I have encountered a problem where I get the correct outcome, but I am uncertain as to whether or not my steps are logically justified. I would really appreciate some input regarding this! The problem is as follows:
Let $f$ and $g$ be $2 \pi$-periodic, piecewise smooth functions having Fourier series $f(x) = \sum_n \alpha_{n}e^{inx}$ and $g(x) = \sum_n \beta_{n}e^{inx}$, and define the convolution of $f$ and $g$ to be $f * g(x) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} f(t)g(x-t)dt$. Show that the complex form of the Fourier series for $f * g$ is
$f * g(x) = \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{inx}$
SOLUTION:
Since $f * g(x) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} f(t)g(x-t)dt$, we have:
$f * g(x) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} \sum_n \alpha_{n}e^{int} \sum_n \beta_{n}e^{in(x-t)}dt$
$ = \frac{1}{2 \pi} \int_{- \pi}^{\pi} \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{int} e^{in(x-t)}dt$
$= \frac{1}{2 \pi}\int_{- \pi}^{\pi} \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{in(t + x - t)}dt$
$= \frac{1}{2 \pi}\int_{- \pi}^{\pi} \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{inx}dt$
$= \frac{1}{2 \pi} \cdot \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{inx} \int_{- \pi}^{\pi} dt$
$= \sum_{- \infty}^{\infty} \alpha_{n} \beta_{n} e^{inx}$
What I am unsure about here is the step where I merge the two sums into one (step 1 - 2 in the solution above). This doesn't intuitively quite make sense to me based on the definition of the Cauchy product, but it gets me to the correct answer. If anyone can explain whether or not this proceude is valid or not, and how this "merging" of sums can be justified, I would be extremely grateful!