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I'm working on Richter's "A new elementary proof of the Prime Number Theorem" paper. I have some doubt about the proof of proposition 3.1 Here's the reference to the paper: https://arxiv.org/abs/2002.03255

What I found not clear is the bound at the beginning of p.7:

Lemma 3.3 (applied with $\sigma = 2$) gives the estimate: $$\bigg|P \cap\left(\frac{8^x}{2^{n+1}}, \frac{8^x}{2^n}\right]\bigg| \leq \frac{8^x}{2^{n+1}x} + O(1).$$

It is clearly not the direct application of Lemma 3.3 and I'm not sure how to get this bound starting from the one that Lemma 3.3 actually gives.

Bonus: it's less relevant to me at the moment but I'm also struggling to understand how the term $O(1/N)$ arises in the first equality in “Proof of Theorem 1.1 assuming Proposition 2.2.” at p.4

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    $\begingroup$ as for the second question my feeling is that it has nothing to do with the Proposition $2.2$ itself but just replacing the index of summation from $n \le [N/q]$ to $n \le N, q |n$ and seeing what error we get given that $|g| \le 1$ and the various definitions of the projections $\mathbb E$ - that's where I would look $\endgroup$
    – Conrad
    Commented 2 hours ago
  • $\begingroup$ That’s my idea too, and I tried to do as you suggest but it didn’t get me the desired O(1/N). $\endgroup$
    – rr_math
    Commented 2 hours ago
  • $\begingroup$ I will look a bit later and see what happens $\endgroup$
    – Conrad
    Commented 2 hours ago

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Here I think the author skips an intermediate step as for $n \le x-1$ actually Lemma $3.3$ indeed applies directly as the $\log \frac{8^x}{2^{n+1}}=3x\log 2 - (n+1)\log 2 \ge 2x\log 2$ while $\beta(2)=2\log 2$ so the given inequality holds, namely: $$\bigg|P \cap\left(\frac{8^x}{2^{n+1}}, \frac{8^x}{2^n}\right]\bigg| \leq \frac{(2\log 2) 8^x}{2^{n+1}x(3x\log 2-(n+1)\log 2)} + O(1) \leq\frac{8^x}{2^{n+1}x} + O(1).$$

For $n \ge x$ we can use that the number of primes is at most the length of the interval giving us a $4^x$ total in $(1,4^x]$ which is $o(8^x/x)$ so the bound obtained is $$\bigg|P \cap\left(8^x,8^{x+1}]\right]\bigg| \geq \frac{8 \times 8^x}{3x+3}-\frac{8^x}{x}(1+o(1)) + O(1) \geq \frac{8^x}{x}$$

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  • $\begingroup$ thank you very much. Any idea about the bonus question? $\endgroup$
    – rr_math
    Commented 2 hours ago
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    $\begingroup$ see my comment above - try that and see if it works $\endgroup$
    – Conrad
    Commented 2 hours ago

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