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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 $f,g \in L^1[-\pi,\pi].$

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What does $f \star g$ mean? –  Graphth Oct 8 '12 at 20:42
    
@Stefan: Let me ask you that do you mean $\epsilon - \delta $ - definition? Ok. –  laovultai Oct 8 '12 at 20:47
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Oh I'm sorry, I was thinking about something entirely else. Please forget my comment (I have deleted it). –  Stefan Oct 8 '12 at 20:51
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Hint: Take a point $x_0 \in \mathbb{R}$ and $(x_n)_n \to x_0$. Use the fact that $f,g \in L^1$ to apply the dominated convergence theorem. Then use the continuity of $f$ or $g$. –  nullUser Oct 8 '12 at 21:09

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up vote 2 down vote accepted

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 $$|f\star g(x_1)-f\star g(x_2)|\leq\int|(g(x_1-t)-g(x_2-t))f(t)|dt\leq \varepsilon\int_{[-\pi,\pi]}|f(s)|ds.$$

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Should there be some intermediate steps? I mean that how you get from $|f(x+t)-f(x)| \leq \epsilon$ last line and how do you get $|f \star g(x_1)-f \star g(x_2)| \leq \epsilon \int_{[-\pi,\pi]}|f(s)|ds$? –  laovultai Oct 16 '12 at 11:31
    
@alvoutila I've added a step. –  Davide Giraudo Oct 16 '12 at 11:39
    
excuse me. How do you get from $\int |(g(x_1-t)-g(x_2-t))f(t) |dt$ to $\leq \epsilon \int_{[-\pi,\pi]} |f(s)|ds$? –  laovultai Oct 24 '12 at 9:18
    
You can write $|g(x_1-t)-g(x_2-t)|\leq\varepsilon$ by uniform continuity. –  Davide Giraudo Oct 24 '12 at 9:43

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