> [!definition] > > Let $f, g$ be [[Measurable Function|measurable functions]] on $\real^n$. The **convolution** of $f$ and $g$, $f * g$ is the function > $ > f * g(x) = \int f(x - y)g(y)\ dy > $ > for all $x$ on which the [[Integral|integral]] exists. > [!theorem] > > Let $f, g$ be measurable functions on $\real^n$, and suppose that all relevant integrals exist, then > 1. $f * g = g * f$ > 2. $(f * g) * h = f * (g * h)$ > 3. For any $z \in \real^n$, $\tau_z(f * g) = (\tau_z f) * g = f * (\tau_z g)$, where $\tau_z f(x) = f(x - z)$ is the translation map. > 4. If $A$ is the [[Topological Closure|closure]] of $\supp{f} + \supp{y}$, then $\supp{f * g} \subset A$. > > *Proof*. Firstly, substituting $z = x - y$ yields > $ > \begin{align*} > f * g(x) &= \int f(x - y)g(y)dy \\ > &= \int f(z)g(x - z)dz = g * f(x) > \end{align*} > $ > Secondly, by the [[Fubini-Tonelli Theorem]] and the statement above, > $ > \begin{align*} > (f * g) * h(x) &= \int\int f(x - y - z)g(y)dy \cdot h(z)dz \\ > &= \int\int f(y)g(x - y - z)h(z) dy \cdot dz \\ > &= \int f(y) \cdot \int g(x - y - z)h(z) dz \cdot dy \\ > &= \int f(y) \cdot (g * h)(x - y)dy \\ > &= f * (g * h)(x) > \end{align*} > $ > Thirdly, > $ > \begin{align*} > \tau_z(f * g)(x) &= \int f(x - z - y)g(y)dy \\ > &= \int \tau_zf(x - y)g(y)dy \\ > &= (\tau_z f) * g(x) > \end{align*} > $ > and lastly, if $x \not\in A$, then for any $y \in \supp{g}$, $x - y \not\in \supp{f}$, so $f(x - y)g(y) = 0$ for all $y$, and $f * g(x) = 0$. > [!theorem] > > Let $p$ and $q$ be Holder conjugates, $f \in L^p$, and $g \in L^q$, then > 1. $f * g$ exists everywhere. > 2. $f * g$ is bounded with $\norm{f * g}_u \le \norm{f}_p \norm{g}_q$. > 3. $f * g$ is uniformly continuous. > 4. If $p, q \in (1, \infty)$, then $f * g \in C_0(\real^n)$. > > *Proof*. The bound for $\norm{f * g}_u$ is based on [[Hölder's Inequality]]. If $p < \infty$, then > $ > \norm{\tau_y(f * g) - f * g}_u = \norm{(\tau_yf - f) * g}_u \le \norm{\tau_y - f}_p \norm{g}_q \to 0 > $ > as $y \to 0$. If $p, q \in (1, \infty)$, then there exists compactly supported functions $\seq{f_n} \subset L^p$, $\seq{g_n} \subset L^q$ such that $f_n \to f$ in $L^p$ and $g_n \to g$ in $L^q$. Since each $f_n * g_n$ and > $ > \begin{align*} > \norm{f_n * g_n - f * g}_u &\le \norm{f_n * g_n - f * g_n}_u + \norm{f * g_n - f * g}_u \\ > &\le \norm{f_n - f}_p\norm{g_n}_q + \norm{f}_p \norm{g_n - g}_q \to 0 > \end{align*} > $ > thus $f * g \in C_0$. > [!definition] Discrete Case > > $ > (p * q)(k) = \sum_{m = -\infty}^\infty p(m)q(k - m) > $ > Let $X$ and $Y$ be [[Probabilistic Independence|independent]] [[Random Variable|random variables]] with the [[Probability Distribution|probability distributions]] $p$ and $q$, then the probability distribution of $X + Y$ is known as the **convolution** of $p$ and $q$, denoted as $p * q$. > > Given the range of $X$, $R_X = \{1, 2, \cdots, m\}$ and the range of $Y$, $R_Y = \{1, 2, \cdots, n\}$, the range of $X + Y$ would be $R_{X+Y} = \{2, \cdots, m + n\}$. Let $k \in R_{X + Y}$. Since a [[Probability Distribution|probability distribution]] is a [[Measure Space|measure]], addition can be converted into set union. > $ > (p * q)(k) = P(X + Y = k) > $ > Since the event $(X + Y = k)$ can occur in many different ways ($(X = 1) \cap (Y = k - 1)$, $(X = 2) \cap (Y = k - 2)$, ... $(X = k - 1) \cap (Y = 1)$), and these events form a [[Partition|partition]] of $X + Y = k$: > $ > \begin{align*} > (p * q)(k) &= P\paren{\bigcup_{m = 1}^{k - 1}(X = m) \cap (Y = k - m)} \\ > &= \sum_{m = 1}^{k - 1} P((X = m) \cap (Y = k - m)) \\ > &= \sum_{m = 1}^{k - 1} P(X = m)P(Y = k - m) \\ > &= \sum_{m = 1}^{k}p_mq_{k - m} > \end{align*} > $ > > Expanding the range of both $X$ and $Y$ to integers between $-\infty$ and $\infty$ by filling the probability distribution with $0$ everywhere else, the formula can be generalised for any ranges containing integers. > [!theorem] > > Let $\phi, \psi \in C_c(\real^d)$, and $R$ be a rectangle covering $\supp{\phi} + \supp{\psi}$. For each $n \in \nat$, let > - $\seqf{R_{n, j}}$ as a disjoint covering of $\supp{\phi} + \supp{\psi}$ with cubes of volume $\Delta_n$. > - $\seqf{y_{n, j}}$ with $y_{n, j} \in R_{n, j}$ for all $n \in \nat, j \in [n]$. > > such that > 1. $\Delta_n \to 0$ as $n \to \infty$. > 2. $\sup_{n \in \nat}n\Delta_n < \infty$. > > Then by continuity, for each $x \in \supp{\phi * \psi}$, > $ > \phi * \psi(x) = \int \phi(y)\psi(x - y)dy = \limv{n}\Delta_n\sum_{j = 1}^n\phi(y_{n, j})\psi(x - y_{n, j}) > $ > Moreover, the convergence is [[Uniform Convergence|uniform]]. > > *Proof*. Let $M \ge 0$ such that $n\Delta_n \le M$ for all $n \in\nat$. Let $\eps > 0$, by [[Uniform Continuity|uniform continuity]], there exists $N \in \nat$ such that $\abs{\psi(x_1) - \psi(x_2)} < \eps/M$ whenever $\abs{x_1, x_2} \le d\Delta_n$[^1] for all $n \ge N$. From here, > $ > \abs{\Delta_n \sum_{j = 1}^n \phi(y_{n, j})\psi(x - y_{n, j}) - \int\phi(y)\psi(x - y)dy} \le \eps n\Delta_n/M \le \eps > $ > for all $n \ge N$. Therefore the convergence is not uniform.