Here we go. Because \( \phi \star 1 = n \), it follows that \( \sum_{n =1}^\infty \Phi(\frac{x}{n}) = \sum_{n \le x} n \equiv F(x) \). Because \( \Phi \) is always positive, the upper bound \( F(x) \ge \Phi(x) \) follows trivially. To obtain a lower bound, calculate

\( \begin{align}

F(x) - 2 F(\frac{x}{2} ) &= \Phi(x) + \Phi(\frac{x}{2}) + \Phi(\frac{x}{3}) + \cdots - 2 \Phi(\frac{x}{2}) -2 \Phi(\frac{x}{4 }) -\cdots \\

&= \Phi(x) - \Phi(\frac{x}{2}) + \Phi(\frac{x}{3}) - \cdots \\

\end{align}

\)

This is an alternating series, thus \( F(x) - 2 F(\frac{x}{2} ) \le \Phi(x) \). If for simplicity I do not make a distinction between \( x \) and \( \lfloor x \rfloor \), then \( F(x) = \frac{1}{2} x ( x +1 ) \) and \( F(x) - 2 F(\frac{x}{2} ) = \frac{x^2}{4} \). One has thus

$$ \frac{x^2}{4} \le \Phi(x) \le \frac{1}{2} x ( x +1 ) $$

__Remarks__

- The proof is not complete because I have not made a distinction between \( x \) and \( \lfloor x \rfloor \)
- It is known that \( \Phi(x) \sim \frac{3}{\pi^2} x^2 \), see here for example.
- The calculations were performed in Mathematica 10 Home Edition.

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