Let $A(x)$ be defined for all $x >0$ and assume that
$$T(x) = \sum_{n\le x} A(x/n) = ax\log x + bx + o\left(\frac{x}{\log x}\right),~~~~ \text{as}~~x \rightarrow \infty$$
where $a$ and $b$ are constants. Prove that:
$$A(x)\log x + \sum_{n\le x} A(x/n) \Lambda (n) = 2ax\log x + o(x\log x), ~~~~ \text{as}~~x \rightarrow \infty$$ (Here $\Lambda(n)$ is the Mangoldt function). Verify that Selberg’s formula of Theorem 4.18 is a special case.
I am having difficulty in seeing how Selberg's formula
$$\psi(x) \log x + \sum_{n\le x} \psi(x/n) \Lambda (n) = 2x\log x + O(x)$$ ($\psi(x) = \sum_{n\le x} \Lambda (n)$) follows from the first part of the problem: Putting $A(x) = \psi(x)$, $a=1$ and $b=-1$ we see the assumptions of the problem are satisfied but
Selberg's formula gives an estimate of $O(x)$ whereas the result of this problem gives an (weaker) error estimate of $o(x\log x)$. Derivation of Selberg's formula (as derived in Theorem 4.18 in Apostol's book) uses the stronger error estimate of $O(\log x)$ for the sum $\sum_{n \le x} \psi(x/n)( = x\log x -x +O(\log x))$(Theorem 4.11), so I am not sure how the weaker assumption of $o(x/\log x)$ leads to the stronger estimate of $O(x)$.
What am I missing here? Any help is appreciated.