# Does $z$ depend on $x$ in the proof of Theorem 6.20 in Folland?

The theorem states:

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 $$f \in L^p$$ and $$g \in L^q$$, let $$Tf(y) = \int_0^\infty K(x,y)f(x)\,dx, \quad Sg(x) = \int_0^\infty K(x,y)g(y)\,dy.$$ Then $$Tf$$ and $$Sg$$ are defind a.e., and $$\|Tf\|_p \leq C\|f\|_p$$ and $$\|Sg\|_q \leq C\|g\|_q$$.

The proof proceeds as follows:

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

The doubt I have is rather elementary. In the very last equality, how can we pull out $$z$$ if $$z$$ itself depends on $$x$$?

The proof then proceeds:

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

How does the Minkowski inequality for integrals imply the existence of $$Tf$$?

I understand the rest of the proof so I have omitted it.

In

$$\|f_z\|_p = \Bigg[\int_0^\infty |f(yz)|^p\,dy\Bigg]^{1/p} = \Bigg[\int_0^\infty |f(x)|^pz^{-1}\,dx\Bigg]^{1/p} = z^{-1/p}\|f\|_p.$$ the change of variable $$x=yz$$ is performed. After that, $$x$$ is the variable of integration, and $$z$$ is a constant. Therefore, it call be pulled out of the integral.

The existence of $$Tf$$ is just a consequence of the bounding of the map $$x \mapsto \lvert K(x,1) f(x) \rvert$$ by a map which is integrable.

• So the implicit dependence of $x$ within $z$ is not a problem? Commented Jul 10, 2022 at 8:14
• @CBBAM No, it is not. What is important is the variable of integration. Commented Jul 10, 2022 at 8:16

When estimating $$\|f_z\|_p$$, the $$z$$ is fixed, and $$x$$ is there used as a variable of integration unrelated to $$z$$.

As for the use of Minkowski's inequality, it's in the inequality below: \begin{align} \|Tf\|_p &=\bigg[\int_0^\infty\bigg|\int_0^\infty |K(z,1) f_z(y)|\,dz\bigg|^p\,dy\bigg]^{1/p}\\[0.3cm] &\leq \int_0^\infty\bigg(\int_0^\infty|K (z,1)|^p\,|f_z(y)|^p\,dy \bigg)^{1/p}\,dz \\[0.3cm] &= \int_0^\infty |K (z,1)|\,\bigg(\int_0^\infty\,|f_z(y)|^p\,dy \bigg)^{1/p}\,dz \\[0.3cm] &= \int_0^\infty |K (z,1)|\,\|f_z\|_p\,dz\\[0.3cm] &= \int_0^\infty |K (z,1)|\,\|f\|_p\,z^{-1/p}\,dz. \\[0.3cm] &=C\,\|f\|_p. \end{align}

• Thank you! So the implicit dependence of $x$ within $z$ is not a problem? If we rewrite $z$ as $z(x)$, wouldn't it need to be seen as a function as opposed to a fixed constant? Commented Jul 10, 2022 at 8:15
• If it makes you nervous, you can call it something else. The author introduces a function $f_{\rm something}$, and proceeds to calculate its $p$-norm in terms of something. Then he uses that calculation in the integral where $z$ appeared as a substitution for $x$. Commented Jul 10, 2022 at 8:27
• And that's the point, $z$ appeared in a substitution. There is no $x$ anymore. Commented Jul 10, 2022 at 8:28