A proof of the inequality $latex \varepsilon_{m}\le 2\varepsilon_{moy}$

Given n\ge 14 a positive integer, let’s call the difference between two consecutive potential typical primality radii of n a « Goldbach gap ». There are exactly N_{1}(n) such Goldbach gaps. Let’s denote by \eta(n) the number of distinct Goldbach gaps, and let’s write \varepsilon_{i}, 1\le i\le \eta(n) the i-th such Goldbach gap in the increasing order. Thus \varepsilon_{1}=\varepsilon_{min} and \varepsilon_{\eta(n)}=\varepsilon_{max}. Moreover, let’s denote by w_{i} the multiplicity of \varepsilon_{i}. One has \displaystyle{\sum_{i}^{\eta(n)}w_{i}\varepsilon_{i}=P_{ord_{C}(n)}} and \displaystyle{\sum_{i}^{\eta(n)}w_{i}=N_{1}(n)}. Writing p_{i}:=\dfrac{w_{i}}{N_{1}(n)}, one gets \varepsilon_{moy}=\displaystyle{\sum_{i=1}^{\eta(n)}p_{i}\varepsilon_{i}}. Let’s define s_{i} for i ranging from 1 to \eta(n) so that s_{1}=\dfrac{1}{2}, s_{\eta(n)}=\dfrac{1}{2} and s_{i}=0 if i\not\in\{1,\eta(n)\}. One has \vert\varepsilon_{moy}-\varepsilon_{m}\vert=\vert\displaystyle{\sum_{i}^{\eta(n)}\varepsilon_{i}(p_{i}-s_{i})\vert}=\vert(p_{1}-\dfrac{1}{2})\varepsilon_{1}+(p_{\eta(n)}-\dfrac{1}{2})\varepsilon_{\eta(n)}+\displaystyle{\sum_{i=2}^{\eta(n)-1}p_{i}\varepsilon_{i}}\vert. Hence \vert\varepsilon_{moy}-\varepsilon_{m}\vert\le\displaystyle{\sum_{i=2}^{\eta(n)-1}p_{i}\varepsilon_{i}}, and \vert\varepsilon_{moy}-\varepsilon_{m}\vert\le \varepsilon_{moy}. Therefore \varepsilon_{m}\le 2\varepsilon_{moy}.

A proof of the upper bound r0(n)=O(log^4 n)

Given a positive integer n\ge 14, let \varepsilon_{min} (respectively \varepsilon_{max}) be the minimal (respectively maximal) distance between two consecutive potential typical primality radii of n. Moreover, let \varepsilon_{m}:=\dfrac{\varepsilon_{min}+\varepsilon_{max}}{2} and \varepsilon_{moy}:=\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}. One has \varepsilon_{m}\le 2\varepsilon_{moy} (this inequality will be proved in the next article).

Using the inequality of arithmetic and geometric means, one can write the following inequality: \varepsilon_{min}.\varepsilon_{max}\le \varepsilon_{m}^{2}. As  \varepsilon_{m}\le 2\varepsilon_{moy}, one gets \varepsilon_{min}.\varepsilon_{max}\le 4\varepsilon_{moy}^{2}. So that \varepsilon_{max}\le\dfrac{4}{\varepsilon_{min}}\left(\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}\right)^{2}. Obviously 2r_{0}(n)\le \varepsilon_{max}, hence r_{0}(n)\le\dfrac{2}{\varepsilon_{min}}\left(\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}\right)^{2}. As \varepsilon_{min}\ge 2, one finally gets r_{0}(n)\le\left(\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}\right)^{2}\lesssim\dfrac{\log^{4} n}{4.C_{J}^{2}}, where by x\lesssim y we mean x\le (1+o(1)) y and where C_{J}=0.66016... is the twin prime constant. Hence there exists n_{0} such that n\ge n_{0}\Rightarrow r_{0}(n)\le n-1. Therefore every large enough even integer is the sum of two primes.