Spring 2014 - Problem 2

Lp spaces

Let $f \in L^1\p{\R, \diff{x}}$ and $\beta \in \p{0, 1}$ . Prove that

\int_\R \frac{\abs{f\p{x}}}{\abs{x - a}^\beta} < \infty

for (Lebesgue) a.e. $a \in \R$ .

It suffices to show that

\int_{-R}^R \int_\R \frac{\abs{f\p{x}}}{\abs{x - a}^\beta} \,\diff{x} \,\diff{a} < \infty

for every $R > 0$ . By Fubini-Tonelli (everything is non-negative),

\begin{aligned} \int_{-R}^R \int_\R \frac{\abs{f\p{x}}}{\abs{x - a}^\beta} \,\diff{x} \,\diff{a} &= \int_\R \abs{f\p{x}} \int_{-R}^R \frac{1}{\abs{x - a}^\beta} \,\diff{a} \,\diff{x} \\ &= \int_\R \abs{f\p{x}} \int_{x-R}^{x+R} \frac{1}{\abs{a}^\beta} \,\diff{a} \,\diff{x}. \end{aligned}

From here, there are three cases: If $x + R < 0$ , then

\int_{x-R}^{x+R} \frac{1}{\abs{a}^\beta} \,\diff{a} \leq \int_{-R}^0 \frac{1}{\abs{a}^\beta} \,\diff{a} \leq \int_{-R}^R \frac{1}{\abs{a}^\beta} \,\diff{a}.

Similarly, if $x - R > 0$ , then

\int_{x-R}^{x+R} \frac{1}{\abs{a}^\beta} \,\diff{a} \leq \int_0^R \frac{1}{\abs{a}^\beta} \,\diff{a} \leq \int_{-R}^R \frac{1}{\abs{a}^\beta} \,\diff{a}.

Finally, if $x - R \leq 0 \leq x + R$ , then $x \leq R$ , so $\br{x - R, x + R} \subseteq \br{-2R, 2R}$ , which means

\int_{x-R}^{x+R} \frac{1}{\abs{x}^\beta} \,\diff{x} \leq \int_{-2R}^{2R} \frac{1}{\abs{x}^\beta} \,\diff{x}.

Hence, in every case,

\int_{x-R}^{x+R} \frac{1}{\abs{x}^\beta} \,\diff{x} \leq \int_{-2R}^{2R} \frac{1}{\abs{x}^\beta} \,\diff{x} = 2\int_0^{2R} \frac{1}{\abs{x}^\beta} \,\diff{x} = \frac{2\p{2R}^{1-\beta}}{1 - \beta} < \infty.

Hence,

\int_{-R}^R \int_\R \frac{\abs{f\p{x}}}{\abs{x - a}^\beta} \,\diff{x} \,\diff{a} \leq \int_\R \abs{f\p{x}} \int_{x-R}^{x+R} \frac{1}{\abs{a}^\beta} \,\diff{a} \,\diff{x} \leq \frac{2\p{2R}^{1-\beta}}{1 - \beta} \norm{f}_{L^1} < \infty,

which completes the proof.