Research proof index

Every zero $s$ of the Riemann zeta function in the upper half-plane lies in the critical strip: if $\zeta(s)=0$ and $\operatorname{Im}s>0$, then $0\le\operatorname{Re}s\le1.$

The Riemann zeta zero-counting function has a crude polynomial bound: there is a constant $A>0$ such that, for every $T\ge2$, $|N_\zeta(T)|\le A T^{3/2}.$

Let $Z(s)$ be the project's entire zeta surrogate, obtained by removing the pole of $\zeta(s)$ at $s=1$. There is a constant $C>0$ such that, for every $R\ge1$, the total multiplicity of the zeros of $Z$ in $|s|\le R$ is bounded by $C(1+R)^{3/2}.$

A closed-disc form of the Borel–Carathéodory theorem. Let $R,M>0$, let $f$ be analytic on $|z|\le R$, assume $f(0)=0$, and suppose $\operatorname{Re}f(z)\le M$ throughout that disc. If $r<R$ and $|z|\le r$, then $|f(z)|\le \frac{2Mr}{R-r}.$

For the Selberg sieve that removes primes up to $z$ from an interval $[x,x+y]$, with $x,y>0$ and $z\ge1$, the remainder in the expected count of multiples of every nonzero integer $d$ is uniformly bounded: $|r_d|\le5$.

Suppose sequences $f,g:\mathbb N\to\mathbb R$ satisfy $f(n)=O(g(n))$ as $n\to\infty$, and $f(n)=0$ whenever $g(n)=0$. Then a single constant bounds $|f(n)|$ by that constant times $|g(n)|$ for every natural number $n$, not only for all sufficiently large $n$.