Existence of a countable linear combination with positive coefficients

Question

Consider a (Hausdorff and complete) locally convex topological vector space $V$ and a countable subset $(v_k)_{k=1}^\infty \subset V$ of non-zero vectors.

$(*)$ Under what conditions on this subset are we guaranteed the existence of a sequence of positive real numbers $(\alpha_k)_{k=1}^\infty$, such that the series $\sum_{k=1}^\infty \alpha_k v_k$ converges (that is, the sequence of its partial sums) converges to a vector $v \in V$?

The positivity of the coefficients, $\alpha_k > 0$ for each $k$, is the crucial aspect of the question.

On the one hand, if $V$ is a Banach space, the answer is always Yes. We can always choose $\alpha_k = c_k / \|v_k\|$, where $(c_k)_{k=1}^\infty$ is any summable sequence of positive numbers.

On the other hand, consider the space $c_{00}$ of finite sequences of real numbers, turned into a complete locally convex space in the usual way. The the subset $(v_k)_{k=1}^\infty$, where the $v_k$ is the sequence that is zero everywhere except the $k$-th place, is an example for which such a sequence of $(\alpha_k)_{k=1}^\infty$ does not exist. This is clear because here a necessary and sufficient condition for the convergence of $\sum_{k=1}^\infty \alpha_k v_k$ is that only finitely many $\alpha_k$ are non-zero.

So maybe I could ask the question in a slightly different way. Is there a well-studied (joint?) property of the space $V$ and the subset $(v_k)_{k=1}^\infty$ that is sufficient (hopefully also necessary) for the answer to $(*)$ to be Yes? I was hoping that the answer would be related to one of the standard theorems of functional analysis, but I haven't recognized the right one yet.

Answer 1

As a partial answer, note the following necessary (and often sufficient) criteria.

Let $E$ be a (real or complex) topological vector space. Let us say that a sequence $\{x_n\}_{n=1}^\infty$ in $E$ can be made summable/bounded/to converge to zero if there is a sequence $\{\alpha_n\}_{n=1}^\infty$ of strictly positive real scalars such that $\{\alpha_n x_n\}_{n=1}^\infty$ is summable/bounded/converges to zero.

Proposition. Consider the following statements:

$\text{(a)}$. $\{x_n\}_{n=1}^\infty$ can be made summable;

$\text{(b)}$. $\{x_n\}_{n=1}^\infty$ can be made to converge to zero;

$\text{(c)}$. $\{x_n\}_{n=1}^\infty$ can be made bounded.

In every topological vector space $E$ one has $$\text{(a)} \Longrightarrow \text{(b)} \Longleftrightarrow \text{(c)}.$$ If $E$ is locally convex and sequentially complete, then $$\text{(a)} \Longleftrightarrow \text{(b)} \Longleftrightarrow \text{(c)}. $$

Proof. $\text{(a)} \Longrightarrow \text{(b)}$. Suppose that $\sum_{k=1}^\infty \alpha_k x_k$ converges to $y \in E$. Let $s_n := \sum_{k=1}^n \alpha_k x_k$ denote the $n$-th partial sum. Then $\lim_{n\to\infty} s_n = \lim_{n\to\infty} s_{n+1} = y$, so by continuity of addition $$ \lim_{n\to\infty} \alpha_{n+1} x_{n+1} = \lim_{n\to\infty} s_{n+1} - s_n = y - y = 0. $$

$\text{(b)} \Longrightarrow \text{(c)}$. Every convergent sequence is bounded.

$\text{(c)} \Longrightarrow \text{(b)}$. If $\{\alpha_n x_n\}_{n=1}^\infty$ is bounded, then $\lim_{n\to\infty} \alpha_n \beta_n x_n = 0$ for every sequence $\{\beta_n\}_{n=1}^\infty$ of scalars converging to zero (see e.g. Rudin, Theorem 1.30).

Assume now that $E$ is locally convex and sequentially complete. $\text{(b)} \Longrightarrow \text{(a)}$. Suppose that $\lim_{n\to\infty} \alpha_n x_n = 0$. Define $s_n := \sum_{k=1}^n \frac{\alpha_k}{2^k} x_k$. Use local convexity to prove that $\{s_n\}_{n=1}^\infty$ is Cauchy, and use sequential completeness to conclude that the series $\sum_{k=1}^\infty \frac{\alpha_k}{2^k} x_k$ converges. $\Box$

Some remarks:

1. Each of the properties $\text{(a)}$, $\text{(b)}$ and $\text{(c)}$ carries over to all weaker (i.e. coarser) topologies, even those with different dual spaces. Interestingly, in locally convex spaces, property $\text{(c)}$ only depends on the dual pair.

2. If the topology of $E$ is given by a metric $d$, then every sequence $\{x_n\}_{n=1}^\infty$ satisfies $\text{(b)}$. To prove this, note that for every $z \in E$ we have $\lim_{k\to\infty} \frac{1}{k}z = 0$ (by continuity of scalar multiplication). Equivalently: for every $\varepsilon > 0$ there exists some $K_\varepsilon \in \mathbb{N}_1$ so that $d(0,\frac{1}{k} z) < \varepsilon$ for all $k \geq K_\varepsilon$. Using this, we choose a sequence $\{\alpha_n\}_{n=1}^\infty$ of strictly positive real scalars so that $d(0,\alpha_n x_n) < \frac{1}{n}$ for all $n\in\mathbb{N}_1$. Then $\lim_{n\to\infty} \alpha_n x_n = 0$.

3. The answer of Robert Israel follows from the previous remark and the Proposition.

4. There are also non-metrizable tvs where every sequence satisfies $\text{(b)}$ (take a metrizable space and pass to a weaker topology). Likewise, there exist non-Fréchet tvs where every sequence satisfies $\text{(a)}$. Interesting follow-up question: if every sequence satisfies $\text{(b)}$ (resp. $\text{(a)}$), then is there necessarily a finer topology which is metrizable (resp. Fréchet)?

Answer 2

It's true in a Fréchet space. If $\|\cdot\|_k$, $k = 0,1,2,\ldots$, is a family of seminorms that induces the topology, choose $\alpha_j = c_j/\max\{\| v_j \|_k : k \le j\}$ for a summable sequence of positive numbers $c_j$.