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Let $\{x_n\}$ be a sequence in a Hilbert space $H$. If $\sum_n \|x\|< \infty$, how to show that $\sum x_n$ is convergent in $H$?

There is no doubt that $x_n \rightarrow 0$ as $n \rightarrow \infty$ (right?) since we have that \begin{align*} \sum_n \|x\|< \infty &\iff \|x_n\|\rightarrow 0 \text{ as } n \rightarrow \infty \\ &\iff \langle x_n,x_n\rangle \rightarrow 0 \text{ as } n \rightarrow \infty \\ &\iff x_n \rightarrow 0 \text{ as } n\rightarrow \infty \end{align*}

(it is a property of the inner product that $\langle x,x\rangle=0 \implies x=0$)

What to do next? Can I use the completeness of $H$ somehow?

Michael Hardy
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Ingvar
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3 Answers3

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The series $\sum_n ||x_n||$ is convergent if and only if the partial sum $\sum_{k=1}^n||x_k||$ is a cauchy sequence hence $$\forall \epsilon >0\ \exists\ n_0\quad \forall p\ge q\ge n_0,\ ||\sum_{k=p}^q x_k||\le\sum_{k=p}^q||x_k||<\epsilon $$ and hence $\sum_{k=1}^n x_k$ is a Cauchy sequence in a complete space. Conclude.

2

Try to show that $y_k = \sum_{i=1}^k x_i$ is a Cauchy sequence.

gerw
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Yes, you can use completeness; a normed vector space X is complete iff every absolutely-summable --meaning $\Sigma ||x_n||$ converges-- is summable, i.e., if $\Sigma x_n$ converges in norm, i.e., if there is an $x$ in $X$ with $||x- \Sigma x_i||\rightarrow 0$. Basically, if $\Sigma ||x_n||< \infty$ , there is a pos. integer N so that $\Sigma _{n=N}^{\infty}||x_n||< \epsilon$. Then, for $j,k > N $, we have, for the partial sums $S_n, S_m$, that $$ ||S_n-Sm||=||\Sigma_ {x=m}^{n}x_k||\leq \Sigma_{k=m}^{\infty} ||x_k||< \epsilon$$ , so you have a Cauchy sum in $X$ , which then converges by completeness of $X$.

user99680
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