Net convergence and norm-convergence in Hilbert spaces

Question

Let $\mathcal H$ be a Hilbert space which is not necessarily separable. Given a sequence of element $h_n$ indexed by $\mathbb N$, we say their sum converges in norm if the sequence $\{\sum_{n=1}^k h_n \}$ converges in norm. This is the usual notion of a convergent sum in a Hilbert space.

However, there is also the concept of convergence as a net, which applies also to uncountable sums. Given some index set $I$ and elements $h_i$, $i\in I$, we let $\mathcal F$ be the collection of all finite subsets of $I$, ordered by inclusion. This makes $\mathcal F$ into a directed. set. For each $F\in \mathcal F$, we define

$$h_F = \sum\{ h_i : i\in F\}.$$

This is a finite sum, so it is an element of $\mathcal H$. The set of sums $h_F$ forms a net in $\mathcal H$. We say the sum over $I$ of the $h_i$ converges if the net converges. (See here for more information on nets.)

Given a set of elements of $\mathcal H$ indexed by $\mathbb N$, one naturally wonders about how net convergence and norm convergence relate.

I would like to prove the following.

If $\{ h_n\}$ is a sequence in Hilbert space and $\sum\{ h_n: n\in \mathbb N\}$ converges to $h$ as a net, then $h_n$ converges to $h$ in norm. The converse is false.

The problem and definitions are taken from chapter 1, section 4 of Conway's A Course in Functional Analysis, second edition.

Answer 1

If the sum converges to $h$ as a net, that means for every $\varepsilon > 0$, you have a finite set $F_\varepsilon \subset \mathbb{N}$ such that for every finite $F \supset F_\varepsilon$ you have

$$\left\lVert \sum_{n \in F} h_n - h\right\rVert < \varepsilon.$$

In particular, for all $N \geqslant \max F_\varepsilon$, you have

$$\left\lVert\sum_{n=1}^N h_n - h\right\rVert < \varepsilon,$$

that means $\sum h_n$ converges in norm.

Conversely, let $\mathcal{H} = \mathbb{R}$ and $h_n = \frac{(-1)^{n-1}}{n}$. It is well-known that the sequence of partial sums $s_k =\sum_{n=1}^k h_n$ converges in norm (to $\log 2$), but for every finite set $F_\varepsilon$, there exists a finite subset $F \supset F_\varepsilon$ with $$\sum_{n\in F} h_n > K$$ for every $K \in \mathbb{R}$.

Answer 2

Suppose that $\{h_n\}$ converges to $h$ as a net and fix $\epsilon>0$. Then by the definition of net convergence, there is some finite set $F\subset I$ such that $\|h_G-h\|<\epsilon$ for every $G\ge F$. If $F$ has maximal element $N$, then for every $k> N$, we have

$$\left \|\sum_1^k h_n - h\right \|<\epsilon.$$

This implies the sequence converges in norm.

To show the converse is false, we produce a counterexample. Let $\mathcal H=\mathbb R$ and let $\{h_n\}$ be the alternating harmonic series. It is well known this converges in norm with sum $\ln 2$. Consider the neighborhood $[\ln 2 -1,\ln 2+1]$. Suppose there exists $F$ such that $\|h_G-h\|<1$ for every $G\ge F$. By taking $F$ and adding a large number of odd indices to the index set, we can make $h_G$ grow arbitrarily large. This is a contradiction, and we have exhibited a set that converges in norm but not as a net.