A conjectural convergence condition for a weakened Elliott-Halberstam conjecture

Question

For $a$ and $q$ positive integers such that $a\lt q$ and $(a,q)=1$, let $\pi(x;q,a)$ be the number of primes $p\equiv a\pmod q$ below $x$. One can show that $\pi(x;q,a)\sim \dfrac{\pi(x)}{\varphi(q)}$ where $\pi(x)$ is the number of primes below $x$ and $\varphi(n)$ the number of positive integers $k$ less than $n$ such that $(k,n)=1$. Let's denote by $E(x,q)$ the quantity defined by $E(x,q):=\displaystyle{\max_{(a,q)=1}\vert\pi(x;q,a)-\dfrac{\pi(x)}{\varphi(q)}\vert}$.

Let's denote by $EH(\theta_{m})$ the following assertion:$\forall \theta\lt \theta_{m},\forall A>0, \exists C_{\theta,A},\displaystyle{\sum_{1\le q\le x^{\theta}}E(x,q)\le C_{\theta,A}\dfrac{x}{(\log x)^{A}}}$.

The still unproven Elliott-Halberstam conjecture is $EH(1)$.

Let's denote by $\mathcal{E}_{\theta}(x)$ the quantity $\displaystyle{\sum_{1\le q\le x^{\theta}}\dfrac{\varphi(q)E(x,q)}{\pi(x)}}$ and by $\theta_{s}$ the quantity $\displaystyle{\sup\{\theta, \lim_{x\to\infty} \mathcal{E}_{\theta}(x)\lt\infty\}}$.

Does $EH(\theta_{s})$ hold?

Thanks in advance.

Answer 1

It is easy to see that $\theta_s\leq 1/2$, hence $EH(\theta_{s})$ holds by the Bombieri-Vinogradov theorem.

To see the claim $\theta_s\leq 1/2$, assume that $1/2<\theta<1$, and let $1\leq q\leq x^\theta$. If $x$ and $q$ are fixed for a moment such that $E(x,q)<1/2$, then for $(a,q)=1$ the prime count $\pi(x;q,a)$ fluctuates less than $1$, hence it is constant. If, furthermore, $q$ itself is prime, then $ \sum_{(a,q)=1}\pi(x;q,a) = \pi(x)-1 $ shows that $\varphi(q)=q-1$ divides $\pi(x)-1$. That is, for fixed $x$, all but $O(x^\epsilon)$ primes $1\leq q\leq x^\theta$ satisfy $E(x,q)\geq 1/2$, and this implies that $$ \sum_{1\le q\le x^{\theta}}\dfrac{\varphi(q)E(x,q)}{\pi(x)} \gg \frac{\pi(x^\theta)}{\pi(x)}x^\theta\gg x^{2\theta-1}.$$ The right hand side tends to infinity, hence so does the left hand side.