Riemann functional equation and Hamburger’s theorem.

In this post I shall give a proof of functional equation of Riemann zeta function by poisson summation formula and then a proof to a converse theorem.

The Riemann zeta function $\displaystyle\zeta(s)$ , defined for $\displaystyle \hbox{Re}(s) > 1$ by &fg=000000

$\displaystyle \zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^s}$

and extended meromorphically to other values of $s$ by analytic continuation, obeys the functional equation

$\displaystyle \xi (s) = \xi(1-s),$

where

$\displaystyle \xi (s) := \pi^{-s/2} \Gamma(s/2) \zeta(s)$ ,

and the Gamma function $\displaystyle \Gamma(s)$ is defined for $\hbox{Re}(s) > 1$ by

$\displaystyle \Gamma(s) := \int_0^\infty e^{-t} t^s\ \frac{dt}{t}$

A very important tool in number theory is the Poisson summation. We shall use it to prove the functional equation.
Suppose $f,\hat f$ are in $\displaystyle L^1(\mathbb{R})$ and have bounded variation. Then

$\displaystyle \sum_{m\in \mathbb{Z}}f(m)=\sum_{n\in\mathbb{Z}}f(n)$

where both sums converge absolutely.
Proof: Consider

$\displaystyle F(x)=\sum_{n \in \mathbb{Z}} f(x+n)$ .

This is periodic of period one. It has the absolutely convergent Fourier series expansion

$\displaystyle F(x)=\sum c_n e(nx)$ where $c_n=\int_0^1 F(x)e(-nx)dx=\int _{-\infty}^\infty f(x)e(nx)dx=f(n)$

Therefore

$\displaystyle F(0)=\sum_{m\in \mathbb{Z}}f(m)=\sum_{n\in\mathbb{Z}}f(n)$

Applying Poisson summation to the function $\displaystyle f(x)=e^{-\pi x^2/y}$ we get a beautiful result

$\displaystyle \sum_{m \in \mathbb{Z}} e^{-\pi m^2/y}=\sqrt y\sum _{n \in \mathbb{Z}}e^{-\pi n^2y}$

We shall now use the Poisson summation to prove the functional equation for the Riemann zeta function.
The basic idea is that we use the Poisson summation formula and take Mellin transform on both sides.

$\displaystyle \sum_{m} f(m)=\sum_{n}\hat f(n)$

In fact, we have

$\displaystyle \sum f(nx)= \frac{1}{|x|} \sum \hat f(n/x)$ .

Now taking the Mellin transform we have

$\displaystyle \sum \frac{1}{n^{s}}M_f= \sum\frac{1}{n^{1-s}}M_{\hat f}$ .

Mellin transform is like the Fourier transform for the multiplicative group $(\mathbb{R},\times)$ .

$\displaystyle M(f)(s)= \int_0^\infty f(x) x^s \frac{dx}{x}$ .

Here $x^s$ is the muliplicative character and $\frac{dx}{x}$ is the invariant Haar measure of the multiplicative group of $\mathbb{R}$ .

So $\displaystyle \zeta(s)M_f=\zeta(1-s)M_{\hat f}$

Now this is a functional equation for the zeta function. So for any Schwartz function we get a functional equation for zeta function.

In particular take $f$ to be the gaussian function which is its self Fourier dual, we get the functional equation in terms of the Gamma function. The Gamma function appears here, as it is the Mellin transform for the additive character $e^{-t}$ .

$\displaystyle \Gamma(s)=\int_0^\infty e^{-t} t^s dt/t$ .

Therefore we have

$\displaystyle \Gamma(s/2){\pi}^{-s/2} n^{-s}= \int_0^\infty x^{s/2} e^{-{\pi} n^2 x} dx/x$

by making the substitution $t=-\pi n^2 x$

Summing over $n$ , we have

$\displaystyle {\pi}^{-s/2}\Gamma(s/2) \sum_{n=1}^\infty \frac{1}{n^s} =\sum \int_0^\infty x^{s/2} e^{-{\pi} n^2 x} dx/x.$

On the left-hand side we find the Dirichlet series defining $\zeta(s)$ . In view of its convergence, the latter formula is valid only for $\mbox{Re}(s) > 1$ .

