# Riemann Z function, bounds on number of non-trivial zeros along horizontal lines, rather than vertical ones

Concerning the non-trivial zeros of the Riemann Zeta function, one can find quite a lot of literature on:

• the rate of growth of the number of zeros along the vertical critical line,

• the zero-free regions of the critical strip

• bounds on the number of hypothetical non-trivial zeros inside the critical strip, but off the critical line (though none have been found as of today, and never will if the RH is true)

However, I was unable to find any work concerned with estimates of the maximum number of hypothetical non-trivial zeros which may possibly lie on a same horizontal section of the critical strip (i.e. just the horizontal lines for fixed t values). Of course, it is well known that the functional equation implies that any hypothetical zero (1/2-a+it) must be symmetrically mirrored on the other side of the critical line by a zero (1/2+a+it). I imagine that such a set of hypothetical zeros would need to be discrete, as zeros of holomorphic functions are isolated, but I have no hint whatsoever about its cardinality (instinctively, we might feel that at most there are probably finitely many).

But perhaps some of you reading this question might know better.

-
Perhaps I am missing something, but since these zeros would all have to lie in the critical strip, their real paerts are bounded, if you now fix the imaginary part, you have a bounded set. So any inifinite set would have an accumaltion point. –  quid Mar 26 '13 at 18:07
I think that from standard asymptotics on the number of zeros of height up to T you get a bound of O(log T). Apparently even on RH is is only known that the multiplicity of 1/2 + iT as a zero is O(log T/log log T). So I guess you shouldn't expect too much better than this O(log T) bound, –  Ben Green Mar 26 '13 at 18:27
The number of (nontrivial) zeros with imaginary part between $0$ and $T$ is $\frac T{2\pi}\log\frac T{2\pi}-\frac T{2\pi}+O(\log T)$, so the number of zeros with imaginary part in $[T,T+1)$ is $O(\log T)$, with average about $\frac1{2\pi}\log T$. This includes roots both on and off the critical line. –  Emil Jeřábek Mar 26 '13 at 18:31

It $t$ is not an ordinate of a zero of $\zeta(s)$, define $$S(t) = \frac{1}{\pi} \arg \zeta(1/2+it) = -\frac{1}{\pi} \Im \int_{1/2}^\infty \frac{\zeta'}{\zeta}(\sigma+it) d\sigma$$ and define $$S(t)= \lim_{\delta\to 0} \frac{1}{2}\Big(S(t+\delta) + S(t-\delta)\Big)$$ otherwise. Then the number $N(T)$ of zeros of $\zeta(s)$ in the strip $0<\Im s \le T$ is $$N(T) = \frac{T}{2\pi}\log \frac{T}{2\pi e} +\frac{7}{8}+S(T)+O(\frac{1}{T})$$ where the big-$O$ term is actually continuously differentiable. For a proof, look either in Titchmarsh's book on the zeta-function or in Montgomery & Vaughan's "Multiplicative Number Theory, I."

By continuity, the quantity you are looking for is precisely $$\lim_{\delta\to 0} \Big(S(t+\delta) - S(t-\delta)\Big).$$ Unconditionally, I think Tim Trudgian has the best results for this quantity showing that $$|S(t)| \le 0.111 \log t + 0.275 \log \log t + 2.450$$ for $t>e$ (so your quantity is bounded by essentially twice this amount). This can be sharpened if $t$ is allowed to tend to infinity.

As is mentioned in previous comments/answers, assuming the Riemann hypothesis (RH) you are looking for bounds on the multiplicity of a zero. In this case, Goldston & Gonek showed that $$\lim_{\delta\to 0} \Big(S(t+\delta) - S(t-\delta)\Big) \le \Big(\frac{1}{2}+o(1)\Big) \frac{\log t}{\log \log t}$$ as $t\to\infty$ using the Guinand-Weil explicit formula.

References:

http://arxiv.org/pdf/1208.5846.pdf

http://arxiv.org/pdf/math/0511092v1.pdf

-