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.



A conjectural convergence condition for a weakened Elliott-Halberstam conjecture
It is easy to see that $\theta\_s\leq 1/2$, hence $EH(\theta\_{s})$ holds by the [Bombieri-Vinogradov theorem](http://en.wikipedia.org/wiki/Bombieri%E2%80%93Vinogradov_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.


