Lebesgue integrable function involving distance function

Question

$f$ is a bounded measurable function with compact support in $\mathbb{R}, \int_\mathbb{R}f(x)dx=0$

$M_f(x)=\sup_{t>0}\{|\int_\mathbb{R}f(y)\frac{t}{t^2+(x-y)^2}dy|\}$

Prove $M_f(x)\in L(\mathbb{R})$

My attempt to solve the problem: I saw Lebesgue integral involving distance function. I tried to use Tonelli Theorem but didn't succeed.

Answer 1

In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(\Bbb R)$:

Theorem. Suppose $f\in L^1(\Bbb R)$. Then $M_f$ is inegrable if and only if $f=\sum c_j f_j$ where $\sum|c_j|<\infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_\infty\le1/(b_j-a_j)$ and $\int f_j=0$.

Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1\le c$, which is essentially proved below. The other direction is not so simple

Anyway: Say $|f|\le c_1$ and $f$ is supported in $[-A,A]$. If you can show

(i) $|M_f|\le\pi c_1$

(ii) $|M_f(x)|\le c_2/x^2$ for $|x|\ge 2A$

it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $\int f=0$ you have $$\int f(y)\frac t{t^2+(x-y)^2}dy= \int_{-A}^A f(y)\left(\frac t{t^2+(x-y)^2}-a_{t,x}\right)dy$$for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|\le A$ and $|x|\ge 2A$ then $|x-y|\ge|x|/2$.

Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|\le A$ and $x>2A$. s noted above, $x-y\ge x-x/2=x/2$. Now $$\begin{align}\left|\frac t{t^2+(x-y)^2}-\frac t{t^2+x^2}\right|&=t\frac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)} \\&\le t\frac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.\end{align}$$ Since $A^2\le Ax/2$ this gives $$\begin{align}\left|\frac t{t^2+(x-y)^2}-\frac t{t^2+x^2}\right|&\le c\frac{tx}{(t^2+(x-y)^2)(t^2+x^2)} \\&\le c\frac{tx}{(t^2+x^2)^2}.\end{align}$$Now $x/(t^2+x^2)\le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)\le t/t^2=1/t\le 1/x$, whhile if $t<x$ then $t/(t^2+x^2)\le x/x^2=1/x$; so we have $$\left|\frac t{t^2+(x-y)^2}-\frac t{t^2+x^2}\right|\le\frac{c}{x^2}\quad(x>2A, |y|\le A).$$Similarly for $x<-2A$; hence $$M_f(x)\le\frac{c||f||_1}{x^2}\quad(|x|>2A),$$as required.