On the right-hand side we may interchange summation and integration, justified by absolute convergence.
Thus we obtain

$\displaystyle \pi^{-s/2}\Gamma(s/2)\zeta(s)=\int_0^\infty x^{s/2}\sum_{n=1}^\infty e^{-\pi n^2 x} dx/x$ .

Therefore we have

$\displaystyle \pi^{-s/2}\Gamma(s/2)\zeta(s)=\int_0^\infty x^{s/2} \frac{\theta(x)-1}{2},$

where

$\displaystyle \theta(x)= \sum_{n=-\infty}^\infty e^{-\pi n^2 x}$ .

But by Poisson summation we have

$\theta(x)=\frac{1}{\sqrt x}\theta(\frac{1}{x})$ .

Hence we have

$\displaystyle \pi^{-s/2}\Gamma(s/2)\zeta(s)= \frac{1}{s(s-1)} + \int_1^\infty x^{\frac{-s+1}{2}}+x^{-s/2 -1} \frac{\theta(x)-1}{2} dx$

by splitting the integral as $\displaystyle \int_0^1 + \int_1^\infty$ and making the change of variables $\displaystyle x \to 1/x$ .

Now the right hand side integral converges because $\displaystyle \theta(x)$ decays exponentially and is invariant under the change $\displaystyle s \mapsto 1-s$ . This proves the functional equation.

Thus to prove the functional equation we used the transformation property of theta function under the change $x \to 1/x$ and then took a Mellin transform to get the functional equation. Functions like theta function which have such nice transformation properties are called called modular forms. So this links the subject of Riemann zeta like functions to the study of modular forms. In fact people think that most L-functions arise as above from some modular form or even more general objects called automorphic forms.

We have proved the functional equation of Riemann zeta function. Now we may ask for what all function satisfy such a functional equation. Assuming that the function is a meromorphic continuation of some Dirichlet series $\sum a_n n^{-s}$ and some growth conditions we can prove that $\zeta(s)$ is the only function that satisfies the functional equation.

To prove this we just invert the process we followed to prove the functional equation. The last step was to take a Mellin transform of theta function. So we first take an inverse Mellin transform( similar to inverse Fourier transform) and then get a identity like that of theta function identity. Then show that $\zeta(s)$ is unique function which satisfies the identity.( actually unique upto multiplication by scalar).

Hamburger’s Converse theorem.

Hamburger’s theorem states that if $G(s)$ is an entire function of finite order, $P(s)$ a polynomial, $\displaystyle f(s) = G(s)/P(s)$ , and $\displaystyle f(s) =\sum_{n=1}^{\infty}\frac{a_n}{n^s}$ the series converging absolutely for $\displaystyle \mbox{Re}(s) > 1$ , and
$\displaystyle g(1-s)=\sum_{n=1}^{\infty}\frac{b_n}{n^{1-s}}$ , the series converging absolutely for $\mbox{Re} s <-\beta<0$ , then the functional equation

$\displaystyle \pi^{-s/2}\Gamma(s/2)f(s)=\pi^{\frac{1-s}{2}}\Gamma(\frac{1-s}{2})g(1-s)$

implies that

$\displaystyle f(s)=g(s)=a_1\zeta(s)$ .

The following was a proof given by Siegel.

Proof:
For

$\displaystyle x>0$ , $\displaystyle \phi(x):=\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}\pi^{-s/2}\Gamma(s/2)f(s)x^{-s/2} ds$

$\displaystyle =\sum_{n=1}^{\infty}a_n\int_{2-i\infty}^{2+i\infty}\Gamma(s/2)(\pi n^2x)^{-s/2}ds.$

$\displaystyle =\sum_{n=1}^{\infty}a_n e^{-\pi n^2 x}$

We also have

$\displaystyle \phi (x)=\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}g(1-s)\Gamma(\frac{1-s}{2})\pi^{\frac{1-s}{2}}x^{-s/2}ds.$

We have $f(s)$ bounded on $Re(s)=2$ and $g(1-s)$ bounded on $Re(s)=-1-\alpha$ .

By Stirling’s formula we have

$\displaystyle \frac{\Gamma(s/2)}{\Gamma((1-s)/2)}\ll |t|^{\sigma-1/2}$ $|t| \to \infty$ .

