> [!definition] > > Let $\phi \in L^1(\real^d)$ be an [[Integrable Function|integrable function]], then $\phi$ is a **mollifier** if $\int \phi = 1$. For any $t > 0$, denote > $ > \phi_t(x) = \frac{1}{t^d}\phi\paren{\frac{x}{t}} > $ Let $F$ is an input, and $\phi$ be a mollifier. If $\phi$ has certain properties, then $\phi_t * F$ may inherit some of them, with $\phi_t * F \to F$ as $t \to 0$ in some sense. The following table describes the interactions: | Input Type | Mollifier Type | Mollified Function Type | Convergence | | ----------------------------------------------------- | ------------------------ | ----------------------- | ------------------------------------------- | | [[Probability\|Probability Measure]] | Non-negative | $L^1$ | [[Weak Convergence of Measures\|Weak]] | | Probability Measure | Non-negative, $UC$ | $UC$ | Weak | | [[Locally Integrable\|Locally Integrable Function]] | Bounded, Bounded Support | $\loci$ | [[Almost Everywhere]] (on [[Lebesgue Set]]) | | $L^p$ Function | All | $L^p$ | $L^p$ | | [[Continuity\|Continuous Function]] | Bounded Support | $C$ | [[Pointwise Convergence\|Pointwise]] | | [[Uniform Continuity\|Uniformly Continuous Function]] | Bounded Support | $UC$ | [[Uniform Convergence\|Uniform]] | | Bounded Continuous Function | All | $BC$ | Pointwise | | Bounded Uniformly Continuous Function | All | $BC \cap UC$ | Uniform | | [[Space of Test Functions\|Test Function]] | $\cd$ | $\cd$ | $\cd$ | | [[Distribution]] | $\cd$ | $C^\infty$ | $\cd'$ | | [[Sobolev Space\|Sobolev Function]] | $C_c^\infty$ | $C^\infty$ | $W^{k, p}_{\text{loc}}$ | | $W^{k, p}$, compactly supported | $C_c^\infty$ | $C_c^\infty$ | $W^{k, p}$ | Here, multiple versions of the convolution are at play: - Against functions: the standard [[Convolution of Functions|convolution of functions]]. - Against measures: view $\phi(x) dx$ as a measure, use [[Convolution of Measures|convolution of measures]], and extract the resulting density function. - Against distributions: [[Convolution of Distribution|convolution of distributions]]. > [!theorem] > > Let $\mu$ be a probability measure on $\real^d$ and $\phi \in L^1$ be a mollifier. > 1. $\phi_t \to \delta_0$ weakly. > 2. $\phi_t * \mu \to \mu$ weakly. > 3. If $\phi \in UC$, then $\phi_t * \mu$ is continuous. > > *Proof*. Suppose that $\phi \in UC$, then $\phi_t \in UC$ for all $t > 0$, and it's sufficient to show continuity for $\phi$. In this case, > $ > \abs{\phi * \mu(x) - \phi * \mu(y)} \le \int \abs{f(x - z) - f(y - z)}d\mu(z) \to 0 > $ > as $\abs{x - y} \to 0$, independent of $x$ and $y$ by uniform continuity. > [!theorem] > > Let $f \in L^p$ ($p \in [1, \infty)$), and $\phi \in L^1$ with $\int \phi = 1$, then $\phi_t * f \to f$ in $L^p$ as $t \to 0$. > > *Proof*. Over a change of variables with $y = tz$, > $ > \begin{align*} > \phi_t * f(x) - f(x) &= \int [f(x - y) - f(x)]\phi_t(y)dy \\ > &= t^d\int [f(x - tz) - f(x)]\frac{\phi(z)}{t^d}dz \\ > &= \int [f(x - tz) - f(x)]\phi(z)dz \\ > &= \int [\tau_{tz}f(x) - f(x)]\phi(z)dz > \end{align*} > $ > By [[Minkowski's Inequality]], > $ > \begin{align*} > \norm{\phi_t * f - f}_p &= \braks{\int \paren{\int [\tau_{tz}f(x) - f(x)]\phi(z)dz}^p dx}^{1/p} \\ > &\le \int \norm{\tau_{tz}f - f}_p \abs{\phi(z)}dz > \end{align*} > $ > Since $\norm{\tau_{tz}f - f}_p \le 2\norm{f}_p$, $2\norm{f}_p \abs{\phi}$ dominates the function for all $z$. By the dominated convergence theorem, the norm converges to $0$ as $t \to 0$. > [!theorem] > > Let $f$ be a continuous function and $\phi \in L^1$ be a mollifier. > 1. If $f$ is bounded or $\phi$ has bounded support, then $\phi_t * f \to f$ pointwise > 2. Suppose that $f$ is uniformly continuous. If $f$ is bounded or $\phi$ has bounded support, then $\phi_t * f \to f$ uniformly. > > *Proof*. For any $x, y \in \real^d$, > $ > \begin{align*} > \abs{\phi_t * f(x) - f(x)} &= \int \phi_t(z)f(x - z) - \phi_t(z)f(x)dz \\ > &\le \int \abs{\phi_t(z)} \cdot \abs{f(x - z) - f(x)}dz \\ > &= \int \abs{\phi(z)} \cdot \abs{f(x - tz) - f(x)}dz > \end{align*} > $ > Suppose that $\phi$ has bounded (compact) support, then by continuity, $\abs{\phi_t * f(x) - f(x)} \to 0$ as $t \to 0$. If $f$ is also uniformly continuous, then this convergence does not depend on $x$. > > Suppose that $f$ is bounded and let $\eps > 0$, then there exists $R \ge 0$ such that $\int_{B(0, R)^c} \abs{\phi} < \eps/(2\norm{f}_u)$. So > $ > \begin{align*} > \int \abs{\phi(z)} \cdot \abs{f(x - tz) - f(x)}dz &\le \int_{{B(0, R)}} \abs{\phi(z)} \cdot \abs{f(x - tz) - f(x)}dz \\ > &+ 2\norm{f}_u\cdot \int_{B(0, R)^c} \abs{\phi} \\ > &= \int_{{B(0, R)}}\abs{\phi_t(z)}\abs{f(x - z) - f(x)}dz + \eps > \end{align*} > $ > By the same argument, $\abs{\phi_t * f(x) - f(x)} \to 0$ as $t \to 0$. If $f$ is uniformly continuous, then this convergence does not depend on $x$. > [!theorem] > > Let $\psi \in \cd$ be a [[Space of Test Functions|test function]], then $\phi_t * \psi \to \psi$ in $\cd$ as $t \to 0$. > > *Proof*. Suppose that $\supp{\phi} \subset \ol{B(0, R)}$, then > $ > \supp{\phi_t * \psi} \subset \supp{\psi} + \ol{B(0, tR)} > $ > so for sufficiently small $t$, $\phi_t * \psi$ are supported on a common compact set. From here, by the continuous result, $\phi_t * \psi \to \psi$ uniformly as $t \to 0$. Since $\partial^\alpha(\phi_t * \psi) = \phi_t * (\partial^\alpha \psi)$, the uniform convergence holds on all derivatives, so $\phi_t * \psi \to \psi$ in $\cd$. > [!theorem] > > Let $F \in \cd'$ be a distribution, then $\phi_t * F \to F$ in $\cd'$ as $t \to 0$. > > *Proof*. Let $\psi \in \cd$, then > $ > \angles{\phi_t * F, \psi} = \anglesn{F, \psi * \td{\psi_t}} \to \angles{F, \psi} > $ > so $\phi_t * F \to F$ in $\cd'$.