# Dirichlet series and Dirichlet convolution

Let $$f$$ and $$g$$ be an arithmetic functions, and let $$f*g$$ be the Dirichlet convolution of $$f$$ and $$g$$.

As known from fundamental analytic number theory, the Dirichlet series generating function is: $$DG(f;s)=\sum_{n=1}^\infty \frac{f(n)}{n^s}$$.

Hence, The multiplication of Dirichlet series can be written as: $$DG(f;s)DG(g;s)=DG(f*g;s)$$, but as we know, Riemann series theorem says that if the series is conditionally convergence, then any permutation of the series may generate another sum.

My question is, if $$DG(f;s)$$ $$DG(g;s)$$ are conditionally convergence, how can we definde $$DG(f;s)DG(g;s)$$?

• if a Dirichlet series converges anywhere, it will also converge absolutely at an abscissa $+1$ from the conditional convergence one (at least of course), so one defines the Dirichlet product where $f,g$ are both absolutely convergent and then extend it as much as possible by analytic continuation which is unique Oct 28 '20 at 13:28
• Also @Conrad $\sum_{n=1}^\infty (-1)^nn^{-s}$ converges for $\Re(s) > 0$ while $\sum_{n=1}^\infty (\sum_{d| n}(-1)^{d}(-1)^{n/d})n^{-s}$ does not. Anywhere $DG(f,s),DG(g,s),DG(f\ast g,s)$ converge we have $DG(f,s)DG(g,s)=DG(f\ast g,s)$ (analytic continuation stuff plus $DG(f,s)=\lim_{h\to 0^+} DG(f,s+h)$) Oct 28 '20 at 13:49

This question can be completely answered by the following surprising theorem.

Theorem: Let $$D(f,s)$$, $$D(g,s)$$ be the Dirichlet series corresponding to $$f$$ and $$g$$. If $$D(f,s)$$ and $$D(g,s)$$ conditionally converge at $$s=0$$, then the convolution Dirichlet series $$D(f*g,s)$$ must be conditionally converges at $$s=\frac{1}{2}+it$$ for each $$t\in \mathbb{R}$$. And the constant "$$\frac{1}{2}$$" is optimal as concerns $$\text{Re}\hspace{0.05cm}s$$.

The proof can be found in the book Diophtine apprximation and Dirichlet series,pp114 theorem 4.3.4 for second Edition.

I really like its proof, which used hyperbola method of Dirichlet to prove the convergence, and used Banach-Steinhuas theorem and Kronecker's lemma to prove the optimality.

Other interesting consequences can also be found in section 4.3 of this book.

• Your username just reminded me of Landau's lemma, another helpful result in this area :D Nov 13 '21 at 6:20