Thus $\displaystyle g(1-s)\ll |t|^{3/2}$ on $\mbox{Re}(s)=2$ as $|t| \to \infty.$ Thus the Phragmen-Lindelof principle justifies moving the line of integration to $\sigma=-1-\alpha$ and we have

$\displaystyle \phi (x)=\frac{1}{2\pi i}\int_{-1-\alpha-i\infty}^{-1-\alpha+i\infty}g(1-s)\Gamma(\frac{1-s}{2})\pi^{-\frac{1-s}{2}}x^{-s/2}ds + \sum_{j=1}^k R_j,$

where $R_j,j=1,2,..,k$ are residues of the poles at the poles $s_1,s_2,...,s_k$ .
Now $\displaystyle \sum R_j=:R(x)$ is of the form $\displaystyle \sum x^{{s_i}/2}P_j(\mbox{logx})$ where $\displaystyle P_j$ are polynomials.

Therefore we have

$\displaystyle \phi(x) = \frac{2}{\sqrt x} \sum_{n=1}^{\infty}b_n e^{-\pi n^2/x}+ R(x)$ .

Hence

$\displaystyle \sum_{n=1}^{\infty}a_n e^{-\pi n^2 x}= \frac{2}{\sqrt x} \sum_{n=1}^{\infty}b_n e^{-\pi n^2/x}+ R(x)$

Now we try to prove that this identity forces $a_i=const.$

Multiplying with $\displaystyle e^{-\pi t^2x^2},$ $t>0$ and integrating over $(0,\infty)$ with respect to $x$ we get

$\displaystyle\frac{t}{\pi}\sum\frac{a_n}{t^2+n^2}-\frac{t}{2}\int_0^\infty R(x)e^{-\pi t^2x}dx=\sum_{n=1}^{\infty}b_n e^{-2\pi nt}.$

The integral can be evaluated as sum of terms of the form

$\displaystyle Q(t,a,b):=\int_0^\infty x^a \mbox{logx}^b e^{-2\pi t^2x}$

where $b$ are integers and Re $(a)>-1$ .

Hence, we have

$\displaystyle \sum_{n=1}^\infty a_n\{\frac{1}{t-in}-\frac{1}{t+in}\}-t\frac{\pi}{2}Q(t,a,b)=\pi\sum_{n=1}^{\infty}b_ne^{-2\pi nt}.$

The left-hand side is a meromorphic function in $t$ with poles at $t =\pm in$ for $n \in \mathbb{N}$ . The right-hand side is periodic with period $i$ and, by analytic continuation, the function on the left-hand side is also periodic. Hence, the residues at $in$ and $i(n + 1)$ are equal. Thus, $a_n = a_{n+1}$ for all $n\in \mathbb{N}$ and Hamburger’s theorem is proved

Another way to prove the result is the following. One can using Fubini theorem prove that for any Schwartz functions $\displaystyle f,g$ $\displaystyle \frac{M(f,s)}{M(\hat f,1-s)}=\frac{M(g,s)}{M(\hat g,1-s)}$ . Just taking the denominators to other side and directly change of variables and interchange of integrals justified by Fubini’s theorem proves this result. So we can actually replace the Gamma function in the functional equation with any Schwartz function. Then taking the Inverse Mellin transform gives you that

$\displaystyle \sum a_n f(nx) =\sum \frac{1}{x}b_n \hat f(n/x) + R(x)$

$R(x)$ corresponds to residue term. So we get that for any Schwartz function f we have

$\displaystyle \sum a_n f(nx) =\sum \frac{1}{x}b_n \hat f(n/x) + R_f(x)$

There we now have to prove a uniqueness of Poisson summation kind of result saying that if $\displaystyle a_n, b_n$ are such that above identity holds for all Schwartz function f , then in fact $a_n=b_n=c$ i.e., the identity is just the poisson summation identity.

To finish off, we take some nice Schwartz functions supported completed near the integer points. If we take f to be supported near 0, then left side becomes $a_0$ . If we take to be the shifted version of this f now centered at 1 we get the left side as $a_1$ . But since Fourier transform does not change by integral shifts the right side remains the same. So we have $a_0=a_1$ . This way we take shifts to all integers points so that we get $a_n =a_0$ for all $\displaystyle n \in \mathbb{N}$