> [!theorem] > > Let $X$ be a [[Compactness|compact]] [[Hausdorff Space|Hausdorff space]], $(Y, \cm, \mu)$ be a [[Measure Space|measure space]], and $p, q \in [1, \infty]$ be [[Hölder's Inequality|Hölder conjugates]]. Let > $ > T: L^q(Y, \cm, \mu) \to C(X, [0, 1]) > $ > be a [[Bounded Linear Map|bounded linear map]]. If $p \in (1, \infty)$ or $\mu$ is $\sigma$-finite, then $T$ is [[Compact Operator|compact]] if and only if there exists a map $K: X \times Y \to \complex$ such that: > 1. $(Tf)(x) = \int_Y K(x, y)f(y) d\mu(y)$ for all $x \in X$. > 2. $K(x, \cdot) \in L^p(Y)$ for all $x \in X$. > 3. The mapping $X \to L^p(Y)$ given by $x \mapsto K(x, \cdot)$ is [[Continuity|continuous]]. > > where $K$ is known as the **kernel** of $T$. > > The converse holds regardless of assumptions on $\mu$ and $p$. > > Moreover, if $\nu$ is a finite [[Borel Measure|Borel measure]], then for any $r \in [1, \infty]$, $C(X, [0, 1])$ continuously embeds itself into $L^r(X, \cb(X), \nu)$. Thus if $T: L^q(Y) \to C(X, [0, 1])$ is compact, then so is $T$ viewed as a map from $L^q(Y)$ to $L^r(X)$. # Integral to Compact > [!theorem] > > Let $K: X \times Y \to \complex$ be a mapping, and $p, q \in [1, \infty]$ be Hölder conjugates. Suppose that > 1. $K(x, \cdot) \in L^p(Y)$ for all $x \in X$. > 2. The mapping $X \to L^p(Y)$ given by $x \mapsto K(x, \cdot)$ is continuous. > > then the integral operator given by > $ > T: L^q(Y) \to C(X, [0, 1]) \quad f \mapsto \int_X K(x, y)f(y)d\mu(y) > $ > is compact. > > *Proof*. Since the mapping $x \mapsto \norm{K(x, \cdot)}_p$ is continuous, it admits a maximum $M$. Let $f \in L^q(X, \cm, \mu)$, then by Hölder's Inequality, > $ > \abs{(Tf)(x)} \le \norm{K(x, \cdot)}_p \cdot \norm{f}_q \le M \norm{f}_q > $ > Let $x \in [0, 1]$ and $\eps > 0$, then as $x \mapsto K(x, \cdot)$ is continuous, there exists a neighbourhood $U$ of $x$ such that $\norm{K(x, \cdot) - K(x', \cdot)}_p < \eps$ whenever $x' \in U$. From here, using Hölder again gives > $ > \begin{align*} > \abs{Tf(x) - Tf(x')} &\le \int_X \abs{K(x, y) - K(x', y)} \cdot \abs{f(y)} d\mu(y) \\ > &\le \norm{K(x, \cdot) - K(x', \cdot)}_p \cdot \norm{f}_q < \eps \norm{f}_q > \end{align*} > $ > Let $C \ge 0$, then for any $g \in L^q$ such that $\norm{g}_q \le C$, > $ > \norm{(Tg)(x) - Tg(x')} < \eps \norm{g}_q \le C\eps > $ > whenever $x' \in U$. Thus if $\norm{g}_q \le C$, then there exists a neighbourhood $U$ of $x$, independent of the choice of $g$, such that $\norm{Tg(x) - Tg(x')} < C\eps$ for all $x' \in U$. Therefore the family is equicontinuous and uniformly bounded. By the Arzelà–Ascoli theorem, it is pre-compact. Thus $T$ is compact. # Compact to Integral > [!theorem] > > Let > $ > T: L^q(Y) \to C(X, [0, 1]) > $ > be a compact operator. If $p \in (1, \infty)$ or $\mu$ is $\sigma$-finite, then there exists $K: X \times Y \to \complex$ such that > 1. $K(x, \cdot) \in L^p(Y)$ for all $x \in X$. > 2. The mapping $X \to L^p(Y)$ given by $x \mapsto K(x, \cdot)$ is [[Continuity|continuous]]. > 3. $(Tf)(x) = \int_{Y} K(x, y)f(y)d\mu(y)$ for all $f \in L^q(Y)$, $x \in X$. > > *Proof*. Firstly, since the map $f \mapsto (Tf)(x)$ is bounded, there exists $K(x, \cdot) \in L^p(Y)$ such that > $ > (Tf)(x) = \int K(x, y)f(y)d\mu(y) > $ > Since $T$ is compact, by the Arzelà–Ascoli theorem, the family $\bracs{Tf:f \in L^p(Y), \norm{f} = 1}$ is equicontinuous, so for any $x \in X$, > $ > \norm{K(x, \cdot) - K(y, \cdot)} = \sup_{\norm{f} = 1}\abs{\angles{f, K(x, \cdot) - K(y, \cdot)}} \to 0 > $ > as $y \to x$.