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?

  • $\begingroup$ How do you know that $f\ast g$ is differentiable? $\endgroup$ – Giuseppe Negro Sep 20 '15 at 23:19
  • $\begingroup$ When I come to think about it, I guess I don't have a way to know that. Because I need to know that either f or g are differentiable as well as absolutely integrable and this is not known. So what would be a good direction to go to in this proof? $\endgroup$ – YaG32 Sep 20 '15 at 23:31
  • $\begingroup$ I don't know the solution. Maybe you could try the following. The partial Fourier series $S_N[f\ast g](x)$ can be expressed as a convolution with the Dirichlet kernel. Therefore: $$S_N[f\ast g](x)=f\ast g\ast D_N(x).$$ You could try showing that this expression is uniformly Cauchy as $N\to \infty$. $\endgroup$ – Giuseppe Negro Sep 21 '15 at 0:31
  • $\begingroup$ The same question has been asked a few weeks ago, but atm I can't find it. In the meantime, try to use Plancherels formula and use how convolution acts on the Fourier side. $\endgroup$ – PhoemueX Sep 21 '15 at 7:32

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 $\mathbb{R}/2\pi\mathbb{Z}$.

  • $\begingroup$ I hope you dont mind me fixing up the typos. $\endgroup$ – AD. Sep 22 '15 at 13:31
  • 1
    $\begingroup$ Of course not. I approved them as soon as I was informed about them. $\endgroup$ – Julián Aguirre Sep 22 '15 at 13:33
  • $\begingroup$ @YaG32 Can you see how this shows continuity? $\endgroup$ – AD. Sep 22 '15 at 13:35
  • $\begingroup$ The series is uniformly convergent. $\endgroup$ – Julián Aguirre Sep 22 '15 at 13:42

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