Identity between absolutely convergent generalized Dirichlet series

The following identity between generalized Dirichlet series, both absolutely convergent in the whole complex plane, occurs when investigating functions of the Selberg class of degree $d=0$: $$\sum_{n=1}^\infty a(n)\left(\frac{Q^2}{n}\right)^s=\omega Q\sum_{n=1}^\infty\overline{a(n)}n^{s-1}.$$ To prove that $$F(s)=\sum_{n=1}^\infty \frac{a(n)}{n^s}$$ is actually a Dirichlet polynomial, using an uniqueness theorem for generalized Dirichlet series one derives from the above identity that $Q^2/n$ is an integer for all n with $a(n)\neq 0$. But how?

If we set $b(n):=\omega Q \overline{a(n)}$, $\lambda_n=\log(n/Q^2)$ and $\nu_n=\log(n)$ we got $$f(s):=\sum_{n=1}^\infty a(n)e^{-\lambda_n s}=\sum_{n=1}^\infty b(n)e^{-\nu_n(1-s)}=:g(1-s)$$

From the uniqueness theorem I know we would get, that if $f(s)=g(s)$ in some half-plane, then $\lambda_n=\nu_n$ and $a(n)=b(n)$ for all $n$, which is exactly what I need. But in fact, we just got $f(1/2)=g(1/2)$.

We use the

Theorem: Let $(\lambda_k)$ a sequence of distinct real numbers without accumulation point, and $(\alpha_k)$ a sequence of complex numbers, such that the series $$f(s) = \sum_{k = 1}^{\infty} \alpha_k e^{\lambda_k s}\tag{1}$$ converges absolutely on the line $\operatorname{Re} s = c$, where $c\neq 0$, and $f(s) = 0$ for all $s$ with $\operatorname{Re} s = c$. Then $\alpha_k = 0$ for all $k$.

With that uniqueness theorem, we consider

$$\sum_{n = 1}^{\infty} a(n)\biggl(\frac{Q^2}{n}\biggr)^s - \sum_{n = 1}^{\infty} \frac{\omega Q \overline{a(n)}}{n} n^s \equiv 0,\tag{\ast}$$

and write the left hand side in the form of the theorem, letting $(\lambda_k)$ be an enumeration of

$$\{\log n : n \in \mathbb{N}\} \cup \{2\log Q - \log n : n \in \mathbb{N}\}$$

and setting

$$\alpha_k = \begin{cases}a(m) - \dfrac{\omega Q\overline{a(n)}}{n} &, \lambda_k = \log n = 2\log Q - \log m \\ \quad -\dfrac{\omega Q \overline{a(n)}}{n} &, \lambda_k = \log n \\ \qquad a(m) &, \lambda_k = 2\log Q - \log m\end{cases}$$

where we choose the first matching condition. The hypotheses of the theorem are satisfied, since the two series in $(\ast)$ converge absolutely on the entire plane by assumption, and since $Q^2/n \to 0$ and $n\to +\infty$, the sequence $(\lambda_k)$ has no accumulation point. Hence we have $\alpha_k = 0$ for all $k$, and by the definition of $\alpha_k$ this yields $a(m) \neq 0 \implies Q^2/m \in \mathbb{N}$.

Proof (of the theorem): For $\lambda \in \mathbb{R}$, consider the integral

$$I(\lambda) := \frac{1}{2\pi i}\int_{c - i\infty}^{c + i\infty} \frac{e^{\lambda s}}{s^2}\,ds.$$

The integrand is an entire meromorphic function with only one pole at $0$, and the residue there is $\lambda$. By the residue theorem, we have

$$I(\lambda) = \begin{cases} \lambda &, \lambda \geqslant 0 \\ 0 &, \lambda < 0 \end{cases}\quad \text{for } c > 0,\text{ and}\quad I(\lambda) = \begin{cases}\; 0 &, \lambda \geqslant 0 \\ -\lambda &, \lambda < 0\end{cases}\quad\text{for } c < 0.$$

Since the series in $(1)$ converges absolutely we can interchange summation and integration, and since $f(s)$ vanishes identically on the integration contour we obtain

\begin{align} 0 &= \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{f(s)e^{-\lambda s}}{s^2}\,ds \\ &= \sum_{k = 1}^{\infty} \frac{\alpha_k}{2\pi i}\int_{c - i\infty}^{c + i\infty} \frac{e^{(\lambda_k - \lambda)s}}{s^2}\,ds \\ &= \sum_{c(\lambda_k - \lambda) > 0} \alpha_k \lvert \lambda_k - \lambda\rvert \end{align}

for every $\lambda \in \mathbb{R}$. Varying $\lambda$ between two adjacent (in order, not with respect to index) $\lambda_k$ shows that

$$\sum_{\lambda_k > \lambda}\alpha_k = 0\qquad\text{resp.}\qquad \sum_{\lambda_k < \lambda}\alpha_k = 0\tag{2}$$

for every $\lambda\in \mathbb{R}$. Choose $m\in \mathbb{N}$ and consider $(2)$ for a $\lambda$ slightly smaller than $\lambda_m$ and a $\lambda$ slightly greater. It follows that $\alpha_m = 0$. Since $m$ was arbitrary, the theorem is proved.

• I'm impressed, it seems you found some time to think about it eventually. So the thing is solved for me, thanks again! Apr 24 '17 at 7:19
• When checking the details, I didn't need the condition $\sum|\alpha_k|(1+|\lambda_k|)<\infty$... Apr 24 '17 at 9:35
• My calculation just yields $\sum_{\lambda_k=\lambda_m}\alpha_k=0$ for all $m$. So I would rather insert the condition $\lambda_i\neq\lambda_j$ for all $i\neq j$ in order to get $\alpha_m=0$ for all $m$. Apr 24 '17 at 11:03
• And as if by magic the proposed changes just happended :D Apr 24 '17 at 11:06
• Yes, that isn't needed. I added it to have absolute convergence of $\sum \alpha_k(\lambda_k - \lambda)$, but the convergence (not necessarily absolute) follows from the dominated convergence theorem and that suffices for our needs. Didn't see that at 2 a.m., so I inserted the condition. And yes, distinctness of the $\lambda_k$ is necessary, I forgot to write that last night. It sprang to mind after I went to bed, but it wasn't important enough to get out of bed again. Apr 24 '17 at 11:07