## PNT II

I apologize for not having posted in roughly a week, but this past week has been a bit hectic. The week before spring break always seems to be. Anyway, I thought I would move us a bit further along in our goal of proving the Prime Number Theorem. In the post PNT I, we showed that the zeta function ${\zeta(s)}$ has no zeroes for ${\text{Re} \ s = \sigma > 1}$, and in this post, we will show that ${\zeta(s)}$ has no zeroes for ${\sigma = 1}$. The proof, which I believe was first given by Vallee-Poussin, is rather elegant.

We first need an inequality for ${\zeta(s)}$, which is a consequence of an elementary trigonometric identity.

Lemma 1

$\displaystyle \zeta(\sigma)^{3}\left|\zeta(\sigma + it)\right|^{4}\left|\zeta(\sigma + 2it)\right| \geq 1, \indent \forall \sigma > 1$

Proof:Recall the identity ${3 + 4\cos(\theta) + \cos(2\theta) = 2(1+\cos(\theta))^{2}}$. Using the Euler product for ${\zeta(s)}$, we have

$\displaystyle \zeta(s) = \exp\left(\sum_{p} -\log(1-p^{-s})\right) = \exp\left(\sum_{p}\sum_{n=1}^{\infty}\frac{p^{-ns}}{n}\right)$

$\displaystyle = \exp\left(\sum_{p}\sum_{n=1}^{\infty}\frac{p^{-n\sigma}\cos(nt\log p) + ip^{-n\sigma}\sin(nt\log p)}{n}\right)$

$\displaystyle = \exp\left(\sum_{p}\sum_{n=1}^{\infty}\frac{p^{-n\sigma}\cos(nt\log p)}{n}\right)\exp\left(i\sum_{p}\sum_{n=1}^{\infty}\frac{p^{-n\sigma}\sin(nt\log p)}{n}\right)$

We see that ${\left|\zeta(s)\right| = \exp(\sum_{p}\sum_{n=1}^{\infty}n^{-1}p^{-n\sigma}\cos(nt\log p))}$. Hence,

$\displaystyle \zeta(\sigma)^{3}\left|\zeta(\sigma + it)\right|^{4}\left|\zeta(\sigma + 2it)\right|$

$\displaystyle =\exp\left(\sum_{p}\sum_{n=1}^{\infty}\frac{p^{-n\sigma}}{n}\left[3 + 4\cos(nt \log p) + \cos(2nt \log p)\right]\right)$

$\displaystyle = \exp\left(\sum_{p}\sum_{n=1}^{\infty}\frac{p^{-n\sigma}}{n}2(1 +\cos(nt \log p))^{2}\right),$

which is ${\geq 1}$ since the argument is ${\geq 0}$. $\Box$

Theorem 2 ${\zeta(s) \neq 0}$ for ${s = 1+it, t \neq 0}$.

Proof:Suppose ${\zeta(s) = 0}$ for some ${1 + it}$ with ${t \neq 0}$. Since ${\zeta(s)}$ is analytic at ${s = 1 +it}$, we can expand ${\zeta(s)}$ in a power series about ${s = 1+it}$ in some open disk ${D(1+it; r)}$. So for ${0 <\left|\delta\right| < \frac{r}{2}}$,

$\displaystyle \left|\zeta(1+\delta + it)\right|= \left|\sum_{n=1}^{\infty}a_{n}\delta^{n}\right| \leq \left|\delta\right|\sum_{n=1}^{\infty}\left|a_{n}\right|\left(\frac{r}{2}\right)^{n-1} = C\left|\delta\right|$

where we use the fact that a power series converges absolutely strictly within its radius of convergence. Recall that from the meromorphic extension of ${\zeta(s)}$ to the half plane ${\left\{\sigma > 0\right\}}$ that we obtain the estimate ${\zeta(1+\delta) = O(\delta^{-1})}$ as ${\delta \downarrow 0}$. Since ${\zeta(s)}$ is analytic at ${s = 1 + 2it}$ and therefore ${\zeta(1+\delta + 2it)}$ is bounded as ${\delta \downarrow 0}$, we obtain

$\displaystyle 1 \leq \zeta(1+\delta)^{3}\left|\zeta(1 + \delta + it)\right|^{4}\left|\zeta(1+\delta +2it)\right| \lesssim \left|\delta\right|^{-3}\left|\delta\right|^{4} = \left|\delta\right|,$

which yields a contradiction for ${\left|\delta\right|}$ sufficiently small. $\Box$