---- > [!theorem] Theorem. ([[convolution as regularization]]) > 1. *(on the circle - Math 396)* 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]]. > > 2. *(on $\mathbb{R}^{n}$)* The convolution of $f \in L^{p}(\mathbb{R}^{n})$, $1 \leq p \leq \infty$, and $g \in C_{c}^{0}(\mathbb{R}^{n})$ satisfies $f * g \in C^{0}(\mathbb{R}^{n})$. In fact, $f * g$ is [[uniformly continuous]] ([[continuous iff uniformly continuous on compact spaces|owing to]] $g$ being UC). > 3. *(on $\mathbb{R}^{n}$ - transferring derivative onto the more regular argument)* If $g \in C^{k}_{c}(\mathbb{R}^{n})$, then $f * g \in C^{k}(\mathbb{R}^{n})$ with $\partial_{}^{\alpha} |_{x} (f * g) = f * \partial_{}^{\alpha}|_{x} g$ > for $\alpha$ a [[derivative|multiindex]],$|\alpha|=k$. ("Put the derivatives onto the argument that's more regular.") > > 4. *(on $\mathbb{R}^{n}$ - weakly transferring derivative onto the less regular argument)* [[Sobolev space|If]] in fact $f \in W^{k,p}(\mathbb{R}^{n})$ for some $k \geq 0$, $1 \leq p < \infty$, and $g \in C_{c}^{k}(\mathbb{R}^{n})$, then in fact $\partial_{}^{\alpha}(f * g)=\partial_{}^{\alpha}f * g \text{ for any } |\alpha| \leq k,$ > where the [[derivative]] $\partial_{}^{\alpha }f$ is understood [[weak derivative|weakly]] as necessary. Thus, $f * g \in C^{k}(\mathbb{R}^{n}) \cap W^{k,p}(\mathbb{R}^{n})$.[^1] > > > >5. *(summing supports)* $\operatorname{supp}(f * g) \subset \overline{ \operatorname{supp}(f) + \operatorname{supp}(g)}$. *4 was skipped in lecture but done here* [^1]: $f * g \in C^{k}(\mathbb{R}^{n})$ by $(3)$. $f * g \in W^{k,p}(\mathbb{R}^{n})$ by $(4)$ together with [[Young's convolution inequality]] ($q=1, r=p$): for any $|\alpha| \leq k$, $\|\partial_{}^{\alpha} (f * g)\|_{L^{p}(\mathbb{R}^{n})} \leq \|\partial_{}^{\alpha} f\|_{L^{p}(\mathbb{R}^{n})} \|g\|_{L^{1}(\mathbb{R}^{n})}<\infty.$ > [!proof]- Proof of **1**. (Convolution on the circle 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 $k \to \infty$. **Proof of 2. (Convolution with a compactly supported function on $\mathbb{R}^{n}$ creates a uniformly continuous function)** We sketch $(2)$ for the case $k=0$. It is mostly just integral estimates, there aren't really any tricks although admittedly the below is a bit rushed/sloppy. Let $f \in L^{p}(\mathbb{R}^{n})$, $1 \leq p \leq \infty$, $g\in C_{c}^{0}(\mathbb{R}^{n})$. ([[continuous iff uniformly continuous on compact spaces|Note]] $g$ is [[uniformly continuous]].) We have to show $f * g$ is uniformly [[continuous]]. Equivalently, we will show $\sup_{x \in \mathbb{R}^{n}}|(f * g)(x+z) - (f * g)(x) |\xrightarrow{?} 0 \text{ as } z \to 0.