# An integral operator inequality

This problem is from exercise 30 on page 196 of Folland's "Real Analysis". Let $K$ be a non-negative measurable function on $[0,\infty)$ and let

$\phi(s) = \displaystyle\int_0 ^\infty K(x)x^{s-1} dx$.

I am trying to prove that if $f$ and $g$ are non-negative measurable functions and $p$ and $q$ are conjugate exponents, then

$\displaystyle\int_0 ^\infty \int_0 ^\infty K(xy)f(x)g(y) dxdy \leq \phi(1/p) \left(\int_0 ^\infty x^{p-2} f(x)^p dx\right)^{1/p} \left(\int_0 ^\infty g(x)^q dx\right)^{1/q}$.

So far, I have been attempting various combinations of Hölder's inequality and the theorem given in the text for a function $K$ on $\mathbb R^2$ which is homogenous of degree $-1$, but have not met with much success. Any suggestions?

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First perform the change of variable $(x,y)\mapsto(xy,y)=(u,v)$. The left-hand side becomes $$\int K(u)\int \frac{f(u/v)}{v}g(v)\,dv\,du.$$ Now apply Hölder to the inner integral, to get \begin{align*} \int \frac{f(u/v)}{v}g(v)\,dv &\leq \left(\int\frac{f(u/v)^p}{v^p}\,dv\right)^{1/p} \left(\int g(v)^q\,dv\right)^{1/q} \\ &= \left(u^{1-p}\int f(t)^p t^{p-2}\,dt\right)^{1/p} \left(\int g(v)^q\,dv\right)^{1/q}; \end{align*} passing from the first line to the second involves another change of variable (namely, $v\mapsto u/v=t$). Altogether then, \begin{align*} \int K(u)\int \frac{f(u/v)}{v}g(v)\,dv\,du &\leq \int K(u)\left(u^{1-p}\int f(t)^p t^{p-2}\,dt\right)^{1/p} \left(\int g(v)^q\,dv\right)^{1/q}\,du \\ & = \left(\int K(u)u^{1/p-1}\,du\right) \left(\int f(t)^p t^{p-2}\,dt\right)^{1/p} \left(\int g(v)^q\,dv\right)^{1/q}, \end{align*} and that's what you want. Correct me if I'm wrong, but Folland's Theorem 2.47 should serve to justify all of the substitutions I did above.