> [!theorem] > > Let $p \in [1, \infty]$, and $K$ be a [[Lebesgue Measurable Function|Lebesgue measurable]] function on $(0, \infty)^2$ such that > 1. $K(\lambda x, \lambda y) = \lambda^{-1}K(x, y)$ for all $\lambda > 0$. > 2. $\int_0^\infty x^{1/p}\abs{K(x, 1)}dx = C < \infty$. > > Let $q$ be the [[Hölder's Inequality|Hölder conjugate]] to $p$. For $f \in L^p$ and $g \in L^q$, define > $ > Tf(t) = \int_0^\infty K(x, y)f(x)dx \quad Sg(x) = \int_0^\infty K(x, y)g(y)dy > $ > then $Tf$ and $Sg$ are defined [[Almost Everywhere|a.e.]], with $\norm{Tf}_p \le C\norm{f}_p$, and $\norm{Sg}_q \le C\norm{g}_q$. > > *Proof*. Let $z = x/y$, then > $ > \begin{align*} > \int \abs{K(x, y)f(x)}dx = y\int \abs{K(yz, y)f(yz)}dz = \int\abs{K(z, 1)f(zy)}dz > \end{align*} > $ > where $f_z(u) = f(zy)$. Since > $ > \norm{f_z}_p = \braks{\int \abs{f(yz)}^pdy}^{1/p} = \braks{z^{-1}\int \abs{f(yz)}^pdy}^{1/p} = z^{-1/p}\norm{f}_p > $ > By [[Minkowski's Inequality]], > $ > \norm{Tf}_p \le \int \abs{K(z, 1)}\norm{f_z}_pdz = \norm{f}_p\int\abs{K(z, 1)}z^{-1/p}dz = C\norm{f}_p > $ > On the other hand, if $u = 1/y$, then > $ > \begin{align*} > \int \abs{K(1, y)}y^{-1/q}dy &= \int \abs{K(y^{-1}, 1)}y^{-1-(1/q)}dy \\ > &= \int \abs{K(u, 1)}y^{-1/p}dy > \end{align*} > $ > so $Sg$ works out as well.