Why does zeta have infinitely many zeros in the critical strip?

Question

I want a simple proof that $\zeta$ has infinitely many zeros in the critical strip.

The function $$\xi(s) = \frac{1}{2} s (s-1) \pi^{\tfrac{s}{2}} \Gamma(\tfrac{s}{2})\zeta(s)$$ has exactly the non-trivial zeros of $\zeta$ as its zeros ($\Gamma$ cancels all the trivial ones out). It also satisfies the functional equation $\xi(s) = \xi(1-s)$.

If we assume it has finitely many zeros, what analysis could get a contradiction?

I found an outline for a way to do it here but I can't do the details myself: https://mathoverflow.net/questions/13647/why-does-the-riemann-zeta-function-have-non-trivial-zeros/13762#13762

Answer 1

Hardy proved in 1914 that an infinity of zeros were on the critical line ("Sur les zéros de la fonction $\zeta(s)$ de Riemann" Comptes rendus hebdomadaires des séances de l'Académie des sciences. 1914).
Of course other zeros could exist elsewhere in the critical strip.

Let's exhibit the main idea starting with the Xi function defined by : $$\Xi(t):=\xi\left(\frac 12+it\right)=-\frac 12\left(t^2+\frac 14\right)\,\pi^{-\frac 14-\frac{it}2}\,\Gamma\left(\frac 14+\frac{it}2\right)\,\zeta\left(\frac 12+it\right)$$ $\Xi(t)$ is an even integral function of $t$, real for real $t$ because of the functional equation (applied to $s=\frac 12+it$) : $$\xi(s)=\frac 12s(s-1)\pi^{-\frac s2}\,\Gamma\left(\frac s2\right)\,\zeta(s)=\frac 12s(s-1)\pi^{\frac {s-1}2}\,\Gamma\left(\frac {1-s}2\right)\,\zeta(1-s)=\xi(1-s)$$ We observe that a zero of $\zeta$ on the critical line will give a real zero of $\,\Xi(t)$.

Now it can be proved (using Ramanujan's $(2.16.2)$ reproduced at the end) that : $$\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}\cos(x t)\,dt=\frac{\pi}2\left(e^{\frac x2}-2e^{-\frac x2}\psi\left(e^{-2x}\right)\right)$$ where $\,\displaystyle \psi(s):=\sum_{n=1}^\infty e^{-n^2\pi s}\ $ is the theta function used by Riemann

Setting $\ x:=-i\alpha\ $ and after $2n$ derivations relatively to $\alpha$ we get (see Titchmarsh's first proof $10.2$, alternative proofs follow in the book...) : $$\lim_{\alpha\to\frac{\pi}4}\,\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh(\alpha t)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$ Let's suppose that $\Xi(t)$ doesn't change sign for $\,t\ge T\,$ then the integral will be uniformly convergent with respect to $\alpha$ for $0\le\alpha\le\frac{\pi}4$ so that, for every $n$, we will have (at the limit) : $$\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh\left(\frac {\pi t}4\right)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$

But this is not possible since, from our hypothesis, the left-hand side has the same sign for sufficiently large values of $n$ (c.f. Titchmarsh) while the right part has alternating signs.
This proves that $\Xi(t)$ must change sign infinitely often and that $\zeta\left(\frac 12+it\right)$ has an infinity of real solutions $t$.

Probably not as simple as you hoped but a stronger result! $$-$$

From Titchmarsh's book "The Theory of the Riemann Zeta-function" p. $35-36\;$ and $\;255-258$ :

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Answer 2

It is known that $$ \xi(s)=\frac12\prod_{\xi(s)=0}\left(1-\frac s\rho\right),$$ i.e. $\xi$ would turn out to be a polynomial of degree $n$, say. Then we conclude that $\ln \xi(s)\sim n\ln s$ as $\mathbb R \ni s\to \infty$, but it is known that $\ln \xi(s)\sim s\ln s$.

Answer 3

In 1912 Harald Bohr proved that there are infinitely many zeros of $\zeta(s)$ in $0\leq \sigma \leq 1$ by contradiction. He assumes that there are no zeros in the critical strip and deduces from this that that $1/|\zeta(s)|$ is bounded, contradicting his (earlier extrinsic) result that there exist arbitrarily small values of $\zeta(s)$ for a given range of $\sigma.$

This is as I recall simpler than Hadamard's proof and far simpler than the proof of Hardy's stronger result. As far as I know there is no English version but the Danish is accessible with online translation tools. As I recall the theorem makes use of Caratheodory's theorem to establish a bound.

If I make it back to my lending library I will try to post a better synopsis but for anyone who is interested here is the cite: Et nyt for, at den Riemann'ske Zetafunktion $\zeta(s) = \zeta(\sigma + it)$ har uendelig mange Nulpunkter idenfor Parallelstrimlen $0\leq \sigma\leq 1.$ B9, Harald Bohr, Collected Math. Works, Vol. I.

