Analysis Problem of the Day 23

Today’s problem appeared as Problem 2 on the UCLA Spring 2014 Analysis Qual:

Problem 2: For $f \in L^1(\mathbb{R})$ and $\beta \in (0,1),$ show that $\int_{\mathbb{R}} \frac{|f(x)|}{|x-a|^\beta}dx < \infty$ for almost every $a \in \mathbb{R}.$

Solution: This problem might seem quite tricky at first, but it has a clever solution. The key thing to recognize is that the statement cannot hold true for all $a \in \mathbb{R},$ meaning that there should be some Lebesgue integral machinery involved.

In particular, consider the expression as a function of $a,$ i.e. $g: \mathbb{R} \to [0,\infty]$ given by

$g(a) := \int_{\mathbb{R}} \frac{|f(x)|}{|x-a|^\beta}dx.$

Then, by Tonelli, we may do a double integral and switch the order of integration, i.e.

$\int_{\mathbb{R}} g(a) da = \int_{\mathbb{R}} \int_{\mathbb{R}} \frac{|f(x)|}{|x-a|^\beta}dx da = \int_{\mathbb{R}} \int_{\mathbb{R}} \frac{|f(x)|}{|x-a|^\beta}da dx = C_\beta\|f\|_1$

for $C_\beta = \int \frac{1}{|x-a|^\beta} dx.$ Since $g \in L^1,$ it follows that $g(a) < \infty$ for almost every $a \in \mathbb{R},$ and we are done.

Remark: More generally, for $f\in L^1, g=g(x,y)$ where $\sup_{x \in \mathbb{R}} \int |g(x,y)| da < \infty,$ one has $(Tf)(y):=\int f(x)g(x,y)dx < \infty$ for a.e. $a.$ This shows that as an integral operator with kernel $g,$ the operator is a map $T: L^1 \to L^1.$ The generalization of this fact is known as Young’s inequality for integral operators, which states that if $\sup_{x} \|g\|_{L^q(dy)} < \infty, \sup_{y} \|g\|_{L^q(dx)} < \infty,$ then $T: L^p \to L^r$ for $\frac{1}{p}+\frac{1}{q} = \frac{1}{r}+1.$

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