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Write $\psi(x) = \sum_{n\le x} \Lambda(n)$. The classical omega theorem says that

$\psi(x) - x = \Omega_{\pm}(x^{1/2})$.

Question: How often does this hold? For example, what do we know about the size of the set

$\{ n\le x: \psi(x) - x > c x^{1/2} \}$

for some $c$? Ditto for $\psi(x) - x < c x^{1/2}$. What about replacing e.g. $x^{1/2}$ by $x^{\alpha}$ for some fixed $0 < \alpha < 1/2$? What's a good reference for such results? Thanks!

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  • $\begingroup$ Not to be confused with the Omega lemma! $\endgroup$
    – David Roberts ♦
    Commented Mar 24, 2022 at 0:50

1 Answer 1

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For any $\varepsilon>0$, there exist $c(\varepsilon)>0$ and $X_0(\varepsilon)>0$ such that for any $X>X_0(\varepsilon)$ we have $$\sup_{X\leq x\leq X^{1+\varepsilon}}\frac{\psi(x)-x}{\sqrt{x}\log\log x}>c(\varepsilon)\qquad\text{and}\qquad \inf_{X\leq x\leq X^{1+\varepsilon}}\frac{\psi(x)-x}{\sqrt{x}\log\log x}<-c(\varepsilon).$$ More precisely, Ingham (1935) proved a stronger result for the case when the real parts of the zeta zeros have a maximum, while Pintz (1980) proved a stronger result for the case when the real parts of the zeta zeros do not have a maximum. There might be even stronger results in the literature, please check.

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  • $\begingroup$ Thanks for the references. Ingram's paper make reference to a theorem of Bohr on diophantine approximation. I'm new to this subject; could you give me an English reference of Bohr's result, please? Thanks! $\endgroup$
    – W Sao
    Commented Apr 5, 2022 at 2:04
  • $\begingroup$ @WSao I recommend that you study Bohr's book "Almost periodic functions". Consider $f(x):=\sum_{n=1}^N\max\left(0,1-q\left\|c_n x\right\|\right)$, where $c_n:=\gamma_n/(2\pi)$. This is a sum of continuous periodic functions, hence it is almost periodic (as shown by Bohr). In particular, every sufficiently long interval contains $U$ such that $\sup_x|f(x+U)-f(x)|<1$. Take such a $U$. From $f(0)=N$ it follows that $f(U)>N-1$, so none of the $N$ terms in the definition of $f(x)$ vanishes at $U$. That is, $\|c_n U\|<1/q$ holds for $n\in\{1,\dotsc,N\}$. This is the claim that Ingham uses. $\endgroup$
    – GH from MO
    Commented Apr 7, 2022 at 5:19

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