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Spring 2017 - Problem 1

Lp spaces

Let $K \subseteq \R$ be a compact set of positive measure and let $f \in L^\infty\p{\R}$ . Show that the function

F\p{x} = \frac{1}{\abs{K}} \int_K f\p{x + t} \,\diff{t}

is uniformly continuous on $\R$ . Here $\abs{K}$ denotes the Lebesgue measure of $K$ .

Solution.

By the change of variables $t \mapsto x + t$ , we can write

F\p{x} = \frac{1}{m\p{K}} \int_{x + K} f\p{t} \,\diff{t} = \frac{1}{m\p{K}} \int_\R f\p{t}\chi_{x + K}\p{t} \,\diff{t} = \frac{1}{m\p{K}} \int_\R f\p{t}\chi_K\p{t - x} \,\diff{t}.

Fix $\epsilon > 0$ . Then given $x, y \in \R$ ,

\begin{aligned} \abs{F\p{x} - F\p{y}} &\leq \frac{1}{m\p{K}} \int_\R \abs{f\p{t}}\abs{\chi_K\p{t - x} - \chi_K\p{t - y}} \,\diff{t} \\ &= \frac{\norm{f}_{L^\infty}}{m\p{K}} \int_\R \abs{\chi_K\p{t} - \chi_K\p{t + \p{x - y}}} \,\diff{t} && (t \mapsto t - x) \\ &= \frac{\norm{f}_{L^\infty}}{m\p{K}} \norm{\chi_K - \tau_{x-y}\chi_K}_{L^1}, \end{aligned}

where $\tau_{x-y}g\p{t} = g\p{t + \p{x-y}}$ in the translation operator. However, we know that translation is continuous in $L^1$ , so there exists $\delta > 0$ such that if $\abs{x - y} < \delta$ , then $\norm{\chi_K - \tau_{x-y}\chi_K}_{L^1} < \epsilon$ . This only depends on $\abs{x - y}$ and not $x$ and $y$ themselves, and so $F$ is uniformly continuous on $\R$ .