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.

  • $\begingroup$ 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. $\endgroup$ Apr 8, 2020 at 18:47
  • $\begingroup$ 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. $\endgroup$
    – user208649
    Apr 8, 2020 at 19:05
  • $\begingroup$ 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) $\endgroup$
    – reuns
    Apr 9, 2020 at 0:28

1 Answer 1


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 $1$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .