<Home>Qualifying Exams><Analysis>Fall 2015 - Problem 3

Fall 2015 - Problem 3

Lp spaces

Let $f \in L^1_{\mathrm{loc}}\p{\R^d}$ be such that for some $0 < p < 1$ , we have

\abs{\int f\p{x} g\p{x} \,\diff{x}} \leq \p{\int \abs{g\p{x}}^p \,\diff{x}}^{1/p},

for all $g \in C_0\p{\R^d}$ . Show that $f\p{x} = 0$ a.e. Here $C_0\p{\R^d}$ is the space of continuous functions with compact support on $\R^d$ .

Solution.

Let $K$ be a compact set and $\epsilon > 0$ . Since $K$ is bounded, there exists $R > 0$ such that $K \subseteq B\p{0, R}$ . Since $f \in L^1\p{B\p{0, R}}$ , the map $E \mapsto \int_{B\p{0, R} \cap E} \abs{f\p{x}} \,\diff{x}$ defines a measure absolutely continuous with respect to the Lebesgue measure. Hence, there exists $\delta > 0$ such that if $m\p{E} < \delta$ , then

\int_{B\p{0, R} \cap E} \abs{f\p{x}} \,\diff{x} < \epsilon.

By regularity of the Lebesgue measure, there exists an open set $K \subseteq U \subseteq B\p{0, R}$ such that $m\p{U \setminus K} < \delta$ , and so $\int_{U \setminus K} \abs{f\p{x}} \,\diff{x} < \epsilon$ . Shrinking $\delta$ further, we may assume that $\delta < \epsilon$ . Hence,

\begin{aligned} \abs{\int_K f\p{x} \,\diff{x}} &= \abs{\int_K f\p{x}g\p{x} \,\diff{x}} \\ &= \abs{\int_U f\p{x}g\p{x} \,\diff{x} - \int_{U \setminus K} f\p{x}g\p{x} \,\diff{x}} \\ &\leq \abs{\int_U f\p{x}g\p{x} \,\diff{x}} + \int_{U \setminus K} \abs{f\p{x}} \,\diff{x} \\ &\leq \p{\int \abs{g\p{x}}^p \,\diff{x}}^{1/p} + \epsilon \\ &\leq m\p{U}^{1/p} + \epsilon \\ &\leq \p{m\p{K} + \delta}^{1/p} + \epsilon \\ &\leq \p{m\p{K} + \epsilon}^{1/p} + \epsilon. \end{aligned}

Sending $\epsilon \to 0$ , we see that the given inequality also works when $g$ is the characteristic equation on a compact set. We will now emphasize the fact that $L^p$ does not satisfy the triangle inequality for $0 < p < 1$ : let $Q$ be a cube with side length $L$ , and for each $N \in \N$ , decompose $Q$ into almost disjoint cubes $Q_n$ with side lengths $\frac{L}{N}$ . Then

\begin{aligned} \abs{\int f\p{x} \chi_Q\p{x} \,\diff{x}} \leq \sum_{n=1}^{N^d} \abs{\int f\p{x} \chi_{Q_n} \,\diff{x}} &\leq \sum_{n=1}^{N^d} m\p{Q_n}^{1/p} \\ &= \sum_{n=1}^{N^d} \frac{L^d}{N^{d/p}} \\ &= \frac{L^d}{N^{d\p{\frac{1}{p}-1}}} \xrightarrow{N\to\infty} 0, \end{aligned}

since $\frac{1}{p} - 1 > 0$ . Indeed, since $0 < p < 1$ , the right-hand side shrinks instead of growing as $N \to \infty$ . Thus, we have shown that $\int_Q f\p{x} \,\diff{x} = 0$ for any cube. Hence, by the Lebesgue differentiation theorem, we see that $f\p{x} = 0$ almost everywhere.