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I found in my research regarding the Riemann $\zeta$ function (RZF) a fantastic result of Littlewood. If $\{\gamma_n\}$ is an increasing sequence of the imaginary parts of the zeros of $\zeta$ on the critical line in the upper complex half-plane, then $$ \lim\limits_{n\to\infty} |\gamma_{n+1}-\gamma_n|=0 ~~.$$

I am trying to understand this very well, so I am trying to replicate the proof step-by-step to ensure I understand everything. Specifically, I am working through the proof of this theorem given in article 9.12 of Titchmarsh (1986). However, the specific result which is proven in Titchmarsh, and also originally in Littlewood (1924), is

For every large $T$, $\zeta(s)$ has a zero $\beta+i\gamma$ satisfying $$ |\gamma-T|<\dfrac{A}{\log\log\log T} ~~. $$

Unfortunately, before I can even begin to work through the proof, it is not immediately obvious to me how this statement is equivalent to the statement above about the limit of the difference tending to zero. In the Titchmarsh theorem, $A$ is (I suppose) some absolute constant, but the Titchmarsh theorem says to me only that the difference between $\gamma$ and $T$ is always small because $\log\log\log T$ is large for very large $T$.

Could the community please help me make the connection between the inequality form of the theorem given by Titchmarsh and the most popular modern statement of the result given above?

My guess is that in the limit $T\to\infty$, the difference between $\gamma$ and $T$ has to be "less than 0," and that the less than gets turned into "$\leq$" in the limit. Then, because any finite patch of the critical strip bounded above by $T$ on the imaginary axis has at most finitely many zeros in it, the distance between the zeros has to go to zero. If this is right, could someone help me formalize my statement? If I'm wrong, please help! THANKS!!!

Jimmy Rankerman
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Note first that you quote quite an imprecise form of the (unconditional) result which is that if $\gamma_n$ is the sequence of the positive imaginary parts of the critical zeroes of RZ (so not nececsarily zeroes on the critical line) ordered increasingly, then:

$\lim\limits_{n\to\infty} |\gamma_{n+1}-\gamma_n| =0$

(without RH you cannot assume these are all on the critical line and also you need to take the sequence consisting of all the upper plane critical zeroes).

To show how this follows from the result of Littlewood showing that the spacing between any $T$ and the critical roots of $\zeta$ satisfies $|\gamma-T|<\dfrac{A}{\log\log\log T}$ we can proceed by contradiction:

Assume $\lim\limits_{n\to\infty} |\gamma_{n+1}-\gamma_n| \ne 0$. This means there is $m(n) \to \infty, a>0$, $|\gamma_{m(n)+1}-\gamma_{m(n)}| > 2a$.

Pick $T_n=\frac{\gamma_{m(n)+1}+\gamma_{m(n)}}{2}, T_n \to \infty$, and then trivially $|\gamma-T_n| >a$ for all critical zeroes of $\zeta$ contradicting Littlewood's result for all $n$ large enough s.t. $\dfrac{A}{\log\log\log T_n} < a$

Conrad
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  • Thanks! Maybe you will see this in an important paper one day. – Jimmy Rankerman Nov 19 '19 at 23:08
  • Can you tell me why you choose the $m(n)$ notation instead of just using $n$ directly? If $T_n$ is directly associated to $\gamma_n$, then I don't see what you're doing with some separate $m$. Thanks again though, your answer helped me a lot! – Jimmy Rankerman Nov 27 '19 at 16:05
  • The negation of the property of a limit being zero is precisely what I have written (there is a subsequence that in absolute value is bounded above by a positive constant); we still could have subsequences of the gammas that have the limit here going to zero, so we cannot write that inequality for all $n$ – Conrad Nov 27 '19 at 16:20