proof that the Fourier series of $ f\ast g $ uniformly converge.

Question

Let   $f,g$  be   $2\pi$-periodic piecewise continuous functions.
proof that the Fourier series of $ f\ast g $ uniformly converge.
Where $ f\ast g $ denotes the convolution operator between $f$ and $g$.

What I have so far:

• to show that the Fourier series of $ f\ast g $ uniformly converges I need to show that $ f\ast g $ is continuous and that $ f\ast g(\pi) = f\ast g (-\pi)$. finally I need to show that the derivative of $ f\ast g $ is piecewise continuous.
• $ f\ast g(-\pi) = \int_{-\pi}^{\pi}f(-\pi-t)g(t)dt= \int_{-\pi}^{\pi}f(\pi-t)g(t)dt= f\ast g(\pi)$ where the second equality holds by $2\pi$-periodicity of $f$.

how do I know that $ f\ast g $ is continuous and that the derivative of $ f\ast g $ is piecewise continuous?

Answer 1

Let $\hat f(n)$ and $\hat g(n)$ be the (complex) coefficients of $f$ and $g$. By Plancherel's formula $$ \sum|\hat f(n)|^2=\int_{\mathbb{T}}|f(x)|^2\,dx<\infty,\quad\sum|\hat g(n)|^2=\int_{\mathbb{T}}|g(x)|^2\,dx<\infty. $$ We also know $$ \widehat{f\ast g}(n)=\hat f(n)\,\hat g(n). $$ Then, by the Cauchy-Schwarz inequality, $$ \sum|\widehat{f\ast g}(n)|=\sum|\hat f(n)\,\hat g(n)|\le\Bigl(\sum|\hat f(n)|^2\Bigr)^{1/2}\Bigl(\sum|\hat g(n)|^2\Bigr)^{1/2}<\infty. $$ Where $\mathbb{T}$ denotes the circle $\mathbb{R}/2\pi\mathbb{Z}$.