I want to show that for the $2\pi$-peroidc, integrable function, the convolution operation improves the smoothness, the convolution for my cases is only defined over the complex unit circle, that is
$$ (f *g)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(\tau)g(x - \tau)\,d\tau$$
Now, suppose that $f,g$ are both $2\pi$-peroidc, integrable function defined over the complex unit circle, with $f \in C^{m}$, and $g \in C^{n}$. Let $h(x) = (f *g)(x)$, I want to show that $h \in C^{m+n}$. My approach is simple, begin with a bases cases that $f \in C^{1}$, and $g \in C^{0}$. Then, using the commutativity of convolution, we have $$ \begin{align} h'(x) = (f *g)'(x) = (g *f)'(x) &= \frac{d}{dx} \bigg[\frac{1}{2\pi}\int_{-\pi}^{\pi} g(\tau) f(x - \tau)\,d\tau \bigg]\\ &= \frac{1}{2\pi}\int_{-\pi}^{\pi} g(\tau) \frac{d}{dx}f(x - \tau)\,d\tau\\ &= \frac{1}{2\pi}\int_{-\pi}^{\pi} g(\tau) f'(x - \tau)\frac{d}{dx}(x - \tau)\,d\tau\\ &= \frac{1}{2\pi}\int_{-\pi}^{\pi} g(\tau) f'(x - \tau)\,d\tau \end{align} $$Since differentiation is over $x$, the dummy variable $\tau$ inside the integral is not touched, and $f$ and $g$ are always interchangeable. So, the first order derivative for $h(x)$ exist.
Then, suing the induction to extend to $f \in C^{m}$, and $g \in C^{n}$ cases, would this be vaild approach? All my concern here is the well-definedness of the derivative of the convolution. I direct give the derivative of the convolution but do not show that it is well defined, would this be a problem? Thanks for any help.