# The necessity of absolute convergence in the convergence of the Cauchy product of series?

The Mertens' theorem claims that

Suppose $$\sum_{n=0}^\infty a_n,\sum_{n=0}^\infty b_n$$ are two convergent series of complex numbers, convergent to $$A,\beta$$ respectively. If $$\sum_na_n$$ converges absolutely, then the Cauchy product $$\sum_{n=0}^\infty c_n$$ converges to $$A\beta$$, where $$c_n=\sum_{k=0}^na_kb_{n-k}$$.

I wonder whether the absolute convergence of $$\sum_na_n$$ is generally necessary. I know that there are counterexamples where $$\sum_na_n, \sum_nb_n$$ converge conditionally but the Cauchy product $$\sum_nc_n$$ diverges. However, they are too special. I also know that the Cesàro sum of $$(c_n)$$ is $$A\beta$$. The precise question of which I wonder the answer is that:

Given a series $$\sum_{n=0}^\infty a_n$$ of complex numbers. Suppose that for all convergent series $$\sum_{n=0}^\infty b_n$$ of complex numbers, the Cauchy product $$\sum_{n=0}^\infty c_n$$ converges. Does this imply that the series $$\sum_na_n$$ converges absolutely?

Denote by $$\beta_n$$ the partial sum $$\sum_{k=0}^nb_n$$. The previous question is equivalent to the following:

Given a series $$\sum_{n=0}^\infty a_n$$ of complex numbers. Suppose that for all convergent sequences $$(\beta_n)$$ of complex numbers, the convolution sequence $$(\sum_{k=0}^na_k\beta_{n-k})_{n\in\mathbb N}$$ converges as $$n\to\infty$$. Does this imply that the series $$\sum_na_n$$ converges absolutely?

We have some immediate consequences: First we take $$\beta_n=1$$ for all $$n\in\mathbb N$$ to deduce that $$\sum_na_n$$ converges. In this case, we can only test with those $$(\beta_n)$$ such that $$\lim_{n\to\infty}\beta_n=0$$. And the Silverman-Töplitz theorem tells us that the Cesàro mean of $$(\sum_ka_k\beta_{n-k})_n$$ tends to zero, hence what we know is that $$\lim_{n\to\infty}\sum_ka_k\beta_{n-k}=0$$ under the assumption that $$\lim_{n\to\infty}\beta_n=0$$.

My idea to attack is that, given a conditionally convergent series $$\sum_na_n$$, we try to find a sequence $$\beta_n$$ and some $$\epsilon>0$$ such that there are infinitely many $$n$$ such that $$\lvert\sum_ka_k\beta_{n-k}\rvert>\epsilon$$. I don't know how to proceed next.

# Note

There is a more highbrow aspect. Denote by $$c_0\subseteq \ell^\infty$$ the closed subspace of sequences which converge to zero. Mertens' theorem claim that the convolution map $$a*-\colon \ell^\infty\to \ell^\infty$$ restricts to a map $$c_0\to c_0$$ for any absolutely convergent series $$\sum_{n=0}^\infty a_n$$. This hints that the preceeding question might be solved by tools in functional analysis, such as Baire's category theorem.

# Update

I suddenly come up with a possible solution:

As indicated above, our assumption is that $$a*-\colon c_0\to c_0$$ is a well-defined linear operator. To invoke closed graph theorem, assume that $$\beta^{(n)}\to \beta$$ and $$a*\beta^{(n)}\to\gamma$$ in $$c_0$$. The convergence in $$c_0$$ implies the pointwise convergence, therefore $$a*\beta=\gamma$$. By closed graph theorem, the operator $$a*-$$ is continuous, that is to say, there is a constant $$M$$ such that $$\lVert a*\beta\rVert_{\ell^\infty}\le M\lVert \beta\rVert_{\ell^\infty}$$ for all $$\beta\in c_0$$. Then for all $$m\in\mathbb N$$, we take $$(\beta_n)_n\in c_0$$ such that $$\beta_na_{m-n}=\lvert a_{m-n}\rvert$$ and $$\lvert\beta_n\rvert=1$$ for $$n\le m$$, and $$\beta_n=0$$ for $$n>m$$. Then we have $$\sum_{n=0}^m\lvert a_n\rvert\le\lVert a*\beta\rVert_{\ell^\infty}\le M$$. Q.E.D. Is it correct?

## 1 Answer

I want to take the opportunity to write up a proof which depends on Banach-Steinhaus theorem instead of closed graph theorem, because we have a very short and simple proof of that theorem. But anyway, I am still looking for a more elementary constructive proof.

As described in the question, it suffice to show that

Given a sequence $$(a_n)_{n\in\mathbb N}\in{\mathbb C}^{\mathbb N}$$. If for all $$\beta\in c_0$$, the sequence $$a*\beta$$ is bounded (slightly weaker than the convergence) where $$(a*\beta)_n:=\sum_{k=0}^na_k\beta_{n-k}$$, then the series $$\sum_n\lvert a_n\rvert$$ converges.

This should be a standard exercice in functional analysis.

We consider a sequence of linear maps $$(T_m)_{m\in\mathbb N}$$ where each $$T_m\colon c_0\to\ell^\infty$$ is defined by $$(T_m(\beta))_n=\sum_{k=0}^na_k\beta_{n-k}$$ if $$n\le m$$, and $$0$$ otherwise. Each $$T_m$$ is clearly continuous. By assumption, for each $$\beta\in c_0$$, $$\sup_{m\in\mathbb N}\lVert T_m(\beta)\rVert_{\ell^\infty}=\sup_{n\in\mathbb N}\lvert\sum_{k=0}^na_k\beta_{n-k}\rvert<\infty$$, hence by Banach-Steinhaus theorem, there exists $$M\in\mathbb R$$ such that for all $$m\in\mathbb N$$ and all $$\beta\in c_0$$, we have $$\lVert T_m(\beta)\rVert_{\ell^\infty}\le M\lVert\beta\rVert_{\ell^\infty}$$. Then as in question, we take $$\beta$$ such that $$\beta_na_{m-n}=\lvert a_{m-n}\rvert$$ for $$n\le m$$ and $$\beta_n=0$$ for $$n>m$$, we deduce that $$\sum_{n=0}^m\lvert a_n\rvert\le M$$ for all $$m\in\mathbb N$$. Q.E.D.