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Spring 2018 - Problem 3

Lp spaces

Suppose f∈L1(R)f \in L^1\p{\R} satisfies

lim sup⁡ε→0∫R∫R∣f(x)∣∣f(y)∣∣x−y∣2+ε2 dx dy<∞.\limsup_{\epsilon\to0} \int_\R \int_\R \frac{\abs{f\p{x}}\abs{f\p{y}}}{\abs{x - y}^2 + \epsilon^2} \,\diff{x} \,\diff{y} < \infty.

Show that f=0f = 0 almost everywhere.
Solution.

First, assume that f=χEf = \chi_E. Assume otherwise, and that χE≠0\chi_E \neq 0 almost everywhere, i.e., m(E)>0m\p{E} > 0. Then

∫R∫R∣f(x)∣∣f(y)∣∣x−y∣2+ε2 dx dy=∫R∫RχE(x+y)χE(y)∣x∣2+ε2 dx dy(x↦x+y)=∫R1∣x∣2+ε2∫RχE(x+y)χE(y) dy dx.\begin{aligned} \int_\R \int_\R \frac{\abs{f\p{x}}\abs{f\p{y}}}{\abs{x - y}^2 + \epsilon^2} \,\diff{x} \,\diff{y} &= \int_\R \int_\R \frac{\chi_E\p{x + y} \chi_E\p{y}}{\abs{x}^2 + \epsilon^2} \,\diff{x} \,\diff{y} && \p{x \mapsto x + y} \\ &= \int_\R \frac{1}{\abs{x}^2 + \epsilon^2} \int_\R \chi_E\p{x + y} \chi_E\p{y} \,\diff{y} \,\diff{x}. \end{aligned}

Since translation is continuous in L1(R)L^1\p{\R}, we see that

lim⁡x→0∫RχE(x+y)χE(y) dy=lim⁡x→0∫EχE(x+y) dy=∫EχE(y) dy=m(E)>0.\lim_{x\to0} \int_\R \chi_E\p{x + y} \chi_E\p{y} \,\diff{y} = \lim_{x\to0} \int_E \chi_E\p{x + y} \,\diff{y} = \int_E \chi_E\p{y} \,\diff{y} = m\p{E} > 0.

Thus, there exists δ>0\delta > 0 so that if ∣x∣<δ\abs{x} < \delta, then ∫RχE(x+y)χE(y) dy≥m(E)2\int_\R \chi_E\p{x + y}\chi_E\p{y} \,\diff{y} \geq \frac{m\p{E}}{2}. Hence, if ε<1\epsilon < 1,

∫R1∣x∣2+ε2∫RχE(x+y)χE(y) dy dx≥m(E)2∫B(0,δ)1∣x∣2+ε2 dx=m(E)∫0δ1x2+ε dx=m(E)εarctan⁡(δε)≥m(E)εarctan⁡ (δ)→ε→0∞.\begin{aligned} \int_\R \frac{1}{\abs{x}^2 + \epsilon^2} \int_\R \chi_E\p{x + y} \chi_E\p{y} \,\diff{y} \,\diff{x} &\geq \frac{m\p{E}}{2} \int_{B\p{0,\delta}} \frac{1}{\abs{x}^2 + \epsilon^2} \,\diff{x} \\ &= m\p{E} \int_0^\delta \frac{1}{x^2 + \epsilon} \,\diff{x} \\ &= \frac{m\p{E}}{\epsilon} \arctan\p{\frac{\delta}{\epsilon}} \\ &\geq \frac{m\p{E}}{\epsilon} \arctan\,\p{\delta} \xrightarrow{\epsilon\to0} \infty. \end{aligned}

Hence, if f=χEf = \chi_E, then it must be that m(E)=0m\p{E} = 0. For the general case, suppose f≠0f \neq 0, and so there exist a>0a > 0 and E⊆RE \subseteq \R measurable such that ∣f∣≥aχE\abs{f} \geq a\chi_E. Then

∫R∫R∣f(x)∣∣f(y)∣∣x−y∣2+ε2 dx dy≥a2∫R∫RχE(x)χE(y)∣x−y∣2+ε2 dx dy→ε→0∞.\int_\R \int_\R \frac{\abs{f\p{x}}\abs{f\p{y}}}{\abs{x - y}^2 + \epsilon^2} \,\diff{x} \,\diff{y} \geq a^2 \int_\R \int_\R \frac{\chi_E\p{x} \chi_E\p{y}}{\abs{x - y}^2 + \epsilon^2} \,\diff{x} \,\diff{y} \xrightarrow{\epsilon\to0} \infty.

Thus, f=0f = 0 almost everywhere, which was what we wanted to show.