# 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!

Hint: Suppose $x_n$ is a Cauchy sequence in $X$.
Consider the linear functional $L(x) = \lim_n \langle x, x_n \rangle$.

• How can I turn this into a Cauchy sequence in $\mathcal C$ or $\mathcal R$? That is the difficulty here.Can you please explain me further? – ccc Jan 7 '13 at 1:35
• You don't "turn this into a Cauchy sequence in $\mathbb C$ or $\mathbb R$". Can you prove $L(x)$ exists, and that $L$ is a bounded linear functional? If so, take $z$ so that $L(x) = \langle x, z \rangle$. The limit of the Cauchy sequence $x_n$ is going to be $z$. Can you estimate $\|x_n - z\|$? – Robert Israel Jan 7 '13 at 2:27
• @ robert,still its difficult for me to organize the answer for this.Can you please help me? – ccc Jan 7 '13 at 3:44
• Which part are you stuck on? – Robert Israel Jan 7 '13 at 7:19
• The basic idea here is to pic a sequence in $X$ and show that the limit also in $X$.So how can I prove that $lim_n\langle x,x_n\rangle=\langle x,z\rangle$.How to estimate $∥x_n−z∥$? – ccc Jan 7 '13 at 7:53

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.

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?