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In D.J Newman's paper

A simple analytic proof of the prime number theorem

there is the following theorem:

Suppose $|a_n|<1$ and form the Dirichlet series $F(s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$ which clearly converges to an analytic function for $\Re(s)>1$. If, in fact, $F(s)$ is analytic throughout $\Re(s)\geq1$, then $\sum_{n=1}^{\infty}\frac{a_n}{n^s}$ converges throughout $\Re(s)\geq1$

I do not understand what Newman means by "Analytic Throughout" $\Re(s)\geq1$.

He clearly is not supposing that the function must converge for $\Re(s)\geq1$ for it to be analytic since otherwise, the theorem would be useless, and so I can only assume that he means that the function has an analytic continuation over the line $\Re(s)\geq1$. Since analytic functions are defined on open sets I can only assume this must mean that $F(s)$ has an analytic continuation to some set that contains the real numbers. Am I correct?

If I am, what conditions would one need to show that a function has an analytic continuation throughout $\Re(s)\geq1$? Is it sufficient to show that

$$\lim_{x\to1^+}F(x+ti)$$

exists for all $t\in\mathbb{R}$?

Any insights are appreciated.

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He means the function has an analytic continuation from the open half-plane ${\rm Re}(s) > 1$ to the closed half-plane ${\rm Re}(s) \geq 1$. By definition, to say a function is analytic on a closed set means it is analytic on an open set containing that closed set. It is convenient to be able to talk about a function on an open set having an analytic continuation to its boundary without having to always throw in "larger open set containing the closure of the original open set".

Newman is not saying $F(s)$ has an analytic continuation to the "real numbers" but to the vertical line ${\rm Re}(s) = 1$, meaning to an open set that contains that line (and, in the setting of the theorem, it means analytic continuation to an open set containing ${\rm Re}(s) \geq 1$).

There are no simple general conditions that let you check whether a Dirichlet series has an analytic continuation from a half-plane of known convergence to a point on the boundary line. Each class of important examples can require new ideas. In some sense it is like dealing with analytic continuation of a power series from an open disc to its boundary. There is no simple method to check whether a generic power series with radius of convergence 1 has an analytic continuation to a point on the unit circle (assuming the coefficient tend to $0$, a necessary condition for the power series to converge at some point on the unit circle).

To emphasize the subtlety of analytic continuation of Dirichlet series on the boundary of where they are known to converge, one of the consequences of the work by Wiles on Fermat's Last Theorem is that the Dirichlet series defining the $L$-function of an elliptic curve over $\mathbf Q$ has an analytic continuation to all of $\mathbf C$ from its initial "easy" half-plane of absolute convergence ${\rm Re}(s) > 3/2$. Even the analytic continuation of all such Dirichlet series to the line ${\rm Re}(s) = 3/2$ was unknown before his work. (Of course some special cases were known previously.) The upshot is that the hypothesis of analytic continuation in Newman's theorem is a very serious one, and you don't verify it without knowing something significant about the actual example you want to apply it to.

In Newman's proof of the Prime Number Theorem, he wants to apply his theorem to the function $1/\zeta(s)$, which for ${\rm Re}(s) > 1$ has the Dirichlet series representation $\sum \mu(n)/n^s$ with coefficients $\mu(n)$ that are the Moebius function, which is bounded (values are $0$, $1$, and $-1$). Proving $1/\zeta(s)$ has an analytic continuation from ${\rm Re}(s) > 1$ to ${\rm Re}(s) \geq 1$ basically involves showing $\zeta(s)$ has an analytic continuation from ${\rm Re}(s) > 1$ to ${\rm Re}(s) > 0$ except for a simple pole at $s = 1$ (this is done in nearly any analytic number theory book) and then proving $\zeta(s) \not= 0$ for ${\rm Re}(s) = 1$. The nonvanishing of $\zeta(s)$ on the line ${\rm Re}(s) = 1$ (this is "automatic" at $s = 1$ from the pole, which turns into $1/\zeta(s) = 0$ at $s = 1$) is often regarded as the key analytic property of the zeta-function in the proof of the Prime Number Theorem. The proof is not really hard, but it does require a clever idea. It is not something anyone is going to figure out just by starting at the definition of the zeta-function when ${\rm Re}(s) > 1$ or staring at a formula that analytically continues the zeta-function to ${\rm Re}(s) > 0$.

