If $\sum_{s \in S} \|h_s\| < \infty$ then $\sum_{s \in S} h_s$ converges.

Question

Let $H$ be a Hilbert space and $\{h_s: s \in S\}$ be a set of vectors in $H$. Assume that $\sum_{s \in S}\|h_s\| < \infty$. Can we conclude that $\sum_{s \in S} h_s$ converges in $H$ in norm?

Here, the summations are to be understood in the generalised summation sense. I.e. as the limit of the net of finite partial sums where the finite subsets are ordered via inclusion. This is important because this is not equivalent with the usual kind of convergence in the case $S= \mathbb{N}$.

Attempt:

We will show that $\{\sum_{s \in F} h_s\}$ is a Cauchynet in $H$, where $F$ ranges over all finite subsets of $S$. By completeness of $H$, we will be done.

Given finite subsets $F,G$ of $S$, we estimate using the parallellogram identity \begin{align*}&\|\sum_{s \in F} h_s- \sum_{s \in G} h_s\|^2 \\ &= 2\|\sum_{s \in F} h_s\|^2 + 2\|\sum_{s \in G} h_s\|^2 - \|\sum_{s \in F} h_s + \sum_{s \in G} h_s\|^2\\ & \le 2 (\sum_{s \in F}\|h_s\|)^2 + 2(\sum_{s \in G}\|h_s\|)^2 \end{align*}

but this doesn't become arbitrarily small.

Answer 1

Let $X$ be a complete normed space (no need for the inner product structure). Let $S$ be a set and $\{x_s\}_{s\in S}$ a collection of vectors in $X$ such that $\sum_{s\in S}\|x_s\|<\infty$. Then $\sum_{s\in S}x_s$ converges in $X$.

By definition, $\sum_{s\in S}\|x_s\|<\infty$ means that $\sup_{F\subset_{\text{fin}}S}\sum_{s\in F}\|x_s\|:=\sigma<\infty$. What we need to show is that the net $\bigg\{\sum_{s\in F}x_s\bigg\}_{F\subset_{\text{fin}}S}$ converges in $X$. Since $X$ is complete, it suffices to show that this net is Cauchy.

Let $\varepsilon>0$. We can find a finite set $F_0\subset S$ such that $$\sigma-\varepsilon<\sum_{s\in F_0}\|x_s\|<\sigma$$ Now let $F_1,F_2$ be finite subsets of $S$ such that $F_1\geq F_0$ and $F_2\geq F_0$ in the directed set of finite subsets, i.e. $F_0\subset F_1$ and $F_0\subset F_2$, Then (in the next equations, $k_s=0$ if $s\in F_1\setminus F_2$ and $k_s=1$ if $s\in F_2\setminus F_1$).

$$\bigg\|\sum_{s\in F_1}x_s-\sum_{s\in F_2}x_s\bigg\|=\bigg\|\sum_{s\in F_1\triangle F_2}(-1)^{k_s}x_s\bigg\|\leq\sum_{s\in F_1\triangle F_2}\|(-1)^{k_s}x_s\|$$ $$=\sum_{s\in F_1\triangle F_2}\|x_s\|=\sum_{s\in F_1\setminus F_2}\|x_s\|+\sum_{s\in F_2\setminus F_1}\|x_s\|\leq$$ $$\leq\sum_{s\in F_1\setminus F_0}\|x_s\|+\sum_{s\in F_2\setminus F_0}\|x_s\|=\bigg(\sum_{s\in F_1}\|x_s\|-\sum_{s\in F_0}\|x_s\|\bigg)+\bigg(\sum_{s\in F_2}\|x_s\|-\sum_{s\in F_0}\|x_s\|\bigg)<$$ $$<2\bigg(\sigma-\sum_{s\in F_0}\|x_s\|\bigg)<2\varepsilon.$$

As $\varepsilon$ was arbitrary, this proves the net $\{\sum_{s\in F}x_s\}_{F\subset_{\text{fin}}S}$ is Cauchy, thus convergent.

Answer 2

Let me attempt to give an answer.

Let $K := \{s \in S: h_s \ne 0\}$. Then $K$ is at most countable, and $\sum_{s \in S} h_s$ converges if and only if $\sum_{s \in K} h_s$ converges. Hence, we may assume without loss of generality that $S$ is countably infinite (if $S$ is finite, there is nothing to prove).

Now, we use the following lemma:

Lemma: If $S$ is countably infinite, then $\sum_{s \in S} h_s$ converges if and only if $\sum_{n=0}^\infty h_{g(n)}$ converges (in the usual sense) for every bijection $g: \mathbb{N}\to S$.

The proof (in the case of $S = \mathbb{N}$) can be found in:

T.H. Hildebrandt, On unconditional convergence in normed vector spaces, Bull. Amer. Math. Soc. 46 (12) (1940), 959–962, MR0003448.

Let us now see how this helps. Let $g: \mathbb{N}\to S$ be a bijection. Then by the lemma $$\sum_{n=0}^\infty \|h_{g(n)}\| < \infty$$ and thus because the result trivially holds for ordered sums (this is well-known, and an easy consequence of the triangle inequality), we find that $$\sum_{n=0}^\infty h_{g(n)}$$ converges in $H$. By the lemma, it follows that the unordered sum $\sum_{s \in S} h_s$ converges.