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?

derivative"...? $\endgroup$ – paul garrett Dec 15 '14 at 23:05