Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is known that "If $\{x_n\}$ is a sequence in a real Hilbert space $H$ satisfying $$ \langle x_n, x_m\rangle =0 \quad\forall n\ne m, $$ then $\displaystyle\sum_{n=1}^{\infty}x_n$ is convergent if and only if $\displaystyle\sum_{n=1}^{\infty}\|x_n\|^2$ is convergent". I would like to know if $H$ is an inner product space (pre-Hilbert space) the above statement is still true?

share|improve this question
    
You probably don't mean ortho**normal**. –  Alexander Thumm Oct 2 '12 at 9:06
    
@Alexander Thumm: Thank you for your notice. –  blindman Oct 2 '12 at 9:23
add comment

2 Answers

up vote 2 down vote accepted

No. Take the subspace $l^2_{\text{fin}}(\mathbb R)$ of $l^2(\mathbb R)$ of all real-valued sequences with only finitely many nonzero entries. $l^2_{\text{fin}}(\mathbb R)$ is a pre-Hilbert space with the induced inner product and the sequences $\delta_{i,j}$ whose only nonzero entry is $\delta_{i,i} = 1$ form an orthonormal system $\{ \delta_{i,j}\} \in l^2_{\text{fin}}(\mathbb R)$ but obviously $\sum_i 2^{-i} \delta_{i,j} \not \in l^2_{\text{fin}}(\mathbb R)$.

share|improve this answer
    
Dear Sir. I made a mistake in my question $$\displaystyle\sum_{n=1}^{\infty}\|x_n\|$$ should be replaced by $$\displaystyle\sum_{n=1}^{\infty}\|x_n\|^2$$ –  blindman Oct 2 '12 at 15:08
    
The same argument works unaltered. –  Alexander Thumm Oct 2 '12 at 16:23
    
Thank you for your nice solution. –  blindman Oct 2 '12 at 22:47
add comment

If $\{x_n\}$ is an orthonormal system then $\|x_n\|=1$ for all $n$, so that $\sum\|x_n\|$ does not converge.

I think that what you want is the following result:

Let $X$ be a normed space with norm $\|\,\cdot\,\|$ (in particular it can be a pre-Hilbert space.) $X$ is complete if and only if for every $\{x_n\}\subset X$ $$ \sum_{n=1}^\infty\|x_n\|<\infty\implies \sum_{n=1}^\infty x_n\text{ converges.} $$

share|improve this answer
    
Dear Sir. Thank you for your comment. I revised my question. –  blindman Oct 2 '12 at 9:18
add comment

Your Answer

 
discard

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.