The proof of Theorem 3.10 in Introduction to Analytic Number Theory by Apostol goes like this:
Theorem 3.10 If $ h = f * g $, let $$ H(x) = \sum_{n \le x} h(n), F(x) = \sum_{n \le x} f(n), \text{ and } G(x) = \sum_{n \le x} g(n). $$ Then we have $$ H(x) = \sum_{n \le x} f(n) G \left( \frac{x}{n} \right) = \sum_{n \le x} g(n) F \left( \frac{x}{n} \right). $$
PROOF. We make use of the associative law (Theorem 2.21) which relates the operations $ \circ $ and $ * $. Let $$ U(x) = \begin{cases} 0 & \text{ if } 0 < x < 1, \\ 1 & \text{ if } x \ge 1. \end{cases} $$ Then $ F = f \circ U $, $ G = g \circ U $, and we have $$ f \circ G = f \circ (g \circ U) = (f * g) \circ U = H, $$ $$ g \circ F = g \circ (f \circ U) = (g * f) \circ U = H. $$ This completes the proof.
I am curious to understand why use generalized convolution and Dirichlet multiplication here at all.
Since section 3.5, the book has been relying on the following reordering of summation at various places, $$ \sum_{n \le x} \sum_{d | n} f(d) = \sum_{\substack{q, d \\ qd \le x}} f(d) = \sum_{d \le x} \sum_{q \le \frac{x}{d}} f(q). $$ Can we not arrive at a simpler proof with a similar reordering of summation like this: \begin{align*} H(x) & = \sum_{n \le x} h(n) = \sum_{n \le x} (f * g)(n) = \sum_{n \le x} \sum_{d | n} f(d) g\left(\frac{n}{d}\right) = \sum_{\substack{q, d \\ qd \le x}} f(d) g\left(q\right) \\ & = \sum_{d \le x} \sum_{q \le \frac{x}{d}} f(d) g(q) = \sum_{d \le x} f(d) \sum_{q \le \frac{x}{d}} g(q) = \sum_{n \le x} f(n) \sum_{q \le \frac{x}{n}} g(q) = \sum_{n \le x} f(n) G\left(\frac{x}{n}\right). \end{align*} I am trying to understand if a simpler proof using basic arithmetic and reordering of summations is correct or if I am missing something and there is a specific reason why Apostol went for a proof using generalized convolution and Dirichlet multiplication.
The other proof is arguably simpler (it is actually shorter and requires less manipulation). Also, the first few chapters of Apostol make a big deal about using convolutions.
There is a reason for this. As you move to more complicated content, with much "messier" expressions, techniques like this can be incredibly useful, since they allow for much more efficient notation. If anything, I think this proof shows how nice convolutions can be, since it solves the problem without writing down a single sum.
Hence, although it might seem like a more difficult proof if you are not familiarized with convolutions, it is actually a good choice from an educational point of view, since it will get you used to using this kind of technique for more difficult problems. I always find changing the order of summation a bit fiddly (I have to draw the curves and convince myself I have not made any mistakes) whereas convolutions are pretty much automatic