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From this page: https://en.wikipedia.org/wiki/Chebyshev_function#Asymptotics_and_bounds

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

Since the original paper is in German. I read this paper: https://projecteuclid.org/download/pdf_1/euclid.acta/1485887467

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

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

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  • $\begingroup$ But still the result appeared in Wikipedia is true and unconditional as I read in the linked paper. Page 139, Theorem 2.241. $\endgroup$ – Safwane Dec 28 '19 at 12:39
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    $\begingroup$ @Helena: I agree, and this is what item 1 in my post is about. I also added items 2 and 4 in my post. $\endgroup$ – GH from MO Dec 28 '19 at 14:00
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    $\begingroup$ Remark: the result (2) is actually due to Littlewood alone (even though its full proof first appeared in a paper of Hardy and Littlewood). $\endgroup$ – Greg Martin Dec 29 '19 at 5:30
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    $\begingroup$ @GregMartin: Thank you! See the "P.S." section in my post. $\endgroup$ – GH from MO Dec 29 '19 at 6:02
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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 $$F(s)\sim\frac{\zeta'(s)}{s\zeta(s)}\sim\frac{1/\rho_0}{s-\rho_0}\qquad\text{as $s\to \rho_0$}.$$

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

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