Show $\int _{0} ^{\infty} \frac{|f(x)|}{|x-a|^b} dx <\infty$ for a.e. $a\in \mathbb{R}$ if $f$ Lebesgue integrable on $\mathbb{R}$, $b\in (0,1)$

Question

The question: Let $f$ be a Lebesgue integrable function on $\mathbb{R}$ and let $b\in (0,1)$. Prove that $\int _{0} ^{\infty} \frac{|f(x)|}{|x-a|^b} dx <\infty$ for a.e. $a\in \mathbb{R}$.

My progress: I first consider $E_n := \{x\in [0,\infty): \frac{1}{|x-a|^b} \geq n \}$. Then it is quite obvious that $m(E_n) \leq 2\cdot (\frac{1}{n})^{\frac{1}{b}}$.

Then $\int _{0} ^{\infty} \frac{|f(x)|}{|x-a|^b} dx = \int _{E_n\cap [0,\infty)} \frac{|f(x)|}{|x-a|^b} dx+ \int _{E_n^C\cap [0,\infty)}\frac{|f(x)|}{|x-a|^b} dx$. Here is where I am unsure what to do. $\int _{E_n^C\cap [0,\infty)}\frac{|f(x)|}{|x-a|^b} dx$ is finite for $n<\infty$, but how to deal with $\int _{E_n\cap [0,\infty)} \frac{|f(x)|}{|x-a|^b} dx$ is an issue. Any comments are appreciated!

Answer 1

Let $N$ be a positive integer.

By Tonelli's Theorem we have $\int_{-N} ^{N} \int_{-\infty} ^{\infty}\frac {|f(x)|} {|x-y|^{b}} dxdy=\int_{-\infty} ^{\infty}\int_{-N} ^{N}\frac {|f(x)|} {|x-y|^{b}} dydx$. Now $\int_{-N} ^{N}\frac 1 {|x-y|^{b}} dy=\int_{x-N} ^{x+N}\frac 1 {|z|^{b}} dz$. I will let you check that $x \to \int_{x-N} ^{x+N}\frac 1 {|z|^{b}} dz$ is bounded on the real line for $0<b<1$. Conclude that $\int_{-\infty} ^{\infty}\frac {|f(x)|} {|x-y|^{b}} dx<\infty$ for almost every $y \in [-N,N]$. This is true for each $N$.

[Consider, for example, the limit of $(x+N)^{1-b} -(x-N)^{1-b}$ as $ x \to \infty$. We can write $$(x+N)^{1-b} -(x-N)^{1-b}=x^{1-b}[(1+\frac N x)^{1-b}-(1-\frac N x)^{1-b}]= x^{1-b}[(1+\frac {(1-b)N} x)-(1-\frac {(1-b)N} x)+o(\frac 1 x)]\to 0$$

If a continuous function on $\mathbb R$ has finite limits at $\pm \infty$ then it is bounded].

Answer 2

Let $K(x)=\frac{1}{|x|^b}$. Now, define the splitting $K=K_1+K_{\infty}$ such that \begin{align} K_1(x)=\begin{cases} K(x)&\text{if $|x|\leq R$}\\ 0&\text{else} \end{cases},\\\\ K_{\infty}(x)=\begin{cases} 0&\text{if $|x|\leq R$}\\ K(x)&\text{else} \end{cases} \end{align} where $R>0$ is fixed for now. Now, your integral is $|f|*K$ (evaluated at $a$). Using our decomposition, we get $|f|*K_1+ |f|*K_{\infty}$. Notice that since $0<b<1$, we have that $K_1\in L^1$, so the first term being the convolution of two $L^1$ functions is (by Young’s inequality) an $L^1$ function. For the second term, it is the convolution of an $L^1$ function with an $L^{\infty}$ function so it is in $L^{\infty}$. So, $|f|*K\in L^1+L^{\infty}$, i.e both terms separately converge a.e, so the sum also converges a.e.

This sort of splitting is helpful if you want to go beyond just proving the a.e convergence of the integral, and if you want to prove certain operator boundedness result for this Riesz potential (up to scalar multiples); see Stein’s text on singular integrals for more details.