I have arrived at an elementary-looking "result" via a sketchy argument. Having unsuccessfully searched for the statement and its "proof" in the literature, I would like to hear if anyone knows whether or not it is true, and if they might provide me with a reference? The statement is as follows: let $f:\mathbb{N}\rightarrow \mathbb{C}$ and $$F(s)=\sum_{1}^{\infty}\frac{f(n)}{n^s}$$ converge for $\sigma>\sigma_c>0$, then
converges for $\sigma>\sigma_c>0$, and vice-versa.

EDIT 1: "Vice-versa" meaning that the converse implication also holds.

EDIT 2: The integrand is to be taken as an analytic continuation of the series definition of $F$, assuming a-priori that one exists there.

Sketch of proof: I begin by constructing an auxillary Dirichlet series: $$D(s)=\sum_{1}^{\infty}\frac{d(n)}{n^s}$$ where $$d(n)=2\left(f(1)+f(2)+\cdots+\frac{f(n)}{2}\right)\bar{f}(n).$$ Thus
$$\sum_{1}^{n}d(m)=\left|\sum_{1}^{n}f(m)\right|^2$$ is positive. The abscissa of convergence of $D(s)$ is $$\sigma_d=\limsup_{n\rightarrow\infty}\frac{\log\sum_{1}^{n}d(m)}{\log n}=2\limsup_{n\rightarrow\infty}\frac{\log\left|\sum_{1}^{n}f(m)\right|}{\log n}=2\sigma_c,$$ provided $\sigma_c>0$. I argue that the convergence of the Dirichlet series $D(2s)$ is equivalent to the convergence of the integral above as follows: If $\sigma>\sigma_c$, then $$D(2\sigma)=\sum_{1}^{\infty}\frac{d(n)}{n^{2\sigma}}=\sum_{1}^{\infty}\left(\frac{1}{\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\sum_{1}^{\infty}\frac{f(m)}{m^{s}}\frac{n^{s}ds}{s}\right)\frac{\bar{f}(n)}{n^{2\sigma}},$$ where both $D(2s)$ and $F(s)$ converge absolutely if $\sigma>\sigma_c+1$. Since $F(s)$ has no poles for $\sigma>\sigma_c>0$, so for fixed $n$ the (conditionally convergent) integral is unchanged for $\sigma>\sigma_c$. If one can justify interchanging the order of summation and integration, we get
$$D(2\sigma)=\frac{1}{\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}|F(s)|^2\frac{ds}{s}.$$

  • 2
    $\begingroup$ I'm not certain, but this looks pretty similar to something involving Plancherel's identity: see Theorem 5.4 (p.144) of Montgomery and Vaughan's "Multiplicative Number Theory". $\endgroup$ Oct 19, 2011 at 23:01
  • 1
    $\begingroup$ What could the "vice versa" mean? If the integral converges for a particular $\sigma$, then in particular $F(s)$ must be defined there, hence convergent. Unless you're trying to refer to analytic continuations or something? $\endgroup$ Oct 20, 2011 at 8:13
  • $\begingroup$ By "vice-versa" I mean that the converse implication also holds. Yes, I am referring to an analytic continuation in the integrand because there would be no point in stating the converse implication otherwise (accept for the existence of the integral)- of course I should qualify that I assume one exists, that is a good point, thanks. $\endgroup$ Oct 20, 2011 at 9:16
  • $\begingroup$ Your last integral is real, hence it is the same as $\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}|F(s)|^2\left(\frac{1}{s}+\frac{1}{\bar s}\right)ds$, where $\bar s$ is the conjugate of $s$. Now observe that $\frac{1}{s}+\frac{1}{\bar s}=\frac{2\sigma}{|s|^2}$ to see that your condition is equivalent to (3) in my original response. So your conjecture was OK, and Plancherel makes it rigorous. $\endgroup$
    – GH from MO
    Oct 20, 2011 at 12:36
  • $\begingroup$ @Kevin: See my previous comment also the comments to your comments below. $\endgroup$
    – GH from MO
    Oct 20, 2011 at 12:38

1 Answer 1


Something similar is true. Let $$ A(x):=\sum_{n\leq x}f(n),\qquad x>0,$$ then the first condition is equivalent to $$ \forall\sigma>\sigma_c:\ A(x)\ll_\sigma x^\sigma. \tag{1}$$ This clearly implies $$ \forall\sigma>\sigma_c:\ \int_0^\infty |A(x)|^2 x^{-2\sigma}\frac{dx}{x}<\infty, \tag{2}$$ which by $$ F(s)=s\int_1^\infty A(x)x^{-s}\frac{dx}{x},\qquad\Re s>\sigma_c, $$ is equivalent to $$ \forall\sigma>\sigma_c:\ \int_{\sigma-i\infty}^{\sigma+i\infty}|F(s)|^2\frac{ds}{|s|^2}<\infty. \tag{3}$$ This suggests that your second condition should be changed to (3).

Still, the converse (3)=>(1) would be false, for it might well happen for a step function $A(x)$ that (2) holds but (1) fails. On the other hand, (2) and hence also (3) express the fact that (1) holds for most $x$'s.

  • $\begingroup$ Thanks, GH. Sorry for the long delay in responding, it was night-time in England. The equivalence of our (1)'s is clear. However, your statement that (2) does not imply (1) is confusing me: (2) is equivalent to the statement that $|A(x)|^2=O(x^{2\sigma})$, which appears to imply $A(x)=O(x^{\sigma})$? Can you also explain why (3) is equivalent to (2) please, because it is certainly weaker than the existence of the integral in my question. In the meantime I shall add an edit to explain my argument. $\endgroup$ Oct 20, 2011 at 9:09
  • $\begingroup$ @Kevin: (2) is not equivalent to $|A(x)|^2=O(x^{2\sigma})$. For example, consider $A(n):=n^{\sigma+1/5}$ when $n$ is a square and $A(n):=0$ otherwise. Then (2) holds, but (1) fails. If you define $F(s)$ by the formula between (2) and (3) then the equivalence of (2) and (3) follows from Plancherel for the Mellin transform, which is an application of Plancherel for the Fourier transform. Of course Greg Martin has a point: if we stick with your original definition of $F(s)$, then for $\sigma>0$ and $t\in\mathbb{R}$ the existence of $F(\sigma+it)$ implies $A(x)\ll_\epsilon x^{\sigma+\epsilon}$. $\endgroup$
    – GH from MO
    Oct 20, 2011 at 12:06
  • $\begingroup$ Cont.: Clearly the formula between (2) and (3) provides an analytic continuation of your original $F(s)$, so we are on the same page. $\endgroup$
    – GH from MO
    Oct 20, 2011 at 12:10
  • $\begingroup$ Ok, I see my stupid blunder regarding the inequivalence of (2) and (1), and I follow the Plancherel theorem for Mellin transforms- thank you for pointing those out. $\endgroup$ Oct 20, 2011 at 12:37
  • $\begingroup$ Yes, of course we are on the same page, and I am very grateful for your input. Yet I cannot accept your answer at this stage because (as you explained) the form of Plancherel's theorem given is weaker than (1) and the integral I quoted, the (in)equivalence of which is subject of my question. $\endgroup$ Oct 20, 2011 at 12:45

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