If $E$ is a normed $ℝ$-vector space, are we able to show that $(x_i)_{i∈I}⊆E$ is summable $⇔$ $\left(\left\|x_i\right\|_E\right)_{i∈I}$ is summable?

Question

Let

• $I\ne\emptyset$ be a set
• $E$ be a normed $\mathbb R$-vector space

$(x_i)_{i\in I}\subseteq E$ is called summable with sum $x\in E$ $:\Leftrightarrow$ $$\forall\varepsilon>0:\exists J\subseteq I\text{ with }|J|<\infty:\forall K\subseteq I\text{ with }|K|<\infty\text{ and }J\subseteq K:\left\|x-\sum_{k\in K}x_k\right\|_E<\varepsilon\;.\tag1$$ In that case, $$\sum_{i\in I}x_i:=x\;.$$ It's easy to show that $(x_i)_{i\in I}\subseteq[0,\infty)$ is summable $\Leftrightarrow$ $$x:=\sup_{J\subseteq I\::\:|J|<\infty}\sum_{j\in J}x_j<\infty\;.\tag2$$ In either case, $$\sum_{i\in I}x_i=x\;.\tag3$$

Are we able to show that $(x_i)_{i\in I}\subseteq E$ is summable $\Leftrightarrow$ $\left(\left\|x_i\right\|_E\right)_{i\in I}$ is summable?

I know how this can be proven in the case $E=\mathbb R$, but that proof doesn't have a natural generalization for arbitrary $E$.

Answer 1

If $E$ is finite-dimensional, then the result follows from $E = \Bbb{R}$ case.

In general, we only have $(\Leftarrow)$ direction even when $E$ is Banach. (This is the Weierstrass $M$-test.) Consider the space $E = c_0(\Bbb{N})$ of real-valued sequences which converge to $0$. This is a Banach space with respect to the supremum norm

$$\mathbf{a} = (a_n)_{n\in\Bbb{N}} \quad \Rightarrow \quad \| \mathbf{a} \|_{\sup} = \sup_{n\in\Bbb{N}} |a_n|. $$

Now let $\mathbf{a}_k = (a_{k,n})_{n\in\Bbb{N}}$ by

$$ a_{k,n} = \begin{cases} \frac{1}{k}, & k = n \\ 0, & \text{otherwise} \end{cases} $$

For any $\epsilon > 0$, we pick $J = \{1, \cdots, N\}$ with $N > \epsilon^{-1}$. Then for any finite subsets $K, L \subseteq \Bbb{N}$ with $K, L \supseteq J$, we have

$$ \left\| \sum_{k \in K} \mathbf{a}_k - \sum_{l \in L} \mathbf{a}_l \right\|_{\sup} \leq \frac{1}{N} < \epsilon. $$

This tells that the net $\{ \sum_{j \in J} \mathbf{a}_j : J \subseteq \Bbb{N}, |J| < \infty \}$ is Cauchy and hence $\sum_{j \in J} \mathbf{a}_j $ converges in $c_0(\Bbb{N})$. On the other hand, we have $\sum_{k\in\Bbb{N}} \|\mathbf{a}_k\|_{\sup} = \infty$.

Answer 2

Sangchul Lee has shown that we cannot prove the equivalence. However, if $(x_n)_{n\in\mathbb N}\subseteq E$ with $\sum_{n=1}^\infty\left\|x_n\right\|_E<\infty$, then $(x_n)_{n\in\mathbb N}$ is summable as long as $E$ is complete.

Since $\sum_{n=1}^\infty\left\|x_n\right\|_E<\infty$, there is some increasing $(N_k)_{k\in\mathbb N}\subseteq\mathbb N$ with $$\sum_{n=N+1}^\infty\left\|x_n\right\|_E<\frac1k\;\;\;\text{for all }N\ge N_k\text{ and }k\in\mathbb N\;.$$ It's easy to see that $$s_k:=\sum_{n=1}^{N_k}x_n\;\;\;\text{for }k\in\mathbb N$$ is Cauchy and hence $$\left\|x-s_k\right\|_E\xrightarrow{k\to\infty}0$$ for some $x\in E$ by completeness of $E$. Now, if $\varepsilon>0$, let $k_0\in\mathbb N$ with $k_0>\varepsilon$ and $$\left\|x-s_k\right\|_E<\varepsilon\;\;\;\text{for all }k\ge k_0\;.$$ Moreoever, let $J:=\left\{1,\ldots,N_{k_0}\right\}$ and $K\subseteq\mathbb N$ with $|K|<\infty$ and $J\subseteq K$. Then, $$\left\|x-\sum_{k\in K}x_k\right\|_E\le\left\|x-s_{k_0}\right\|_E+\sum_{j\in K\setminus J}\left\|x_j\right\|_E<\varepsilon+\frac1k_0<2\varepsilon$$ and hence $(x_n)_{n\in\mathbb N}$ is summable with $\sum_{n\in\mathbb N}x_n=x$.

Another counterexample for the other direction is the following: If $H$ is a separable $\mathbb R$-Hilbert space and $(e_n)_{n\in\mathbb N}$ is an orthonormal basis of $H$, then $\left(\frac 1ne_n\right)_{n\in\mathbb N}$ is summable, but $$\sum_{n=1}^\infty\left\|\frac1ne_n\right\|_H=\sum_{n=1}^\infty\frac1n=\infty\;.$$