in X$ such that $ L(x)=\langle x,z\rangle $ then $ X $ is a Hilbert space.Any Help is appreciated.
Thanks!
 A: Hint: Suppose $x_n$ is a Cauchy sequence in $X$.
Consider the linear functional $L(x) = \lim_n \langle x, x_n \rangle$.
A: by the hints we know that $<x,x_n>$ is a cauchy sequence in $\mathbb{K}$.
Hence it converges. We know that $L(x) = <x,z> \rightarrow <x,z> = \lim_n <x,x_n>$.
By choosing $L(x_n-z)$ we get $x_n = z$. So every Cauchy sequence in $X$ converges so $X$ is a Hilbert space.
A: Note that the limit $\langle x,x_n \rangle $, converge because 
$$ | \langle x,x_n \rangle- \langle x,x_m \rangle |= |\langle x,x_n-x_m \rangle|\leq\|x\|\|x_n-x_m\| $$
Then $x_n$ cauchy implies that  $\langle x,x_n \rangle $ cauchy( in $\mathbb R$ or $\mathbb C$) 
How above define a continuous linear functional $L(x)=\lim\limits_{n } \langle x,x_n \rangle $.
By hypothesis there is a $z\in X$ such that $L(x)=\langle x,z \rangle $. Even we know in advance that X is Hilbert then we will conclude only that $x_n$ converge to $z$ weakly.
This is not a answer of course. Is a comment that does not fit in the place of comment. 
@RobertIsrael,@ccc Can you give another hint?
