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

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