Lemma from arithmetic functions

Question

Let $f$ arithmetic and $$H(f)=\lim_{x\rightarrow \infty}\frac{1}{x\log x}\sum_{n\leq x}f(n)\log n,$$ Then $H(f)$ exists if and only if $M(f)$ exists, and $M(f)=H(f)$ Where $$M(f)=\lim_{x\rightarrow \infty}{1\over x}\sum_{n\leq x}f(n).$$

I don't know if I have to use the fact that $$M(f)\implies L(f)=\lim_{x\rightarrow \infty}{1\over \log(x)}\sum_{n\leq x}{f(n)\over n}.$$

Answer 1

Let

$$ M_{x} = \frac{1}{x} \sum_{n \leq x} f(n) \quad \text{and} \quad H_{x} = \frac{1}{x \log x} \sum_{n \leq x} f(n) \log n. $$

As suggested by i707107,s comment, we can use a similar argument to show that

\begin{align*} H_{x} &= \frac{1}{x\log x} \int_{1^{-}}^{x} f(t) \log t \, d[t] = \frac{1}{x\log x} \int_{1^{-}}^{x} \log t \, d ( t M_{t} ) \\ &= \frac{1}{x \log x} \left[ \log t \cdot t M_{t} \right]_{1^{-}}^{x} - \frac{1}{x \log x} \int_{1}^{x} M_{t} \, dt \\ &= M_{x} - \frac{1}{x \log x} \int_{1}^{x} M_{t} \, dt. \end{align*}

Since $M_{x} \to M(f)$ as $x \to \infty$, we have

$$ \int_{1}^{x} M_{t} \, dt = O(x) $$

and it follows that $H_{x} \to M(f)$ as $x \to \infty$.