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Spring 2018 - Problem 1

Lebesgue differentiation theorem

Suppose $f \in L^1\p{\R}$ satisfies

\limsup_{h\to0} \int_\R \abs{\frac{f\p{x + h} - f\p{x}}{h}} \,\diff{x} = 0.

Solution.

Define

F\p{x} = \int_{-\infty}^x f\p{y} \,\diff{y},

which is well-defined since $f \in L^1\p{\R}$ . Observe that

\int_{-\infty}^x f\p{y + h} \,\diff{y} = \int_{-\infty}^{x+h} f\p{y} \,\diff{y} = F\p{x + h},

so that

\abs{F\p{x + h} - F\p{x}} \leq \int_{-\infty}^x \abs{f\p{y + h} - f\p{y}} \,\diff{y} \leq \int_\R \abs{f\p{y + h} - f\p{y}} \,\diff{y},

Let $a, b \in \R$ be Lebesgue points. Then for $h > 0$ , the calculation above gives

\begin{aligned} \frac{1}{2h} \abs{\p{\int_{a-h}^{a+h} f\p{y} \,\diff{y} - \int_{b-h}^{b+h} f\p{y} \,\diff{y}}} &= \frac{1}{2h} \abs{\p{F\p{a + h} - F\p{a - h}} - \p{F\p{b + h} - F\p{b - h}}} \\ &\leq \frac{1}{2h} \p{\abs{F\p{a + h} - F\p{a}} + \abs{F\p{a} - F\p{a - h}} + \abs{F\p{b + h} - F\p{b}} + \abs{F\p{b} - F\p{b - h}}} \\ &\leq \frac{1}{2h} \cdot 4 \int_\R \abs{f\p{y + h} - f\p{y}} \,\diff{y} \\ &= 2\int_\R \abs{\frac{f\p{y + h} - f\p{y}}{h}} \,\diff{y}. \end{aligned}

By the Lebesgue differentiation theorem, we get

\abs{f\p{a} - f\p{b}} \leq 2\limsup_{h\to0} \int_\R \abs{\frac{f\p{y + h} - f\p{y}}{h}} \,\diff{y} = 0,

by assumption, i.e., $f\p{a} = f\p{b}$ . Thus, because almost every $x \in \R$ is a Lebesgue point, we see that $f$ must be a constant. Since $f \in L^1\p{\R}$ , it follows that $f = 0$ , which was what we wanted to show.