Question 22 of chapter 4 in Apostol's "Introduction to Analytic Number Theory" asks to show that the following three statements are equivalent: $$\psi(x) \log(x) + \sum_{n \le x} \Lambda(n) \psi\left(\frac{x}{n}\right) = 2x \log(x) + O(x) \tag{1}$$ $$\psi(x) \log(x) + \sum_{p \le x} \log(p) \psi\left(\frac{x}{p}\right) = 2x \log(x) + O(x) \tag{2}$$ $$\vartheta(x) \log(x) + \sum_{p \le x} \log(p) \vartheta\left(\frac{x}{p}\right) = 2x \log(x) + O(x) \tag{3}$$ where the latter two sums are extended over those prime numbers $p$ which are less than or equal to $x$. $\Lambda$ is the Mangoldt function and $\psi$ and $\vartheta$ are the Chebyshev functions; $\psi(x) = \sum_{n \le x} \Lambda(n)$ and $\vartheta(x) = \sum_{p \le x} \log(p)$, where again the latter sum is over those primes $p$ less than or equal to $x$. My plan of attack is to show that $(1)\Rightarrow(2),(2)\Rightarrow(3)$ and $(3)\Rightarrow(1)$. I show that $(1)\Rightarrow(2)$ by using the fact that $$\sum_{p \le x} \Lambda(p) \psi\left(\frac{x}{p}\right) \le \sum_{n \le x} \Lambda(n) \psi\left(\frac{x}{n}\right)$$ I show that $(2)\Rightarrow(3)$ by using the fact that $$\vartheta(x) \le \psi(x)$$ Therefore I need only show that $(3)\Rightarrow(1)$. To give myself ideas on how to do this I'm first attempting to show that $(2)\Rightarrow(1)$ in the hopes that the method for doing that can be used for $(3)\Rightarrow(1)$. What I have tried so far is writing the sum in $(1)$ to look like the sum in $(2)$: \begin{eqnarray} \sum_{n \le x} \Lambda(n) \psi\left(\frac{x}{n}\right) &=& \sum_{p \le x} \sum_{m \le \log_p(x)} \Lambda(p^m) \psi\left(\frac{x}{p^m}\right) \\ &=& \sum_{p \le x} \sum_{m \le \log_p(x)} \log(p) \psi\left(\frac{x}{p^m}\right) \\ &=& \sum_{p \le x} \log(p) \psi\left(\frac{x}{p}\right) + \sum_{p \le x} \sum_{2 \le m \le \log_p(x)} \log(p) \psi\left(\frac{x}{p^m}\right) \end{eqnarray} and try to bound the right sum by taking the maximal summand values and using $\pi(x) < 6 \frac{x}{\log(x)}$ (I'm only using results of this kind that are proved in the chapter, working on the assumption that these results are all I will need). However, this gives a bound that is greater than $O(x)$, and so this technique will not yield the desired result. Any answers or hints would be appreciated (for continuing on my approach, or using another approach).

  • $\begingroup$ Did you think about ($2$)->($1$) then showing ($3$)->($2$)? $\endgroup$ – mathematics2x2life Jan 26 '14 at 4:35
  • $\begingroup$ Well yeah but the problem I'm having is doing either of those. I was hoping that if I saw how to do one of them I could apply it to proving the other implications. $\endgroup$ – Ross Pure Jan 26 '14 at 5:13

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