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I first asked the following question on Mathematics StackExchange (a few weeks ago), since the content of MathOverflow is mostly above my academic level. I didn't want to bother people on this forum with a maybe trivial question. But since I got no answer, I'm trying my luck here.

I'd like some help to prove the following theorem:

Let $\sum_{n \geq 1}\frac{f(n)}{n^s}$ and $\sum_{n \geq 1}\frac{g(n)}{n^s}$ be two Dirichlet series with respective abscissas of absolute convergence $\alpha_f$ and $\alpha_g$ ($\alpha_f, \alpha_g \neq -\infty$). Then the abscissa of convergence of $\sum_{n \geq 1}\frac{f*g(n)}{n^s}$ is:

  • $\max(\alpha_f, \alpha_g)$ if $\alpha_f \neq \alpha_g$ ;
  • less than or equal to $\alpha$ if $\alpha_f = \alpha_g = \alpha$

where $f*g$ refers to the Dirichlet convolution of $f$ and $g$.

The result is to be proved in an exercise from this book: exercise 9, page 259. I need help to prove the theorem in the case $\alpha_f \neq \alpha_g$.

After searching the literature, I found similar theorems stated in a few sources, but never saw any proof.

What I have managed to prove up to now:

  • If $f$ and $g$ are positive real-valued functions, the result is obvious.
  • Suppose $\alpha_f < \alpha_g$. Then, if $$\sum_{n=2}^{+\infty}\frac{|f(n)|}{n^{\alpha_g}} < |f(1)|$$ the result is also true.

I tried to experiment with particular functions not satisfying either of these strong conditions, but I fail to see why the result is true in the general case.

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  • $\begingroup$ Here is a link to the Math.StackExchange question: math.stackexchange.com/questions/1043825/… $\endgroup$ – Ricardo Andrade Dec 26 '14 at 23:45
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    $\begingroup$ Why don't you try relating $\alpha_f$ to a growth condition on $f(n)$? $\endgroup$ – Kimball Dec 27 '14 at 4:11
  • $\begingroup$ If $\alpha_f > 0$ (if not, we can still multiply both $f$ ang $g$ by a power function to come down to this case, since abscissae of absolute convergence are finite), then $$\alpha_f = \limsup\frac{\ln\sum_{k=1}^{n}|f(k)|}{\ln(n)}$$ Is is what you refer to as "growth condition" ? $\endgroup$ – edouard.gilles Dec 27 '14 at 16:21
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I think it is false:

Consider a simple series with a zero at $s=2$. For example $$E(s)=1-\frac{1}{2^s}-\frac{12}{4^s}=P(2^{-s}),\quad \text{with} \quad P(x)=1-x-12x^2.$$ We have $E(2)=0$ and $E(1)=-5/2$. Since $$\frac{1}{1-x-12x^2}=\frac{1}{7}\sum_{k=0}^\infty (4^{k+1}+(-1)^k 3^{k+1})x^k,\qquad |x|<\frac14.$$

We have $$E(s)^{-1}=1+\frac{1}{7}\sum_{k=1}^\infty (4^{k+1}+(-1)^k 3^{k+1})\frac{1}{2^{ks}},\qquad \sigma>2.$$ Consider $$A(s)=\zeta(s) E(s)$$ The Dirichlet series $A(s)$ is absolutely convergent for $\sigma>1$, because it is the product of two absolutely convergent Dirichlet series is absolutely convergent. The abscissa of absolute convergence of $A(s)$ is $a=1$ because the function $A(s)$ extends to a meromorphic function and has a simple pole at $s=1$.

Let now $B(s)=E(s)^{-1} \zeta(s)^{-1}$. The Dirichlet series $B(s)$ is absolutely convergent for $\sigma>2$ as product of two absolutely convergent Dirichlet series. But the function $B(s)$ extends to a meromorphic function with a pole at $\sigma=2$. Therefore $b=2$ is is abscissa of absolute convergence.

The product $A(s)B(s)=1$ and its abscissa of absolute convergence is $-\infty$.

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