Analysis Problem of the Day 61

Today’s problem is a generalization of Problem 4 from UCLA’s Fall 2010 Analysis Qual.

Problem 4. Let $T: C_c(\mathbb{R}) \to C_c(\mathbb{R})$ be a linear map such that

$\|Tf\|_\infty \leq \|f\|_\infty, \quad \mu(\{x: |Tf(x)|>\lambda\}) \leq \frac{\|f\|_1}{\lambda}, \quad f \in C_c(\mathbb{R}).$

Show that

$\int_\mathbb{R} |Tf(x)|^2 dx \leq C \int_\mathbb{R} |f(x)|^2 dx$

for some fixed $C>0.$

Solution: Note that by splitting $f \in C_c(\mathbb{R})$ into its positive and negative parts, we may assume $f \geq 0.$ For $\lambda \geq 0,$ we split $f$ into $f=f_1+f_2,$ where $f_1 = \max(f,\frac{\lambda}{2})$ is the part of $f$ below $\frac{\lambda}{2}$ and $f_2 = f-f_1$ is the part of $f$ above $\frac{\lambda}{2}.$ Then, by the layer-cake decomposition, note that

$\int |Tf(x)|^2 dx = 2 \int_0^\infty \lambda \mu(|Tf|>\lambda)d\lambda.$

Moreover,

$\{x:|T(f_1+f_2)(x)|>\lambda\} \subseteq \{x: |Tf_1(x)| > \frac{\lambda}{2}\} \cup \{x: |Tf_2(x)|>\frac{\lambda}{2}\}.$

By construction, $\|f_1\|_\infty \leq \frac{\lambda}{2},$ so the first set on the right is empty, and the measure of the second set is by hypothesis bounded by $\frac{2 \int f_2(x) dx}{\lambda}.$ Finally, by Fubini-Tonelli, we conclude that

$\int |Tf(x)|^2 dx \leq 4 \int_0^\infty \int f_2(x)dx d\lambda = 4 \int \int (f(x)-\frac{\lambda}{2}) \chi_{f(x)>\frac{\lambda}{2}}(x)dx d\lambda$

$\leq 4 \int \int_0^{2f(x)} f(x) d\lambda dx = 8 \int |f(x)|^2 dx,$

and thus the claim is proven.

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