Let $T(f)(y)=\int_0^\infty K(x,y)\cdot f(x)dx$, $\,$ show $\,$ $\Vert T(f) \Vert _p \le C\cdot \Vert f \Vert _p$

Question

Now here is the full statement.

Let $K:(0, +\infty) \times (0, +\infty) \rightarrow \Bbb R$ be a Lebesgue measurable function with $K(kx,ky)=k^{-1}K(x,y)$ for every $k>0$ and let $$\int_0^\infty |K(x,1)|\cdot x^{-\frac 1p}\,dx=C<\infty$$ For some $p\in[1,\infty]$. If $f\in L^p$ show that the function $$T(f)(y)=\int_0^\infty K(x,y)\cdot f(x)\,dx$$ is well defined for almost every $y\in (0,+\infty)$ and satisfies $\Vert T(f) \Vert _p \leq C \Vert f \Vert _p$.

I've been using inequalities for hours and got nowhere. I tried using Hölder's inequality and Tonelli's Theorem in many ways and got frustrated everytime. Any idea would help me a lot.

Answer 1

Observe we have \begin{align} \|T(f)\|_p =& \left(\int^\infty_0 \left|\int^\infty_0 K(x, y)f(x)\ dx \right|^p\ dy\right)^{1/p}\\ =&\ \left(\int^\infty_0 \left|\int^\infty_0 K(u, 1)f(yu)\ du \right|^p\ dy\right)^{1/p}. \end{align} Then by Minkowski's integral inequality (essentially triangle inequality), we have \begin{align} \|T(f)\|_p \leq&\ \int^\infty_0 |K(u, 1)|\left( \int^\infty_0 |f(yu)|^p dy\right)^{1/p}\ du\\ =&\ \int^\infty_0|K(u, 1)|u^{-1/p}\left(\int^\infty_0 |f(x)|^p\ dx \right)^{1/p}\ du. \end{align}

Edit: If $p \in [1, \infty)$, then we know that $C^\infty_0(\mathbb{R})$ is dense in $L^p(\mathbb{R})$. Since the above estimate holds on $C^\infty_0(\mathbb{R})$ then it extends uniquely to all of $L^p(\mathbb{R})$. Since $T(f) \in L^p$ then we know that the set of $y \in \mathbb{R}$ such that $T(f)(y)=\infty$ is measure zero.

In the case $p=\infty$, we see that \begin{align} |T(f)(y)| =&\ \left|\int^\infty_0 K(u, 1) f(yu)\ du\right|\\ \leq &\ \int^\infty_0 |K(u, 1)|\ du\cdot \|f\|_{L^\infty}. \end{align} Hence $T$ is well-defined for all $f \in L^\infty$.