Let $f:\mathbb{R}\rightarrow\mathbb{R}$ and $g:\mathbb{R}\rightarrow\mathbb{R}$ be $2\pi$-periodic functions such that $f$ is bounded, $g$ is differentiable and $g'$ is continuous. How can I show that the convolution $g*f$ is differentiable?
I hope you can help. Thanks!
A fun way is using Fourier analysis. You need to know Parseval; that tells you that $$\sum |\hat f(n)|^2<\infty.$$And since $\widehat{g'}(n)=in\hat g(n)$, Parseval also tells you that $$\sum n^2|\hat g(n)|^2<0.$$Now Cauchy-Schwarz tells you that $$\sum|n||\hat f(n)\hat g(n)|<\infty,$$which implies that $f*g$ has a continuous derivative.
Ok, what if that's too much fun? You can do it directly. A hint is that the derivative of $f*g$ "should" be $f*(g')$. Write down the definition of $$\frac{f*g(x+h)-f*g(x)}{h}$$as an integral, write down the definition of $$f*(g')(x)$$as an integral, and see if you can show the first integral tends to the second integral as $h\to0$.
Hint for that: Since $g'$ is continuous (and periodic) it follows that $(g(x+h)-g(x))/h$ tends to $g'(x)$ uniformly as $h\to0$.
