Need help understanding this proof of a certain inequality of $L^p$ norms.

Question

The following theorem and proof is lifted from Folland (Real Analysis: Modern Techniques and their Applications). I am having trouble understanding one single line of the proof:

Theorem: Let $K$ be a Lebesgue measurable function on $(0, \infty) \times (0, \infty)$ such that $K(\lambda x, \lambda y)=\lambda^{-1}K(x, y)$ for all $\lambda > 0$ and $\int_0^\infty |K(x, 1)|x^{-1/p} dx = C < \infty$ for some $p \in [1, \infty],$ and let $q$ be the conjugate exponent to $p.$ For $f \in L^p,$ let $$Tf(y) = \int_0^\infty K(x, y)f(x) dx.$$ Then $Tf$ and is defined a.e., and $\|Tf\|_p \le C\|f\|_p.$

Proof: Setting $z = x/y,$ we have $$\int_0^\infty |K(x, y)f(x)|dx = \int_0^\infty |K(yz, y)f(yz)|ydz = \int_0^\infty |K(z, 1)f_z(y)|dz$$ where $f_z(y) = f(yz);$ moreover, $$\|f_z\|_p = \left[ \int_0^\infty |f(yz)|^p dy\right]^{1/p} = \left[\int_0^\infty |f(x)|^p z^{-1} dz\right]^{1/p} = z^{-1/p}\|f\|_p.$$

[[Therefore, by Minkowski's inequality for integrals, $Tf$ exists a.e. and $$\|Tf\|_p \le \int_0^\infty |K(z, 1)|\|f_z\|_p dz\Big]\Big] = \|f\|_p\int_0^\infty |K(z, 1)|z^{-1/p}dz = C\|f\|_p.$$

Q.E.D.

I do not understand the last line, between the double brackets [[ ]]. I understand why everything else is true. I have tried using Minkowski's inequality, but I can't get it to give me what I want.... Thank you!

Answer 1

\begin{align*} \|Tf\|_{p}&=\left(\int_{0}^{\infty}\left|\int_{0}^{\infty}K(x,y)f(x)dx\right|^{p}dy\right)^{1/p}\\ &\leq\left(\int_{0}^{\infty}\left(\int_{0}^{\infty}|K(x,y)f(x)|dx\right)^{p}dy\right)^{1/p}\\ &=\left(\int_{0}^{\infty}\left(\int_{0}^{\infty}|K(z,1)f_{z}(y)|dz\right)^{p}dy\right)^{1/p}\\ &\leq\int_{0}^{\infty}\left(\int_{0}^{\infty}|K(z,1)f_{z}(y)|^{p}dy\right)^{1/p}dz\\ &=\int_{0}^{\infty}|K(z,1)|\left(\int_{0}^{\infty}|f_{z}(y)|dy\right)^{1/p}dz\\ &=\int_{0}^{\infty}|K(z,1)|\|f_{z}\|_{p}dz. \end{align*}

Answer 2

Here I did a direct way as follows :

$i) : 1\le p<\infty,$ one has

\begin{align} \|Tf\|_{L^{p}}&\le\bigg(\int_{(0,\infty)}\bigg(\int_{(0,\infty)}|f(s)K(s,t)|~ds\bigg)^{p}dt\bigg)^{1/p}\\ &=\bigg(\int_{(0,\infty)}\bigg(\int_{(0,\infty)}|f(s)\frac{1}{t}K(\frac{s}{t},1)|~ds\bigg)^{p}dt\bigg)^{1/p}\\ &=\bigg(\int_{(0,\infty)}\bigg(\int_{(0,\infty)}\frac{1}{t}|f(s)K(\frac{s}{t},1)|~ds\bigg)^{p}dt\bigg)^{1/p}\\ &=\bigg(\int_{(0,\infty)}\bigg(\int_{(0,\infty)}|f(st)K(s,1)|~ds\bigg)^{p}dt\bigg)^{1/p}\\ &\le\int_{(0,\infty)}\bigg(\int_{(0,\infty)}|f(st)K(s,1)|^{p}~dt\bigg)^{1/p}ds\\ &=\int_{(0,\infty)}|K(s,1)|\bigg(\int_{(0,\infty)}|f(st)|^{p}~dt\bigg)^{1/p}ds\\ &=\int_{(0,\infty)}|K(s,1)|s^{\frac{-1}{p}}\bigg(\int_{(0,\infty)}|f(t)|^{p}~dt\bigg)^{1/p}ds\\ &=\|f\|_{L^{p}}\int_{(0,\infty)}|K(s,1)|s^{\frac{-1}{p}}~ds \end{align}

and

$ii):p=\infty$, we have

\begin{align} |(Tf)(t)|&=\bigg|\int_{(0,\infty)}f(s)K(s,t)~ds\bigg|\\ &\le\int_{(0,\infty)}\big|f(s)K(s,t)\big|~ds\\ &\le\|f\|_{L^{\infty}}\int_{(0,\infty)}|K(s,t)|~ds\\ &=\|f\|_{L^{\infty}}\int_{(0,\infty)}\frac{1}{t}|K(\frac{s}{t},1)|~ds\\ &=\|f\|_{L^{\infty}}\int_{(0,\infty)}|K(s,1)|~ds\\ \end{align}