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This is only a partial attempt of an answer but I think other people might be able to make it definitive.

Following the March 2016 edit, and replacing $X_{i}$ by $i^{-1}\sum_{k=1}^{i}X_{k}$ one may think that the asymptotic Goldbach conjecture, namely that every large enough even integer is the sum of two primes, boils down to saying that the corresponding version of what I denote by $\| x\|_{2}$ converges as $n$ tends to $\infty$. But the asymptotic Goldbach conjecture is equivalent to $\alpha_{n}=o(n)$. So one would get that $\alpha_{n}=o(n)\Longrightarrow\alpha_n=O(\sqrt{n}\log^{2}n)$.

This is only a partial attempt of an answer but I think other people might be able to make it definitive.

Following the March 2016 edit, and replacing $X_{i}$ by $i^{-1}\sum_{k=1}^{i}X_{k}$ one may think that the asymptotic Goldbach conjecture, namely that every large enough even integer is the sum of two primes, boils down to saying that the corresponding version of what I denote by $\| x\|_{2}$ converges as $n$ tends to $\infty$. But the asymptotic Goldbach conjecture is equivalent to $\alpha_{n}=o(n)$. So one would get that $\alpha_{n}=o(n)\Longrightarrow\alpha_n=O(\sqrt{n}\log^{2}n)$.

Edit March 22nd, 2021: Following https://en.m.wikipedia.org/wiki/Chebotarev%27s_density_theorem, it appears that $N_{2}(n)$ counts the number of prime numbers in residue classes $P_{ord_{C}(n)}+r$ where $r$ runs over the integers coprime with $P_{ord_{C}(n)}$. From the section "Relation with Dirichlet's theorem" and the effective error term under GRH, taking the $P_{ord_{c}(n)}$-th cyclotomic field leads to $\alpha_{n}\ll_{\varepsilon}n^{1/2+\varepsilon}$ under GRH, providing an affirmative answer to this nearly 10 year old question.

This is only a partial attempt of an answer but I think other people might be able to make it definitive.

Following the March 2016 edit, and replacing $X_{i}$ by $i^{-1}\sum_{k=1}^{i}X_{k}$ one may think that the asymptotic Goldbach conjecture, namely that every large enough even integer is the sum of two primes, boils down to saying that the corresponding version of what I denote by $\vert\vert x\vert\vert_{2}$ converges as $n$ tends to $\infty$. But the asymptotic Goldbach conjecture is equivalent to $\alpha_{n}=o(n)$. So one would get that $\alpha_{n}=o(n)\Longrightarrow\alpha_n=O(\sqrt{n}\log^{2}n)$.

This is only a partial attempt of an answer but I think other people might be able to make it definitive.

Following the March 2016 edit, and replacing $X_{i}$ by $i^{-1}\sum_{k=1}^{i}X_{k}$ one may think that the asymptotic Goldbach conjecture, namely that every large enough even integer is the sum of two primes, boils down to saying that the corresponding version of what I denote by $\| x\|_{2}$ converges as $n$ tends to $\infty$. But the asymptotic Goldbach conjecture is equivalent to $\alpha_{n}=o(n)$. So one would get that $\alpha_{n}=o(n)\Longrightarrow\alpha_n=O(\sqrt{n}\log^{2}n)$.

This is only a partial attempt of an answer but I think other people might be able to make it definitive.

Following the March 2016 edit, and replacing $X_{i}$ by $i^{-1}\sum_{k=1}^{i}X_{k}$ one may think that the asymptotic Goldbach conjecture, namely that every large enough even integer is the sum of two primes, boils down to saying that the corresponding version of what I denote by $\vert\vert x\vert\vert_{2}$ converges as $n$ tends to $\infty$. But the asymptotic Goldbach conjecture is equivalent to $\alpha_{n}=o(n)$. So one would get that $\alpha_{n}=o(n)\Longrightarrow\alpha(n)=O(\sqrt{n}\log^{2}n)$.

This is only a partial attempt of an answer but I think other people might be able to make it definitive.

Following the March 2016 edit, and replacing $X_{i}$ by $i^{-1}\sum_{k=1}^{i}X_{k}$ one may think that the asymptotic Goldbach conjecture, namely that every large enough even integer is the sum of two primes, boils down to saying that the corresponding version of what I denote by $\vert\vert x\vert\vert_{2}$ converges as $n$ tends to $\infty$. But the asymptotic Goldbach conjecture is equivalent to $\alpha_{n}=o(n)$. So one would get that $\alpha_{n}=o(n)\Longrightarrow\alpha_n=O(\sqrt{n}\log^{2}n)$.