Otherwise there is Hadamard's theorem which is given in Titchmarsh's Theory of the Riemann Zeta Function at p. 30. A better synopsis of this approach than I could give is found in a brief note by Paul Garrett but I think Hagen von Eitzen's answer is also a good distillation of this.

Answer 4

(I think this is an elementary proof)

everybody here knows that the zeros of $\zeta(s)$ are useful mainly for the Riemann explicit formula :

$$\psi(x) = \sum_{p^k \le x} \ln p = x - \sum_{\rho} \frac{x^\rho}{\rho}-\ln(2 \pi)- \frac{1}{2}\ln(1-x^{-2})$$

suppose there is a finite number of non-trivial zeros, then $\frac{d}{dx}\psi(x)$ could not be the distribution : $$\sum_{p \in \mathcal{P}} \sum_{k=1}^\infty \ \delta(x-p^k) \ \ln p $$

Answer 5

Here is a proof which is in line with what Hagen von Eitzen wrote but including more details. Beside the product expansion of $\xi(s)$, I will only use elementary techniques. To simplify things a bit, I will use the function $$\xi(s)=s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$$ instead (dropping the factor $\frac{1}{2}$).

The function $\xi(s)$ has the well known product expansion via the Hadamard factorization theorem: $$\xi(s)=e^{As}\prod_{\rho}\biggl(1-\frac{s}{\rho}\biggr)e^{s/\rho},$$ where the product runs over all nontrivial zeros $\rho$ of $\zeta(s)$ and $A$ is some complex constant independent of $s$. This formula is included in most textbooks on analytic number theory. (I don't know how the formula can be simplified as in the Wikipedia article cited by Hagen von Eitzen. Probably requires some additional work. Anyway, it isn't necessary for our task.)

Now assume there are only finitely many nontrivial zeros. The above product formula can then be rewritten as $$\xi(s)=e^{As}\prod_{\rho}\biggl(1-\frac{s}{\rho}\biggr)\prod_{\rho}e^{s/\rho},$$ where both products $\prod_{\rho}$ are now finite. The product over $(1-\frac{s}{\rho})$ is thus a polynomial in $s$, say $\prod_{\rho}(1-\frac{s}{\rho})=P(s)$ with $P \in \mathbb{C}[X]$. The product over $e^{s/\rho}$ is equal to $e^{Bs}$ with the finite sum $B=\sum_{\rho}\frac{1}{\rho}$. Thus, writing $C=A+B$, we have $$\xi(s) = e^{Cs}P(s)\text{.}$$ Take a number $n \in \mathbb{N}$ and set $s=2n$. Then take from both side the absolute value, the logarithm and divide by $n\log n$. This gives $$\frac{\log\xi(2n)}{n\log n} = \frac{2\text{Re }C}{\log n} + \frac{\log|P(2n)|}{n\log n}\text{.}\tag{$*$}$$ The idea is now to derive a contradiction by letting $n\to \infty$ on both sides. For the right hand side, it is easy to see that it goes to $0$ (to handle the polynomial $P(2n)$, write $P(X)=a_kX^k+...+a_1X+a_0$ to see this). For the left hand side, we show that it tends to $1$ instead. To accomplish this, one can show the following two statements:

$\phantom{aaaaa}$(1)$\phantom{a}$ $\log \Gamma(n) \sim n\log n$ for $n \to \infty$.

$\phantom{aaaaa}$(2)$\phantom{a}$ $\zeta(r)\to 1$ for $r \to \infty$ with $r$ real.

The first statement can be checked by writing $\Gamma(n)=(n-1)!$ and then using only elementary knowledge about sequences and series (see for example some of the answers here: Limit of $\frac{\log(n!)}{n\log(n)}$ as $n\to\infty$.). For the second statement, assume $r>1$ and wirte $\zeta(r)=\sum_{n=1}^{\infty} 1/n^r$. By uniform convergence, you can exchange the sum with the limit $\lim_{r\to\infty}$ which will prove (2).

With the two statements, we can easily show that the left hand side of ($*$) goes to $1$: Using the definition of $\xi(s)$, we can write $$\frac{\log\xi(2n)}{n\log n} \ = \ \frac{\log(2n)}{n\log n}+\frac{\log(2n-1)}{n\log n} + \frac{\log(\pi^{-n})}{n\log n} + \frac{\log\Gamma(n)}{n\log n} + \frac{\log\zeta(2n)}{n\log n}\text{.}$$ The first three terms on the right hand side clearly tend to $0$. Ther term with $\log\zeta(2n)$ tends to $0$ as well, by statement (2) from above. By statement (1), the term with $\log\Gamma(n)$ goes to 1.