= Uniform interval bound from optimal Selberg weights
{title2=$\pi(a+x)-\pi(a)\le(2+o(1))x/\log x$}
For all interval origins $a\ge0$, the number of multiples of any $d$ in $(a,a+x]$ is $x/d+O(1)$ with an absolute error bound. The <Selberg sieve diagonalization> minimizes $\sum_{d,e\le z}\lambda_d\lambda_e/[d,e]$, subject to $\lambda_1=1$, at $1/G(z)$. The <optimal Selberg weights have modulus at most one>, so the total remainder is $O(z^2)$. Restoring the at most $z$ small <primes> gives $\pi(a+x)-\pi(a)\le z+x/G(z)+O(z^2)$. Take $z=\lfloor\sqrt x/(\log x)^2\rfloor$ and use the <Selberg sieve denominator asymptotic>. The resulting $o(1)$ is independent of $a$.
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