4
$\begingroup$

Ikehara's Tauberian theorem for Dirichlet series states that if $$F(s)=\sum_{1}^{\infty}\frac{f(n)}{n^s}$$ with $f(n)\geq 0$ is such that $$F(s)=\frac{G(s)}{s-1}+H(s)$$ for $\sigma>1$ with $G,H$ analytic and $G(1)\neq 0$, then the continuity of $F(1+it)$, $t\neq 0$, implies that $$\frac{1}{x}\sum_{n\leq x}f(n)\sim G(1).$$

I am wondering if there is a converse to this statement, i.e. Do we know that if $F(1+it)$ has other discontinuities at say $t=\pm a_i$, $i=1,2,...$, then $$\frac{1}{x}\sum_{n\leq x}f(n)\not\sim G(1)?$$

If not, I can imagine that there is a converse under stricter assumptions on $F$, say if we assume $F(s)$ is analytic in a neighbourhood of every point of the line $\sigma=1$ except at $t=0$ and $t=\pm a_i$, or under stricter assumptions on the $a_i's$, then $$\frac{1}{x}\sum_{n\leq x}f(n)\not\sim G(1).$$ Do such theorems exist?

$\endgroup$
3
$\begingroup$

We apply the method outlined (Theorem 15.3) in Montgomery & Vaughan 'Multiplicative Number Theory', volume 1, chapter 15.

We assume much stronger assumptions for $F(s)$ on the line $\sigma=1$ that it has simple poles at $s=1$, $s=1+it_0$ and $s=1-it_0$ with $t_0>0$. Also, assume that the Dirichlet series $F(s)=\sum a_n n^{-s}$ has $\sigma_c=1$. Then we have for $A(x)=\sum_{n\le x} a_n$, $$ B(s):=\frac{F(s)}s-\frac{c}{s-1}=\int_1^{\infty} \left(A(x) - c x\right)x^{-s-1} dx $$ holds for $\sigma>1$.

If $A(x)\geq cx$ for $x>X_0$, then for $\sigma>1$, $$ B(s)+\frac12 e^{i\phi} B(s+it_0)+\frac12 e^{-i\phi} B(s-it_0)=\int_1^{\infty} (A(x)-cx) (1+\cos(\phi-t_0\log x)) x^{-s-1}dx \ \ \ \ (*) $$ The LHS of (*) has a simple pole at $s=1$ with residue $$ G(1)-c+\frac12 e^{i\phi} \beta + \frac12 e^{-i\phi} \overline{\beta} $$ where $\beta = \mathrm{Res}_{s=1+it_0} F(s)/s\neq 0$. On the other hand, the RHS does not tend to $-\infty$ as $s\rightarrow 1+$, because the integrand is nonnegative for $x\geq X_0$. We now take $\phi$ so that $$ \frac12 e^{i\phi} \beta + \frac12 e^{-i\phi} \overline{\beta} = -|\beta|. $$ Then $$ G(1)-c-|\beta|\geq 0 $$ yiending $$ c\leq G(1)-|\beta|. $$ Thus, $\liminf\limits_{x\rightarrow\infty} A(x)/x \leq G(1)-|\beta|$. Similarly, we can prove that $\limsup\limits_{x\rightarrow\infty} A(x)/x \geq G(1)+|\beta|$. Then we have the desired result $A(x)/x \not\sim G(1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.