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.


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.


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 1


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.


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