$ By [[Hölder's inequality|Holder]], $\begin{align} \sup_{x \in \mathbb{R}^{n}} |(f * g)(x+z)- (f * g)(x)| &\leq \sup_{x \in \mathbb{R}^{n}} \int_{\mathbb{R}^{n}} |f(y)| \, |T_{z}g(x-y) - g(x-y)| \\ & \leq \sup_{x} \|f\|_{L^{p}} \|(T_{z}g - g)(x - \cdot)\| \\ & \leq \|f\|_{L^{p}} \, \|T_{z}g - g\|_{L^{p'}}. \end{align}$ The task is therefore to show $\|T_{z}g - g\|_{L^{p'}} \xrightarrow{z \to 0}0$. Since $g \in C_{c}^{0}$, it is uniformly continuous, thus $\|T_{z}g - g\|_{L^{\infty}} \xrightarrow{z\to 0}0$. $g$ has compact support in some ball of radius $R$; once $|z|<R$ the support of $T_{z}g- g$ lies in one fixed compact set $K=B_{2R}$. Now by standard facts on $L^{p}$ spaces of finite measure have $\|T_{z}g - g\|_{L^{p'}} \lesssim \|T_{z}g - g\|_{L^{\infty}} \to 0$ and the result follows. (In this case the fact is especially easy: $\|h\|_{L^{r}(K)}^{r} = \int_{K} |h|^{r} \leq \int_{K}\|h\|_{L^{\infty}(K)}^{r}=|K|\|h\|^{r}_{L^{\infty}(K)}$ and take $r$th roots on both sides.) --- *old below* Expanding, have $\begin{align} (f * g)(x+z) - (f * g)(x) &= \int_{\mathbb{R}^{n}} f(y) g(x+z-y) \, dy - \int_{\mathbb{R}^{n}} f(y) g(x-y)\, dy \\ &= \int_{\mathbb{R}^{n}} f(y) \underbrace{ [g(x+z-y) - g(x-y)] }_{ \text{can be made uniformly small, as } g \text{ UC} }\, dy \end{align}$ $g$ has [[compact]] [[support]] in some ball of radius $R$; for small enough $z$, namely $|z|<R$, we can write $\begin{align} |(f * g)(x+z) - (f * g)(x)| &\leq \int_{|x-y|<2R} |f(y)| \, \overbrace{ |g(x+z-y) - g(x-y)| }^{ \leq \varepsilon } \\ & \leq \varepsilon \int_{|x-y|<2R} |f(y)| \,dy \\ & \leq \varepsilon \|f\|_{L^{p}(\{|x-y|<2R\} )} \|1\|_{L^{p'}(\{ |x-y| < 2R \})} \end{align}$ and the result follows. *If we only want ordinary continuity, here is a quicker argument:* Take a [[sequence]] $x_{n} \to x$ in the domain of $f$. Have $\begin{align} |f_{\varepsilon}(x_{n}) - f_{\varepsilon}(x)| &= | (f * \rho_{\varepsilon})(x_{n}) - (f * \rho_{\varepsilon})(x)| \\ & \leq \int_{} |f(y)| \underbrace{ | \rho_{\varepsilon} (x_{n}-y) - \rho_{\varepsilon}(x-y)| }_{ \to 0 \text{ as }n \to \infty }\, dy \end{align}$ using continuity of $\rho_{\varepsilon}$, and thus we are done if we can justify an application of [[Dominated Convergence Theorem|DCT]]. Since $\rho_{\varepsilon}$ has compact support, the integral is WLOG over some compact set $K$. If $1\leq p < \infty$, the majorant $|f(y)|^{} \|\rho_{\varepsilon}\|_{\infty}$ does the trick, where we note $|f(y)|$ is integrable because $L^{p} \hookrightarrow L^{1}$ as $K$ has finite measure. If $p=\infty$ the majorant $\|f\|_{\infty} \|\rho_{\varepsilon}\|_{\infty}$ works. **Proof of 3. (Transferring derivative onto the more regular argument.)** We prove for $k=1$; general case follows by induction. WTS that [[continuously differentiable|when]] $g \in C_{c}^{1}(\mathbb{R}^{n})$., $\partial_{i} |_{x}(f * g)$ exists, equalling $(f * \partial_{i} |_{x} g)$. It will be [[continuous]], so indeed $C^{1}$, by Part 2. The directional difference quotient is $\begin{align} \lim_{t \to 0} \frac{(f * g)(x + t e_{i}) - (f * g)(x) }{t}&=\int_{\mathbb{R}^{n}} f(y)\underbrace{ \frac{ [g(x+te_{i}- y)-g(x-y)]}{t} }_{ \xrightarrow{t \to 0} \partial_{i} |_{x-y}g } \, dy . \end{align}$ We see that if we can justifying passing the [[function limit|limit]] into the [[integral]], we are finished. Indeed we can, by using the [[Dominated Convergence Theorem]] with an appropriate majorant. Specifically, WLOG the integral is over some [[compact]] $K \subset \mathbb{R}^{n}$ since $g \in C_{c}^{1}(\mathbb{R}^{n})$; by the [[mean value theorem]], $|\frac{g(x+t e_{i} - y)-g(x-y)}{t}| \leq \|\partial_{i} g\|_{L^{\infty}}$ and so the majorant $|f(y)| \|\partial_{i} g\|_{L^{\infty}}$ does the trick. **Proof of 4.