Zeros of the derivative of Riemann's $\xi$-function

Question

The Riemann xi function $\xi(s)$ is defined as $$ \xi(s)=\frac12 s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s). $$ It is an entire function whose zeros are precisely those of $\zeta(s)$. Since $\xi$ is real valued on the critical line $s=1/2+it$, there is a zero of the derivative $\xi^\prime$ between each successive pair of zeros of $\xi$, and thus the theorem of Levinson shows that at least $1/3$ (since improved) of the zeros of $\xi^\prime$ lie on the critical line.

In Zeros of the derivative of Riemann's $\xi$-function BAMS v. 80 (5) 1974 pp. 951-954, Levinson adapted his method to show directly that more than $7/10$ of the zeros of $\xi^\prime(s)$ occur on the critical line. In the proof he writes, (with $H(s)=\frac12 s(s-1)\pi^{-s/2}\Gamma(s/2)$, $F(s)$ defined by $H(s)=\exp(F(s))$, and $G(s)$ complicated in terms of $\zeta(s)$ and $\zeta^\prime(s)$)

"… then (3) becomes $$ \xi^\prime(s)=F^\prime(s)H(s)G(s)-F^\prime(1-s)H(1-s)G(1-s) $$ … by Stirling's formula $\arg H(1/2+it)$ changes rapidly and by itself would supply the full quota of zeros of $\xi^\prime(s)$ on $\sigma=1/2$."

This is as close as he comes in the paper to suggesting that all the zeros of $\xi^\prime$ are on the critical line.

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Does this conjecture explicitly appear anywhere in the literature? Is it folklore?

Answer 1

In exercise 1 on page 443 of their book "Multiplicative Number Theory," Montgomery & Vaughan outline a proof of the statement:

"Assuming the Riemann Hypothesis, $\xi'(s)=0 \implies \mathrm{Re}(s)=1/2$."

Assuming RH, let $s=\sigma+it$ and let $\rho=\frac{1}{2}+i\gamma$ denote a zero of $\xi(s)$. The main idea of their argument is that, on RH, it follows from that Hadamard product for $\xi(s)$ that $$ \mathrm{Re} \frac{\xi'}{\xi}(s) = \sum_{\rho} \mathrm{Re}\frac{1}{s-\rho} = \sum_\gamma \frac{\sigma-1/2}{(\sigma-1/2)^2+(t-\gamma)^2}.$$ Now if $\xi'(s)=0$, then the left-hand side of the above expression is zero. On the other hand, the only way that the sum over $\gamma$ vanishes is if $\sigma=1/2$, i.e. $\mathrm{Re}(s)=1/2$.

Answer 2

The Riemann hypothesis implies that the zeros of derivatives of all orders of $\xi$ lie on the critical line.

B. Conrey, Zeros of derivatives of Riemann’s xi-function on the critical line, J. Number Theory 16 (1983), 49-74.

Answer 3

The Riemann hypothesis implies that the function $\Xi(z)=\xi(1/2+iz)$ is in the Laguerre-Pólya class. Therefore it is a limit, uniformly on compact sets, of a sequence of polynomials with real roots. The derivatives are also again in the same class and have therefore only real zeros. It follows that all zeros of $\xi(s)$ and its derivatives will be on the critical line.

I imagine that this is due to Pólya, but have not his Collected Works to confirm.

Given the simplicity of the proof I noticed, I will give a sketch of it. We have $$\Xi(t)=\Xi(0)\prod_{n=1}^\infty \Bigl(1-\frac{t^2}{\alpha_n^2}\Bigr).$$ The Riemann hypothesis is that all $\alpha_n$ are real.

Therefore, assuming RH, $\Xi(t)$ is the limit uniformly in compact sets of $\bf C$ of the polynomials $$P_N(t):=\Xi(0)\prod_{n=1}^N \Bigl(1-\frac{t^2}{\alpha_n^2}\Bigr).$$ We are assuming that all roots of these polynomials are real. Therefore the same will happen with any derivative $P_N^{(k)}(t)$.

By the general Theorems of Complex Analysis $\lim_{N\to\infty}P_N^{(k)}(t)=\Xi^{(k)}(t)$ uniformly in compact sets. By the argument principle any zero of $\Xi^{(k)}(t)$ is limit of zeros of $P_N^{(k)}(t)$. Therefore any zero of $\Xi^{(k)}(t)$ is real. The relation $\Xi(t)=\xi(\frac12+it)$ implies that all the zeros of the derivatives of $\xi(s)$ are in the critical line.

(It is essentially contained in the paper by G. Pólya, Bemerkung zur Theorie der ganzen Funktionen, Collected papers II, 154--162.)

Answer 4

The same argument as in the Gauss-Lucas theorem (adapted to an entire function with appropriate order) works for showing that

If $\xi(1/2+s)$ has no zeros on $|Re(s)| > \alpha$,

then $\xi'(1/2+s)$ has no zeros on $|Re(s)| > \alpha$.

grouping the term $\rho$ with its complex conjugate $\overline{\rho}$,

$\displaystyle \frac{\xi'(s)}{\xi(s)} = \sum_\rho \frac{1}{s-\rho}$ converges conditionally so that

$$\begin{array}{l}\displaystyle\ \ \ \ \xi'(\beta)=0 \\ \displaystyle\text{and }\xi(\beta)\ne 0\end{array}\implies \frac{\xi'(\beta)}{\xi(\beta)} = \sum_\rho \frac{\overline{\beta}-\overline{\rho}}{|\beta-\rho|^2}=0\quad\implies \quad\beta=\frac{\displaystyle\sum_\rho \frac{\rho}{|\beta-\rho|^2}}{\displaystyle\sum_{\rho}\frac{1}{|\beta-\rho|^2}}$$ which is a weighted sum of the zeros of $\xi(s)$, i.e. the zeros of $\xi'(s)$ lie in the convex hull of the zeros of $\xi(s)$.

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Now for $\zeta(s)$ it is different, the zeros of $\zeta'(s)$ are not in the convex hull of the zeros of $\zeta(s)$, since $\zeta(s)$ has a pole at $s=1$ and the trivial zeros sum $\sum_{n=2}^\infty \frac{1}{s+2n}$ diverges.

See zeros of $\zeta'(s)$ and the Riemann hypothesis for a proof that under the RH $\zeta'(s)$ has no zeros on $Re(0,1/2)$. I'd like if someone could derive a similar theorem assuming instead that $\zeta(s)$ has no zeros on $Re(s) > \sigma_0$.

Answer 5

Related proof is in Lagarias paper p.1.

The RH is equivalent to

$$ \Re\left(\frac{\xi'(s)}{\xi(s)}\right) > 0 $$

When $\Re(s) > \frac12$.