Analysis Problem of the Day 84

Today’s problem appeared as Problem 3 on the UCLA Spring 2018 Analysis Qual:

Problem 3. Let $f \in L^1(\mathbb{R}),$ and suppose that

$\limsup_{\epsilon \to 0} \int_{\mathbb{R}} \int_{\mathbb{R}} \frac{|f(x)f(y)|}{|x-y|^2+\epsilon^2}dx dy<\infty.$

Show that $f \equiv 0$ almost everywhere.

Solution: We will simplify the problem by making $f$ a characteristic function of a positive measure set, i.e. let $f(x) =\chi_A(x).$ Intuitively, the integral should blow up due to the singularity on the line $x=y$ and the fact that $\frac{1}{r^2} \not \in L^2(\mathbb{R}^2).$ To that end, it will be convenient to perform the change of variables $u=x+y,v=x-y,$ which has a constant Jacobian. One may then take advantage of the Steinhaus lemma, which states that if $A$ is a set of positive measure, then $A-A$ and $A+A$ will contain some open intervals $(x_1,y_1),(x_2,y_2),$ respectively. Then, one has

$\int_{\mathbb{R}} \int_{\mathbb{R}} \frac{|\chi_A(x)\chi_A(y)|}{|x-y|^2+\epsilon^2}dx dy = C \int_{\mathbb{R}}\int_{\mathbb{R}} \frac{\chi_A(\frac12(u+v))\chi_A(\frac12(u-v))}{v^2+\epsilon^2} du dv$

$\geq \int_{\frac12 x_1}^{\frac12 y_1} \int_{\frac12 x_2}^{\frac12 y_2} \frac{1}{v^2+\epsilon^2} dv du=\frac12 (y_1-x_1)\int_{\frac12 x_2}^{\frac12 y_2} \frac{1}{v^2+\epsilon^2} dv \to \infty$

as $\epsilon \to 0$ by the Monotone Convergence Theorem, as $\frac{1}{v^2}$ is not integrable near $v=0.$ Thus, $A$ cannot be a set of positive measure, i.e. $f \equiv 0$ a.e. Finally, for the general case $f \in L^1(\mathbb{R}),$ note that $\chi_A \leq |f|$ implies $\mu(A)=0$ if and only if $f \equiv 0$ a.e., so if $f$ were a nonzero function satisfying the above condition, one would conclude that the limit supremum is finite for $\chi_A$ for some positive measure set $A,$ which by the above discussion is not the case. Thus, we deduce that if the identity holds for $f \in L^1(\mathbb{R}),$ then $f \equiv 0$ a.e.

Leave a Reply Cancel reply