** Let $\varphi \in C^{\infty}_{c}(\mathbb{R}^{n})$. We claim $\int_{\mathbb{R}^{n}}(f * g) \partial_{i} \varphi \overset{?}{=}-\int_{\mathbb{R}^{n}}(\overbrace{ \partial_{i}f }^{ \text{weak} } * g)\varphi,$ where here $\partial_{i}f$ is the function satisfying $\int_{\mathbb{R}^{n}} f \, \partial_{i} \varphi=-\int_{\Omega} \varphi \partial_{i}f$. We have $\begin{align} \int_{\mathbb{R}^{n}} \partial_{i} \varphi(x) \left[ \int_{\mathbb{R}^{n}} f(y) g(x-y) \, dy \right] dx &= \int_{\mathbb{R}^{n}} \partial_{i} \varphi (x)\left[ \int_{\mathbb{R}^{n}} f(x-z) g(z) \, dz \right] dx \\ &= \int_{\mathbb{R}^{n}} g(z) \, \left[ \int_{\mathbb{R}^{n}} \partial_{i} \varphi(x) f(x-z) dx\right] dz \\ &= \int_{\mathbb{R}^{n}} g(z) \left[ \underbrace{ \int_{\mathbb{R}^{n}} \partial_{i} \varphi(x-z) f(x-z) dx }_{ =-\int_{\mathbb{R}^{n}} \varphi(x) \partial_{i} f(x-z) \,dx}\right] dz \\ &= \int_{\mathbb{R}^{n}} \varphi(x) \left[\underbrace{ \int_{\mathbb{R}^{n}} g(z) \partial_{i} f(x-z) \, dz }_{ \partial_{i} f * g} \right] dx \end{align}$ where [[Fubini's Theorem]] has been used to swap integrals (two times), we used commutativity of [[convolution]] on $\mathbb{R}^{n}$, and we were able to replace $\partial_{i}\varphi(x)$ with $\partial_{i}\varphi(x-z)$ by translation invariance of the [[integral|Lebesgue integral]] (that is not quite justified, not sure what I was thnking, but can do cov with translated test function and end up at the same spot) The result follows. (My solution a bit different in the lecture notes as I proved weak derivative when could have proven classical derivative. Theirs are probably cleaner but this is still correct.) Suppose $f$ has weak $i$-derivative $h$. $\begin{align} \partial_{i} (f * g) &= f * \partial_{i} g \\ &= \int_{\mathbb{R}^{n}} f(y) \partial_{i} g (x-y) \, dy \\ &- \int_{\mathbb{R}^{n}} h (y) g(x-y) \, dy \\ &= \underbrace{ h }_{ \partial_{i}f }* g \end{align}$ this would work if $g \in C_{c}^{\infty}$ which we basically assume in general **Proof of 5.** Write $A=\operatorname{supp}(f)$, $B=\operatorname{supp}(g)$. Let $x \in \operatorname{ supp}(f * g)$. Contrapositively suppose $x \notin A + B$, meaning there are no pairs $(a \in A, b \in B)$ with $x=a+b$. Given $y \in \mathbb{R}^{n}$, either $y \notin A$, in which case $f(y)=0$, or $y \in A$, in which case $x-y \notin B$ and so $g(x-y)=0$. In either case the integrand $f(y)g(x-y)=0$ for all $y \in \mathbb{R}^{n}$, and thus $(f * g)(x)=0$. Thus $x \notin \operatorname{ supp}(f * g)$. ---- #### ----- #### 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 > ```