Show that the mapping $(Tf)(x) = \frac{1}{x}\int_0^{\infty}K(\frac{y}{x})f(y)dy$ is bounded

Question

Let $K$ be a Lebesgue measurable function defined on $(0, \infty)$ and assume $\int_0^{\infty}\vert\,K(y)\,\vert y^{-\frac{1}{p}}dy < \infty$. Pick $f \in L_p(0, \infty) (p > 1)$ and define a linear mapping $T$ by: $$(Tf)(x) = \int_0^{\infty}K(y)f(x y) dy = \frac{1}{x}\int_0^{\infty}K(\frac{y}{x})f(y)dy, x > 0$$

My questions are:

1. $\forall\,f \in L_p(0, \infty), Tf \in L_p(0, \infty)$ (solved)
2. $\|\,T\| \leq \int_0^{\infty}\vert\,K(y)\,\vert y^{-\frac{1}{p}}dy$ (solved) and the equality holds when $K$ is non-negative (not solved)

To prove the first two, by using Minkowski's integral inequality and replace $y$ by $t x$, I tried: $$[\int_0^{\infty}\vert\,\int_0^{\infty}\frac{1}{x}K(t)f(t x)d t\,\vert^p dx]^{\frac{1}{p}} \leq \int_0^{\infty}[\int_0^{\infty}\vert\,\frac{1}{x}K(t)f(t x)\,\vert^p dx]^{\frac{1}{p}} dt = \\\int_0^{\infty}\vert\,K(t)\,\vert[\int_0^{\infty}\vert\,\frac{t}{x}f(x)\,\vert^p\frac{1}{t}dx]^{\frac{1}{p}} d t = \int_0^{\infty}\vert\,K(t)\,\vert t^{-\frac{1}{p}}\underline{[\int_0^{\infty}\vert\,\frac{t}{x}f(x)\,\vert^p dx]^{\frac{1}{p}}} d t$$

Then I can not compare the underlined part and $\|f\|_p$. If my attempt is on the right track, could you anyone provide hints on how to do the desired comparison? Also I was block by the second part of question 2 and the third question. I am not sure if I can apply Minkowski's inequality when $K$ is not absolutely integrable. Here is a similar question but I could not find it helpful ...

Answer 1

It seems to me that the assumption on $K$ should be that $$\int^\infty_0 |K(y)|\,y^{-1/p}\,dy<\infty$$ As you pointed out, the generalized Minkowski inequality gives \begin{align} \left(\int^\infty_0\left|\int^\infty_0 K(y) f(xy)\,dy\right|^p\,dx\right)^{1/p} &\leq \left(\int^\infty_0\left(\int^\infty_0 |K(y)| |f(xy)|\,dy\right)^p\,dx\right)^{1/p}\\ &\leq\int^\infty_0\left(\int^\infty_0 |f(xy)|^p\,dx\right)^{1/p}|K(y)|\,dy\\ &=\|f\|_p\int^\infty_0y^{-1/p}|K(y)|\,dy \end{align}

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This is the version of the general Minkowski inequality that I am using:

Suppose $f:(X\times Y,\mathscr{F}\otimes\mathscr{G})\rightarrow(\mathbb{R},\mathscr{B}(\mathbb{R}))$ is measureable and that where $(X,\mathscr{F},\mu)$ and $(Y,\mathscr{G},\nu)$ are $\sigma$--finite measures. Then, \begin{align} \Big(\int_X\Big|\int_Y f(x,y)\, \nu(dy)\Big|^p\,\mu(dx)\Big)^{\tfrac{1}{p}}\leq \int_Y \Big(\int_X |f(x,y)|^p\,\mu(dx)\Big)^{\tfrac{1}{p}}\,\nu(dy) \end{align} for all $1\leq p<\infty$.

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Comments:

1. Your problem is a particular case of the problem you quoted. Indeed, notice that $$k(x,y):=\frac{1}{y}K(x/y)$$ is homogeneous of order $-1$, that $$\int^\infty_0k(x,1) x^{-1/p}\,dx = \int^\infty_0K(x) x^{-1/p}\,dx,$$ and that $$ Tf(y)=\int^\infty_0k(x,y)f(x)\,dx$$
2. Thus far we have proved that $$\|T\|\leq\int^\infty_0|K(x)|x^{-1/p}\,dx$$ regardless of the sign of $K$.
3. The equality is not straight forward, even in the case $K\geq0$. Consider the particular case $$ K(x)=\mathbb{1}_{(0,1]}(x)$$ which gives the Hardy operator $$Hf(y)=\frac{1}{y}\int^y_0f(x)\,dx$$ Proving that $\|H\|=\frac{p}{p-1}=\int^1_0x^{-1/p}\,dx$ is not trivial. Here is a solution for this particular case