# A question about proof of Ternary Goldbach Conjecture.

Let's recall: Von Mangoldt function $$\Lambda$$ is a following function: $$\Lambda: \mathbb{N}\rightarrow \mathbb{R}$$ $$\Lambda(n) = \left\{ \begin{array}{ll} \log p & \textrm{if n = p^k for some prime p}\\ 0 & \textrm{otherwise}\\ \end{array} \right.$$ Ternary Goldbach Conjecture states that: $$W(N):=\sum _{k_1+k_2+k_3 = N}\chi(k_1)\chi(k_2)\chi(k_3)>0$$ For odd numbers greater than $$5$$. Here $$\chi$$ is characteristic function for prime numbers.

But here is the problem. In proof of this concjecture we see a quite different expression. $$G(N) = \sum_{k_1+k_2+k_3 = N}\Lambda(k_1)\Lambda(k_2)\Lambda(k_3)$$ This sum is proved to be sufficiently big.

My question is:

How are $$W(N)$$ and $$G(N)$$ connected? How to estimate from below the first expression ($$W(N)$$) using just $$G(N)$$?

In analytic number theory, when one studies primes, it is customary to study sums involving $$\Lambda$$ instead of $$\chi$$. Note that $$\Lambda$$ is essentially supported on primes. For any function $$f$$, we have $$\sum_{n\leq x} f(n)\Lambda(n) = \sum_{p\leq x} f(p)\log p + \sum_{\substack{p^k\leq x\\ k\geq 2}} f\left(p^k\right)\log p.$$ If $$f$$ is small, say bounded by 1, the second sum is $$O(\sqrt{x}\log x)$$, which is quite small if the main term is expected to be of size $$x$$. This is a general theme that occurs when studying sums over primes, namely that the sum over prime powers (with the power at least 2) doesn't give a significant contribution to the estimate. Thus summing $$f$$ against $$\Lambda$$ is essentially summing $$f$$ over primes, with the additional weight $$\log p$$ (indeed, sums such as the one above are often regarded in the literature as simply "sums over primes"). To address your question specifically, we have $$G(N) = \sum_{p_1+p_2+p_3=N} (\log p_1)(\log p_2)(\log p_3) + O\Big(\sum_{\substack{p_1^k+p_2^k+p_3^k=N\\ k\geq 2}} (\log p_1)(\log p_2)(\log p_3)\Big)$$ By a trivial estimation, the error term is $$O\left(N^{3/2}(\log N)^3\right)$$. Since the main term is known to be of size $$N^2$$, it follows that for $$N$$ sufficiently large, the sign of $$G(N)$$ is not affected by the error term, and so $$G(N) > 0$$ for $$N$$ sufficiently large. Note that if $$G(N) > 0$$, we certainly must have $$W(N) > 0$$ as well. That is, one does not bound $$W(N)$$ using $$G(N)$$ directly, but rather infers that $$W(N) > 0$$ from the fact that $$G(N) > 0$$, as both are non-negative sums over primes.
• Thank you for the answer .What is $T(N)$? Do you mean $T(N) = W(N)$? May 28, 2021 at 13:12