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.
Does this conjecture explicitly appear anywhere in the literature? Is it folklore?
Best Answer
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. $\mathrm{Re}(s)=1/2$.