Is there a von Koch-type theorem for the generalized Riemann hypothesis?

Question

Helge von Koch proved in 1901 that the Riemann hypothesis is equivalent to the error term in the prime number theorem having the bound $$ \mid\pi(x)-\textrm{li}(x)\mid=O(\sqrt{x} \log x). $$

Q1: Is von Koch's result (and proof) also valid for the generalized Riemann hypothesis (GRH)?

That is, if a and d are coprime, and $\pi(x,a,d)$ denotes the number of primes in the arithmetic progression $a, a+d, a+2d, \dots$, is GRH equivalent to $$ \left|\pi(x,a,d)-\frac{1}{\varphi(d)}\textrm{li}(x)\right|=O(\sqrt{x} \log x)? $$ The Wikipedia page for GRH states that GRH implies this error term, but is short on references and says nothing about whether the given error term implies GRH.

Q2: Does anyone know if von Koch's paper has been translated to English; alternatively a reference to the best modern version of proving his result; or perhaps both?

Answer 1

As GH from MO and Felipe Voloch have already indicated it is standard to show that $\psi(x;q,a) = x/\phi(q) +O(x^{\frac 12+\epsilon})$ for all reduced residue classes $a\pmod q$ is equivalent to GRH for the characters $\pmod q$. I want to make the following small (but amusing) refinement: it is enough to know that $\psi(x;q,1) = x/\phi(q) + O(x^{\frac12+\epsilon})$ and $\psi(x) =x+ O(x^{\frac 12+\epsilon})$ and from these two pieces of information (rather than all $\phi(q)$ residue classes) we can get GRH for all the characters $\pmod q$.

To see this, note that (starting with Re$(s)>1$) $$ -\frac{1}{\phi(q)} \sum_{\chi\neq \chi_0} \frac{L^{\prime}}{L}(s,\chi) = \int_1^{\infty} \frac{s}{x^{s+1}} \Big( \psi(x;q,1) - \frac{\psi(x,\chi_0)}{\phi(q)} \Big) dx, $$ and so by hypothesis, the RHS extends analytically to Re$(s)>1/2$. Therefore $L(s,\chi) \neq 0$ for all $\chi \neq \chi_0$ and Re$(s)>1/2$. Finally $\psi(x)=x+O(x^{\frac 12+\epsilon})$ implies RH; thus GRH follows for all the characters $\pmod q$.

Put differently, this shows that the asymptotic $\psi(x;q,1) = \psi(x)/\phi(q) + O(x^{\frac 12+\epsilon})$ implies $\psi(x;q,a) =\psi(x)/\phi(q)+ O(x^{\frac 12+\epsilon})$ for all $(a,q)=1$.

Answer 2

The bound you state follows from GRH, see Corollary 13.8 in Montgomery-Vaughan: Multiplicative number theory I. Conversely, the bound you state implies GRH. I provide the proof below.

Assume that for all coprime pairs $a$ and $d$ we have $$ E(x,a,d):=\pi(x,a,d)-\frac{\mathrm{li}(x)}{\varphi(d)} = O(\sqrt{x}\log x). $$ Let us introduce $$ \theta(x,a,d):=\sum_{\substack{p\leq x\\p\equiv a\pmod{d}}} \log p \qquad\text{and}\qquad\psi(x,a,d):=\sum_{\substack{n\leq x\\n\equiv a\pmod{d}}} \Lambda(n). $$ Then $$\theta(x,a,d) - \frac{x-2}{\varphi(d)} = \int_{2-}^x \log t\ d E(t,a,d) = E(x,a,d)\log x-\int_2^x\frac{E(t,a,d)}{t}dt,$$ whence $$\psi(x,a,d)=\theta(x,a,d)+O(\sqrt{x}\log x) = \frac{x}{\varphi(d)}+O(\sqrt{x}\log^2 x).$$ As a result, for any nontrivial Dirichlet character $\chi$ modulo $d>1$ we have $$\psi(x,\chi):=\sum_{n\leq x}\chi(n)\Lambda(n)=\sum_{\substack{1\leq a\leq d\\(a,d)=1}}\chi(a)\psi(x,a,d)=O(\sqrt{x}\log^2 x).$$ This implies that the Dirichlet $L$-function $L(s,\chi)$ has no zero in the half-plane $\Re(s)>1/2$, because its logarithmic derivative is analytic there: $$ -\frac{L'(s,\chi)}{L(s,\chi)}=\sum_{n=1}^\infty \frac{\chi(n)\Lambda(n)}{n^s}=\int_{2-}^\infty t^{-s}\ d\psi(t,\chi)=s\int_2^\infty \frac{\psi(t,\chi)}{t^{s+1}}\,dt.$$ Analycity follows from the local uniform convergence of the last integral in the half-plane $\Re(s)>1/2$.