How to show that $f \star g$ is continuous if $f$ or $g$ is continuous? Do you use $\epsilon - \delta $ - approach in the proof? some hint.
I define $(f \star g)(x)=\frac{1}{2\pi} \int_{- \pi}^{\pi} f(y)g(x-y)dy,$ if $f,g \in L^1[-\pi,\pi].$
How to show that $f \star g$ is continuous if $f$ or $g$ is continuous? Do you use $\epsilon - \delta $ - approach in the proof? some hint.
I define $(f \star g)(x)=\frac{1}{2\pi} \int_{- \pi}^{\pi} f(y)g(x-y)dy,$ if $f,g \in L^1[-\pi,\pi].$
Extend $f$ and $g$ by periodicity if necessary to get that $f\star g=g\star f$. So we can assume WLOG that $g$ is continuous on $[-\pi,\pi]$, hence uniformly continuous on this interval.
Fix $\varepsilon>0$, and $\delta>0$ such that if $x\in\Bbb R$, $|s|\leq \delta$ then $|f(x+t)-f(x)|\leq \varepsilon$. Then if $|x_1-x_2|\leq \delta$, we have $|f\star g(x_1)-f\star g(x_2)|\leq\int|(g(x_1-t)-g(x_2-t))f(t)|dt\leq \varepsilon\int_{[-\pi,\pi]}|f(s)|ds.$