# Chebychev Function in Arithmetic Progressions

In the paper "exponential sums with multiplicative coefficients", Maier claims that the Chevbychev function in arithmetic progression satisfies

$\theta(x, r, a) = x/(r-1) +O(x^{1-1/r^\epsilon})$ for sufficently large prime $r$ and arbitrary small $\epsilon>0$.

Is this an error?

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The last item in the proof of Lemma 3.7 is what you ALMOST typed, in that it should be an equals sign (as I edited in) and the hypothesis $x \geq r$ is part of the lemma, and not something you typed. Also the order is x,r,a. – Will Jagy Nov 5 '11 at 22:20

## 3 Answers

I feel like this is an error. It is true that $1-1/r^\epsilon$ is a known upper bound for the real zeros of Dirichlet $L$-functions corresponding to characters $\chi\pmod r$ (for $r$ sufficiently large in terms of $\epsilon$). However, no constant less than $1$ is known to be an upper bound for the real parts of all the zeros; the zero-free region is similar to the classical one for the Riemann $\zeta$ function, so that the best we know for zeros $\beta+i\gamma$ of $L(s,\chi)$ is something like $\beta \le 1-c/\log rt$ for some positive constant $c$. This would give something like $\theta(x;r,a) = x/(r-1) + O_r(x \exp(-c\sqrt{\log x}))$, which is weaker than what is claimed.

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– Will Jagy Nov 6 '11 at 20:03
I see the claimed result in the paper you linked to. I still don't see why the claimed result should be true. But I'm willing to be convinced by a mathematical argument. – Greg Martin Nov 7 '11 at 6:31
No question the claimed result is false, but perhaps Maier doesn't need it. The bound is OK for $x$ not too large, see my response. I think Will Jagy wants someone to read Maier's paper and sort this out. – GH from MO Nov 7 '11 at 18:10

Greg Martin is right: Maier's claim is in error. Precisely, the following can be said (cf. Corollaries 11.15 and 11.20 in Montgomery-Vaughan: Multiplicative number theory I):

Theorem (Siegel-Walfisz with a twist). Let $\epsilon>0$ be fixed. Let $q\geq q_0(\epsilon)$ be sufficiently large in terms of $\epsilon$. There is an absolute constant $c>0$ such that $$\theta(x,q,a)=\frac{x}{\varphi(q)}+O(x^{1-cq^{-\epsilon}})\qquad\text{for}\qquad x\leq e^{q^{2\epsilon}},$$ and $$\theta(x,q,a)=\frac{x}{\varphi(q)}+O(x e^{-c\sqrt{\log x}})\qquad\text{for}\qquad x>e^{q^{2\epsilon}}.$$ The implied constants are absolute.

Perhaps Maier only needs the range $x\leq\exp(q^{2\epsilon})$, so his final conclusions are OK.

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– Will Jagy Nov 6 '11 at 20:03
GH, I see your comment below Greg's answer. It is true it would be nice if someone straightened out the whole business. Wrong claims are an irritant, even in a field about which I know nothing. However, the reason I put three copies of the link is that users are notified of comments below their own answers (in time, anyway), but not typically notified of comments elsewhere, or of edits to the original question. Mostly I was annoyed that the OP mistyped the relation and then failed to give any motivation. – Will Jagy Nov 7 '11 at 20:35
@Will: Thank you! – GH from MO Nov 8 '11 at 16:24

To Will Jagy: Sorry about the typing errors. The motivation is that it can be used to get a zero-free region of L-function better than anything on the literature, as discussed by Greg martin, and everything is elementary number theory.

Maier claims that it holds for all x => r, and said that the proof is a simple, see the paper. So the answer posted by G. H. should be for all x => r.

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The claimed result would be fantastic but it is not known unfortunately. In particular, it is certainly not a "simple consequence of Siegel-Walfisz". I explained what Siegel-Walfisz means and that is the best result known almost 80 years after Siegel-Walfisz. More precisely, Vinogradov improved on the error term a little, but it is nothing like what Maier claims. Just because a famous researcher makes a claim, it does not make it valid. – GH from MO Nov 12 '11 at 22:37