Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question

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

share|cite|improve this answer
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.

share|cite|improve this answer
why -1 ? is it wrong please correct me! – Johan Jan 9 '13 at 9:06

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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.