# Critical strip is strict

It is a well known fact that the Riemann $$\zeta$$-function satisfies a functional equation, has all its non-trivial zeros in the critical strip $$\{ s \in \mathbb{C} \mid 0 \leq \textrm{Re}(s) \leq 1\}$$, and that the zeros in this domain are symmetric in this domain, in the sense that for all $$s$$ in the critical strip: $$\zeta(s)=0 \iff \zeta(1-s)=0$$.

The above fact is proved in most introductions to analytic number theory or $$L$$-series, such as Neukirch, Zagier, Lang etc.

Now I have been told that one can prove that the critical strip is strict, i.e. $$\zeta(s) \neq 0$$ for all $$s \in \{ s \in \mathbb{C} \mid \textrm{Re}(s)=1\}$$, which by the function implies that $$\zeta(s)$$ is also non-zero on the $$\textrm{Re}(s)=0$$ line.

As far as I understand it, the above is quite a profound statement and is equivalent to the prime number theorem(?)

My number theory professor mentioned this in passing. He is gone now and I have been unable to find a proof of the statement in the literature.

So my question is: Does anyone know a textbook proof of the strictness of the critical strip, or at least a place in the literature where it is discussed?

Many thanks.

• Any reasonably comprehensive text on analytic number theory will have this. For instance see chapter 3 of Titchmarsh's The Theory of the Riemann Zeta-Function. Apr 8, 2020 at 18:47
• This is not only in any analytic number theory book, but also in many complex analysis textbooks e.g. Stein and Shakarchi Chapter 7 Theorem 1.2.
– user208649
Apr 8, 2020 at 19:05
• The lack of zeros is not equivalent to the PNT, but it is an important step, although the remaining part is quite technical and non-trivial. Assuming one of the Tauberian theorems it becomes equivalent (thus proving those is useful for the PNT in arithmetic progressions where the aforementioned technical part becomes even more) Apr 9, 2020 at 0:28

Let $$s\in\mathbb{C}$$ such that $$s=\sigma+it$$ with $$\sigma>1$$ and $$t\in\mathbb{R}$$. By Euler's product : $$\zeta(s)=\prod_{p\in\mathcal{P}}\left(1-\frac{1}{p^s}\right)^{-1}$$ Thus $$\ln\zeta(s)=-\sum_{p\in\mathcal{P}}\ln\left(1-\frac{1}{p^s}\right)=\sum_{p\in\mathcal{P}}\sum_{k=1}^{+\infty}\frac{1}{kp^{ks}}=\sum_{p\in\mathcal{P}}\frac{1}{p^s}+\sum_{p\in\mathcal{P}}\sum_{k=2}^{+\infty}\frac{1}{kp^{ks}}$$ But for all $$p\in\mathcal{P}$$, $$\sum_{k=2}^{+\infty}\left|\frac{1}{kp^{ks}}\right|\leqslant\frac{1}{2}\sum_{k=2}^{+\infty}\frac{1}{p^{k\sigma}}\leqslant\frac{1}{2p^2}\sum_{k=0}^{+\infty}\frac{1}{2^k}=\frac{1}{p^2}$$ But $$\sum_{p\in\mathcal{P}}\frac{1}{p^2}<+\infty$$ so by Fubini's theorem : $$\ln\zeta(s)=\sum_{p\in\mathcal{P}}\frac{1}{p^2}+\sum_{k=2}^{+\infty}\sum_{p\in\mathcal{P}}\frac{1}{kp^{ks}}=\sum_{k=1}^{+\infty}\sum_{p\in\mathcal{P}}\frac{1}{kp^{ks}}$$ Moreover, $$\ln|\zeta(\sigma+it)|=\text{Re}(\ln\zeta(\sigma+it))=\sum_{k=1}^{+\infty}\sum_{p\in\mathcal{P}}\frac{\cos(t\ln(p^k))}{kp^{k\sigma}}$$ and $$\forall u\in\mathbb{R},3+4\cos(u)+\cos(2u)=2(1+\cos(u))^2\geqslant 0$$ thus $$3\ln\zeta(\sigma)+4\ln|\zeta(\sigma+it)|+\ln|\zeta(\sigma+2it)|\geqslant 0$$ You finally get $$(\star)\ \ \ \zeta(\sigma)^3|\zeta(\sigma+it)|^4|\zeta(\sigma+2it)|\geqslant 1$$ Now, let us suppose there exists $$t_0\in\mathbb{R}^*$$ such that $$\zeta(1+it_0)=0$$, $$\sigma\mapsto\zeta(\sigma+it_0)$$ is analytic on $$[1,+\infty)$$ and its only zero is $$1$$ thus there exists $$g$$ analytic, $$p\in\mathbb{N}^*$$ and $$\delta>0$$ such that $$\zeta(\sigma+it_0)=(\sigma-1)^pg(\sigma)$$ for all $$\sigma\in[1,1+\delta)$$ and $$g(1)\neq 0$$. Then $$\forall\sigma\in(1,1+\delta),\zeta(\sigma)^3|\zeta(\sigma+it_0)|^4=(\sigma-1)^3\zeta(\sigma)^3(\sigma-1)^{4p-3}|g(\sigma)|$$ $$\lim\limits_{\sigma\rightarrow 1^+}(\sigma-1)\zeta(\sigma)=1$$ and $$\lim\limits_{\sigma\rightarrow 1^+}(\sigma-1)^{4p-3}|g(\sigma)|=0$$ because $$4p-3\geqslant 1$$. You finally get $$\lim\limits_{\sigma\rightarrow 1^+}\zeta(\sigma)^3|\zeta(\sigma+it_0)|^4=0$$. Because of $$(\star)$$ we have $$\lim\limits_{\sigma\rightarrow 1^+}|\zeta(\sigma+2it_0)|=+\infty$$ and $$1+it_0\neq 1$$ which is not because the only pole of $$\zeta$$ is $$1$$.