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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?

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  • $\begingroup$ You probably don't mean orthonormal. $\endgroup$ – Alexander Thumm Oct 2 '12 at 9:06
  • $\begingroup$ @Alexander Thumm: Thank you for your notice. $\endgroup$ – blindman Oct 2 '12 at 9:23
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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 $\sum_i 2^{-i} \delta_{i,j} \not \in l^2_{\text{fin}}(\mathbb R)$.

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  • $\begingroup$ Dear Sir. I made a mistake in my question $$\displaystyle\sum_{n=1}^{\infty}\|x_n\|$$ should be replaced by $$\displaystyle\sum_{n=1}^{\infty}\|x_n\|^2$$ $\endgroup$ – blindman Oct 2 '12 at 15:08
  • $\begingroup$ The same argument works unaltered. $\endgroup$ – Alexander Thumm Oct 2 '12 at 16:23
  • $\begingroup$ Thank you for your nice solution. $\endgroup$ – blindman Oct 2 '12 at 22:47
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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 $\{x_n\}\subset X$ $$ \sum_{n=1}^\infty\|x_n\|<\infty\implies \sum_{n=1}^\infty x_n\text{ converges.} $$

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  • $\begingroup$ Dear Sir. Thank you for your comment. I revised my question. $\endgroup$ – blindman Oct 2 '12 at 9:18

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