Convergence of limits from inside a domain of convergence of a series to a point on the boundary is not sufficient to imply convergence of the series at the boundary point, e.g., consider $\sum_{n \geq 0} (-1)^nz^n$ as $z \rightarrow 1^{-}$ or $\sum_{{\rm odd} \, n \geq 1} (-1)^{(n-1)/2}/n^s = 1 - 1/3^s + 1/5^s - 1/7^s + \cdots$ as $s \to 0^+$. The power series does not converge at $z = 1$, the Dirichlet series does not converge at $s = 0$, but both series have limit $1/2$ as $z \to 1^-$ or as $s \to 0^+$. The value of the power series limit is easier to see since the power series equals $1/(1+z)$ for $|z| < 1$ and that simple formula gives you an analytic continuation for all of $\mathbf C - \{1\}$, which at $z = 1$ is $1/2$. The Dirichlet series has an analytic continuation to $s = 0$ since, well, there's a more complicated formula you can write down that matches the series for ${\rm Re}(s) > 0$ and makes sense and is analytic on a bigger half-plane than ${\rm Re}(s) > 0$. Without seeing such a formula, which I won't write down here, I don't think it is obvious that the limit of that Dirichlet series (it is the $L$-function of the nontrivial character mod $4$) as $s \to 0^+$ is $1/2$. Read an analytic number theory book that discusses analytic continuation of Dirichlet $L$-functions and you'll see how such analytic contiuation is proved. It's not as easy as the case of a geometric series.

A power series converging on the open unit disc is analytic there, but if it has radius of convergence 1 then it need not be analytic at each point on the unit circle to which it converges. (If the radius of convergence of the power series is bigger than $1$ then the situation is different!) In fact, if a series converging on on the open unit circle has an analytic continuation to each point on the unit circle, then by compactness of the closed unit disc the power series has radius of convergence greater than 1. Therefore a power series with radius of convergence 1 that converges on the closed unit disc, such as $\sum z^n/n^2$, is not analytic somewhere on the unit circle even though it converges on the whole unit circle. The series $\sum z^n/n^2$ has a name, the "dilogarithm", and is denoted ${\rm Li}_2(z)$ (you can replace the exponent $2$ in the denominator with $k$ and get ${\rm Li}_k(z)$, hence the notation). It has an analytic continuation from the open unit disc to all of $\mathbf C$ except the point $z = 1$, and on the closed unit disc (including $z = 1$) it is continuous.

The situation with Dirichlet series is more subtle: $\sum_{{\rm odd} \, n \geq 1} (-1)^{(n-1)/2}/n^s$ converges if and only if ${\rm Re}(s) > 0$, but it has no analytic singularity on the imaginary axis. In fact, this series extends analytically to all of $\mathbf C$ (an entire function). Nothing strange happens anywhere on the imaginary axis as far as analytic behavior is concerned. The proof that a power series has an analytiic singularity somewhere on the boundary of its disc of convergence does not carry over to a Dirichlet series and the boundary of its half-plane of convergence because the boundary of a half-plane is not compact, unlike a circle.

I can carry over the dilogarithm example to the setting of Dirichlet series, since every power series $\sum c_kz^k$ could be interpreted as a vertically periodic Dirichlet series by a change of variarbles $z = 1/2^s$, so $|z| = 1/2^{{\rm Re}(s)}$. Then $|z| < 1$ corresponds to ${\rm Re}(s) > 0$ and $\sum c_kz^k = \sum c_k/2^{ks}$. This is a Dirichlet series supported on the powers of $2$. (I could have used $z = 1/3^s$ or other options, but picked one for concreteness.) Since $2^s$ has period $2\pi i/\log 2$, the function $\sum c_k/2^{ks}$ is unchanged when we add an integral multiple of $2\pi i/\log 2$ to $s$. Let's consider $$ f(s) = \sum_{k \geq 1} \frac{1/k^2}{2^{ks}} = {\rm Li}_2(1/2^s). $$ This series converges for ${\rm Re}(s) \geq 0$, just as ${\rm Li}_2(z)$ converges for $|z| \leq 1$. Since ${\rm Li}_2(z)$ extends analytically from $|z| < 1$ to $\mathbf C - \{1\}$, $f(s)$ extends analytically from ${\rm Re}(s) > 0$ to $\mathbf C - A$ where $A = \{s : 1/2^s = 1\} = (2\pi i/\log 2)\mathbf Z$. Therefore $f(s)$ has half-plane of convergence ${\rm Re}(s) \geq 0$, but is not analytic at the points in $A$ on the imaginary axis.

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  • $\begingroup$ Thank you so much! One thing still left unanswered about my question of the case that the limit does exist everywhere on the boundary. You state that convergence at one point on the boundary does not imply that it converges everyone on said boundary, but if it does converge everywhere on the boundary does that mean that it can be analytically continued over that boundary? $\endgroup$ – Milo Moses Jun 18 '20 at 0:49
  • $\begingroup$ The reason I'm being so particular is that I have a specific example in mind, and if this were true it would save me a huge headache. $\endgroup$ – Milo Moses Jun 18 '20 at 1:02
  • $\begingroup$ I added a paragraph at the end showing that what you are asking for is not true. The analogue for power series is already not true, so it would not be realistic to expect the version for Dirichlet series is true either, and in any case I have given a concrete example (or counterexample, depending on your point of view). $\endgroup$ – KConrad Jun 18 '20 at 1:19
  • $\begingroup$ Ok, this makes sense. Thanks again for the clarification. I really appreciate the work that you're doing here on MathOverflow. $\endgroup$ – Milo Moses Jun 18 '20 at 1:26

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