A Generalized Schur’s Test

In Folland’s chapter on $L^p$ spaces, theorem 6.18 is called Schur’s Test. I wish to generalize this test further.

Motivation: To be real honest with you, I never had to use this general test in practice. Folland’s theorem suffices in most cases. So I rather just treat the following theorem as an exercise in manipulations using Holder’s inequality and Fubini’s theorem, which obviously has tons of practical uses in mathematics.

Theorem: Let $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ be two $\sigma$-finite measure spaces and let $K$ be an $\mathcal{M}\otimes\mathcal{N}$-measurable function on $X\times Y.$ Consider any $p\in(1,\infty).$ Say, there exists positive measurable functions $u$ and $w$ on $X$ and $Y$ respectively such that

$\int |K(x,y)|u(x)d\mu(x)\le Bw(y)$ for $\nu$-every $y\in Y$

$\int |K(x,y)|w(y)^{1/(p-1)}d\nu(y)\le Au(x)^{1/(p-1)}$ for $\mu$-every $x\in X$

Then, for any $f\in L^p(\nu),$ the integral

$Tf(x)=\int K(x,y)f(y)d\nu(y)$

converges absolutely for a.e $x\in X,$ the function $Tf$ thus defined is in $L^p(\mu),$ and $||Tf||_{op}\le A^{1-1/p}B^{1/p}.$

Proof: Consider any $f\in L^p(\nu).$ Since $w$ is positive, $w(y)^{-1/p}$ and $w(y)^{-1}$ are defined a.e. Let $q$ be the Holder-conjugate exponent of $p.$ Now, for $\mu$-almost all $x\in X,$ we have by Holder's inequality

$\int |K(x,y)||f(y)|d\nu(y)=\int \left(|K(x,y)|^{1/q}w(y)^{1/p}\right)\left(|K(x,y)|^{1/p}|f(y)|w(y)^{-1/p}\right)d\nu(y)$

$\le\left[\int|K(x,y)|w(y)^{q/p}d\nu(y)\right]^{1/q}\left[\int |K(x,y)||f(y)|^pw(y)^{-1}d\nu(y)\right]^{1/p}$

$=\left[\int|K(x,y)|w(y)^{1/(p-1)}d\nu(y)\right]^{1/q}\left[\int |K(x,y)||f(y)|^pw(y)^{-1}d\nu(y)\right]^{1/p}$

$\le\left[Au(x)^{1/(p-1)}\right]^{1/q}\left[\int |K(x,y)||f(y)|^pw(y)^{-1}d\nu(y)\right]^{1/p}$

$=\left[A^{1/q}u(x)^{1/p}\right]\left[\int |K(x,y)||f(y)|^pw(y)^{-1}d\nu(y)\right]^{1/p}$ since $(p-1)q=p$

By Tonelli's theorem, $x\mapsto\int |K(x,y)||f(y)|d\nu(y)$ is integrable. Now,

$\int\left[\int |K(x,y)||f(y)|d\nu(y)\right]^pd\mu(x)\le\int\left[\left(A^{1/q}u(x)^{1/p}\right)^p\int |K(x,y)||f(y)|^pw(y)^{-1}d\nu(y)\right]d\mu(x)$

$=\int\int A^{p/q}u(x)|K(x,y)||f(y)|^pw(y)^{-1}d\nu(y)d\mu(x)$

$=\int\int A^{p/q}u(x)|K(x,y)||f(y)|^pw(y)^{-1}d\mu(x)d\nu(y)$ (by Tonelli's theorem)

$=\int A^{p/q}|f(y)|^pw(y)^{-1}\left[\int|K(x,y)|u(x)d\mu(x)\right]d\nu(y)$

$\le \int A^{p/q}|f(y)|^pw(y)^{-1}Bw(y)d\nu(y)$

$=A^{p/q}B\int |f(y)|^pd\nu(y)=A^{p/q}B||f||_p^p<\infty$ (as $f\in L^p(\nu)$)

So, $\int |K(x,y)||f(y)|d\nu(y)<\infty$ for $\mu$-almost all $x\in X$ by prop. 2.20 (Folland). So $Tf(x)$ is defined $\mu$-a.e.

By Tonelli's theorem, both $x\mapsto\int G_x^+d\nu$ and $x\mapsto\int G_x^-d\nu$ are measurable, where $G(x,y)=K(x,y)f(y).$ Hence, $Tf$ being the difference of the two measurable functions is also measurable. Now,

$||Tf||_p^p=\int \left|\int K(x,y)f(y)d\nu(y)\right|^pd\mu(x)$

$\implies||Tf||_p^p\le\int\left(\int |K(x,y)||f(y)|d\nu(y)\right)^pd\mu(x)\le A^{p/q}B||f||_p^p$

$\implies||Tf||_p\le A^{1/q}B^{1/p}||f||_p=A^{1-1/p}B^{1/p}||f||_p<\infty$

Since $f\in L^p(\nu)$ is arbitrary, we conclude that $T$ maps $L^p(\nu)$ to $L^p(\mu).$ Furthermore, $||T||_{op}\le A^{1-1/p}B^{1/p}.$