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Fall 2010 - Problem 4

Hardy-Littlewood maximal inequality, operator theory

Let $T$ be a linear transformation on $C_\mathrm{c}^0\p{\R}$ , the space of continuous functions of compact support, that has the following two properties:

\norm{Tf}_{L^\infty} \leq \norm{f}_{L^\infty} \quad\text{and}\quad \abs{\set{x \in \R \mid \abs{Tf\p{x}} > \lambda}} \leq \frac{\norm{f}_{L^1}}{\lambda}.

(Here $\abs{A}$ denotes the Lebesgue measure of the set $A$ .) Prove that

\int \abs{Tf\p{x}}^2 \,\diff{x} \leq C \int \abs{f\p{x}}^2 \,\diff{x}

for all $f \in C_\mathrm{c}^0\p{\R}$ and some fixed number $C$ .

Solution.

First, we have the following identity:

\begin{aligned} \int_\R \abs{Tf\p{x}}^2 \,\diff{x} &= \int_\R \int_0^{\abs{Tf\p{x}}} 2\lambda \,\diff\lambda \,\diff{x} \\ &= \int_\R \int_0^\infty 2\lambda \chi_{\set{0 < \lambda < \abs{Tf\p{x}}}} \,\diff\lambda \,\diff{x} \\ &= \int_0^\infty 2\lambda \int_\R \chi_{\set{0 < \lambda < \abs{Tf\p{x}}}} \,\diff{x} \,\diff\lambda \\ &= \int_0^\infty 2\lambda m\p{\set{\abs{Tf\p{x}} > \lambda}} \,\diff\lambda. \end{aligned}

In the above, we switched the order of integration by applying Fubini-Tonelli, which we can do since everything is non-negative. We will first show the result for non-negative $f$ and extend it to a general $f$ . Consider the decomposition of $f \geq 0$ via a cutoff:

g = f\chi_{\set{f < \frac{\lambda}{2}}} + \frac{\lambda}{2}\chi_{\set{f \geq \frac{\lambda}{2}}} \quad\text{and}\quad h = f - g.

It's clear that $g$ is continuous on $\set{f \neq \frac{\lambda}{2}}$ , since $f$ is continuous. For $x \in \set{f\p{x} = \frac{\lambda}{2}}$ , let $\set{x_n}_n \subseteq \R$ be a sequence which converges to $x$ . If $x_n \in \set{f \geq \frac{\lambda}{2}}$ , then $g\p{x_n} = \frac{\lambda}{2}$ . The remaining elements form a subsequence $\set{x_{n_k}}_k$ and satisfies $g\p{x_{n_k}} = f\p{x_{n_k}}$ . By continuity of $f$ , $g\p{x_{n_k}} \xrightarrow{k\to\infty} f\p{x} = \frac{\lambda}{2}$ , so $g$ is continuous. Since $h$ is a difference of two continuous functions, $h$ is also continuous. Because $f$ had compact support, it follows that $g$ and hence $h$ have compact support as well.

Notice that by linearity of $T$ , $\abs{Tf\p{x}} \leq \abs{Tg\p{x}} + \abs{Th\p{x}}$ , and so

\set{\abs{Tf\p{x}} > \lambda} \subseteq \set{\abs{Tg\p{x}} > \frac{\lambda}{2}} \cup \set{\abs{Th\p{x}} > \frac{\lambda}{2}}.

Otherwise, if both $\abs{Tg\p{x}}, \abs{Th\p{x}} \leq \frac{\lambda}{2}$ , then $\abs{Tf\p{x}} \leq \lambda$ . By construction, $\norm{g}_{L^\infty} = \frac{\lambda}{2}$ , so by the first property of $T$ , $\norm{Tg}_{L^\infty} \leq \norm{Tg}_{L^\infty} = \frac{\lambda}{2}$ , and so $\set{\abs{Tg\p{x}} > \frac{\lambda}{2}} = \emptyset$ , which gives

\set{\abs{Tf\p{x}} > \lambda} \subseteq \set{\abs{Th\p{x}} > \frac{\lambda}{2}}.

Hence,

\begin{aligned} \int_\R \abs{Tf\p{x}}^2 \,\diff{x} &= \int_0^\infty 2\lambda m\p{\set{\abs{Tf\p{x}} > \lambda}} \,\diff\lambda \\ &\leq \int_0^\infty 2\lambda \cdot \frac{\norm{h}_{L^1}}{\lambda/2} \,\diff\lambda \\ &= 4 \int_0^\infty \int_0^\infty \abs{h\p{x}} \,\diff{x} \,\diff\lambda \\ &= 4 \int_0^\infty \int_0^\infty \p{f\p{x} - \frac{\lambda}{2}}\chi_{\set{f \geq \frac{\lambda}{2}}} \,\diff{x} \,\diff\lambda \\ &= 4 \int_0^\infty \int_{\set{f \geq \frac{\lambda}{2}}} f\p{x} - \frac{\lambda}{2} \,\diff{x} \,\diff\lambda \\ &\leq 4 \int_0^\infty \int_{\set{f \geq \frac{\lambda}{2}}} f\p{x} \,\diff{x} \,\diff\lambda \\ &= 4 \int_0^\infty f\p{x} \int_0^{2f\p{x}} \,\diff\lambda \,\diff{x} \\ &= 8 \int_0^\infty \abs{f\p{x}}^2 \,\diff{x}. \end{aligned}

Here, we were able to apply Fubini-Tonelli since $f \geq 0$ . For a general $f$ , we may decompose $f = f^+ - f^-$ where $f^+ = \max\,\set{0, f}$ and $f^- = \max\,\set{0, -f}$ , which are non-negative. Thus,

\begin{aligned} \int \abs{Tf\p{x}}^2 \,\diff{x} &= \int_{\set{f \geq 0}} \abs{Tf^+\p{x}}^2 \,\diff{x} + \int_{\set{f < 0}} \abs{Tf^-\p{x}}^2 \,\diff{x} \\ &\leq 8\int \abs{f^+\p{x}}^2 \,\diff{x} + 8\int \abs{f^-\p{x}}^2 \,\diff{x} \\ &\leq 16 \int \abs{f\p{x}}^2 \,\diff{x}, \end{aligned}

which completes the proof.