[[Noteworthy Uses]]:: *[[Noteworthy Uses]]* [[Proved By]]:: *[[Proved By|Crucial Dependencies]]* ---- > [!theorem] Theorem. ([[convolution creates continuous functions]]) > The [[convolution]] $f * g$ of [[Riemann integral|Riemann integrable]] functions $f,g$ [[function on the (unit) circle|on the circle]] is [[continuous]] [[function on the (unit) circle|on the circle]]. > [!proof]- Proof. ([[convolution creates continuous functions]]) > **Step 1.** We will prove the claim in the [[continuous]] case. Suppose $g$ is [[continuous]]; hence [[uniformly continuous]]. If $f=0$ [[almost-everywhere]] then $(f *g)(x)=0$ for all $x$ and thus $(f * g)$ is [[continuous]]; so suppose this is not the case. Define $c:=\frac{1}{2\pi} \int_{-\pi}^{\pi} |f(y)| \,dy>0$, fix $\varepsilon>0$, and let $\varepsilon'=\frac{\varepsilon}{c}>0$. Then by $gs [[uniformly continuous|uniform continuity]] $\ex \delta>0$ s.t. $|g(s)-g(t)|<\varepsilon$ $\fa s,t$ s.t. $|s-t|< \delta$. Fix $x_{1},x_{2}$ satisfying $|x_{1}-x_{2}|<\delta$. Now $\begin{align} > (f * g)(x_{1})-(f * g)(x_{2}) \leq & \frac{1}{2\pi}\int_{-\pi}^{\pi} |f(y)| \overbrace{|g(x_{1}-y)-(x_{2}-y)|}^{\leq \varepsilon'}\, dy . \\ > = & \frac{\varepsilon'}{2\pi} \int_{-\pi}^{\pi} |f(y)|\,dy \\ > = & \varepsilon'c \\ > = & \varepsilon. > \end{align}$ > Thus $f * g$ is (uniformly) [[continuous]] in this case. > > **Step 2**. Now let $g$ be (merely) [[Riemann integral|integrable]]. We'll use a density argument: by [[on the circle, continuous functions are dense in Riemann integrable functions with respect to the 1-seminorm|this result]], there exist [[continuous]] functions $g_{k}$ [[function on the (unit) circle|on the circle]] s.t. > - $|g_{k}(x)| \leq B$ for all $x$ and for all $k$; > - $\lim_{ k \to \infty }\|g_{k}-g\|_{1}=0$. > > Let's study $f * g_{k}$... > > By **Step 1**, $f * g_{k}$ is [[continuous]]. We want to show that $(f * g_{k})(x) \to (f * g)(x) \text{ uniformly in }x$ > (the [[uniform convergence|uniformity]] [[uniform limit theorem|guarantees the limit is continuous]]). This will complete the proof. Since $(f * g)(x)-(f *g )(x)=\frac{1}{2\pi} \int_{-\pi}^{\pi} f(x-y)\big(g_{k}(y)-g(y)\big)\, dy, $ > calling $fs bound (recall $f$ is [[Riemann integral|Riemann integrable]]) $M$ we get $\begin{align} > |(f * g_{k})(x) - (f * g)(x) | \leq & \frac{1}{2\pi}\int_{-\pi}^{\pi} |f(x-y)||g_{k}(y)-g(y)|\, dy \\ > \leq & \frac{M}{2\pi} \int_{-\pi }^{\pi}|g_{k}(y)-g(y)|\, dy, > \end{align}$ > where we note that this final expression does not depend on $x$. Thus for all $x \in \rr$, $|(f * g_{k})(x)-(f * g)(x)| \leq M \|\delta_{k}-\delta\|_{1}$ > independent of $x$ and we therefore conclude that $f * g_{k} \to f * g$ [[uniform convergence|uniformly]] in $x$ as $k \to \infty$. ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```