# Is the result of Schmidt conditional to RH

A theorem due to Erhard Schmidt states that, for some explicit positive constant $$K$$, there are infinitely many natural numbers $$x$$ such that $$ψ(x)>x+K√x$$

My question is: Is the result of Schmidt conditional to Riemann Hypothesis.

• Even if Schmidt's proof uses RH, it is relatively easy to prove much stronger lower bounds assuming RH is false. – Wojowu Dec 28 '19 at 11:36
• @Wojowu: I know that. I am asking on that result if it is related to RH or its negation. – Safwane Dec 28 '19 at 11:38

1. It is known unconditionally that, as $$x$$ tends to infinity, $$\psi(x)-x=\Omega_{\pm}(x^{1/2}).\tag{1}$$ This is Corollary 15.4 in Montgomery-Vaughan: Multiplicative number theory I.

2. In fact Hardy and Littlewood proved the stronger result $$\psi(x)-x=\Omega_{\pm}(x^{1/2}\log\log\log x).\tag{2}$$ This is Theorem 15.11 in Montgomery-Vaughan: Multiplicative number theory I.

3. Schmidt (1903) proved the analogue of $$(1)$$ for the function $$f(x):=\sum_{k=1}^\infty\frac{1}{k}\pi(x^\frac{1}{k}).$$ His proof is essentially the same as of the above quoted Corollary: if the Riemann Hypothesis is false, then one has a better result, while if the Riemann Hypothesis is true, then one has a precise form of the stated result with an implied constant given in terms of the lowest lying zero of $$\zeta(s)$$. So Schmidt's result is unconditional as well, but it differs slightly from the statement attributed to him in the Wikipedia article.

4. Hardy and Littlewood (1916) attribute $$(1)$$ to Schmidt, and they quote it as Theorem 2.241. Precisely, they say that "This is substantially the well-known result of Schmidt". The stronger statement $$(2)$$ is Theorem 5.8 in their paper.

P.S. As Greg Martin kindly pointed out, $$(2)$$ is really due to Littlewood (1914).

• But still the result appeared in Wikipedia is true and unconditional as I read in the linked paper. Page 139, Theorem 2.241. – Safwane Dec 28 '19 at 12:39
• @Helena: I agree, and this is what item 1 in my post is about. I also added items 2 and 4 in my post. – GH from MO Dec 28 '19 at 14:00
• Remark: the result (2) is actually due to Littlewood alone (even though its full proof first appeared in a paper of Hardy and Littlewood). – Greg Martin Dec 29 '19 at 5:30
• @GregMartin: Thank you! See the "P.S." section in my post. – GH from MO Dec 29 '19 at 6:02

It is elementary, take $$K=1/30 < 1/|\rho_0|$$ where $$\rho_0\approx 1/2+i14.13$$ is the first zero, if $$\psi(x)-x\le Kx^{1/2}$$ for $$x$$ large enough, then $$x-\psi(x)+K x^{1/2}+C\ge 0$$ for all $$x$$, where $$C$$ is a suitable real constant. Let $$F(s):=\int_1^\infty (x-\psi(x)+Kx^{1/2}+C) x^{-s-1}dx= \frac1{s-1}+\frac{\zeta'(s)}{s\zeta(s)}+\frac{K}{s-1/2}+\frac{C}{s}.$$ By the non-negativity of the integrand, it has a singularity at its abscissa of convergence $$\sigma$$. But the RHS is analytic on $$(1/2,\infty)$$, thus $$\sigma=1/2$$. And by the non-negativity again we have, as $$\Re(s) \to 1/2$$, $$|F(s)|\le F(\Re(s))\sim \frac{K}{\Re(s)-1/2}.$$ This contradicts that $$F(s)\sim\frac{\zeta'(s)}{s\zeta(s)}\sim\frac{1/\rho_0}{s-\rho_0}\qquad\text{as s\to \rho_0}.$$

• Showing that $K$ arbitrary large works should be much more difficult – reuns Dec 29 '19 at 2:20