Let \(T\) be a linear transformation on \(C_0(\mathbb{R})\) (the space of continuous functions of compact support) that satisfies: \[ \|Tf\|_{L^\infty} \le \|f\|_{L^\infty}, \qquad \text{and} \qquad \left| \{x \in \mathbb{R} : |Tf(x)| > \lambda\} \right| \le \frac{\|f\|_{L^1}}{\lambda} \] for all \(\lambda > 0\). Prove that there exists a fixed constant \(C > 0\) such that \[ \int_{-\infty}^{+\infty} |Tf(x)|^2\,dx \le C \int_{-\infty}^{+\infty} |f(x)|^2\,dx \] for all \(f \in C_0(\mathbb{R})\).
Solution
1. Use the distribution-function identity.
For any measurable \(h\), \[ \int_{\mathbb R}|h(x)|^2\,dx =2\int_0^\infty \lambda\,m\{|h|>\lambda\}\,d\lambda. \] This is the layer-cake formula applied to \(|h|^2\). Hence \[ \|Tf\|_2^2 =2\int_0^\infty \lambda\,m\{|Tf|>\lambda\}\,d\lambda. \]
2. Decompose \(f\) by continuous truncation at each level.
Fix \(\lambda>0\). Define \[ f_{2,\lambda}(x)= \begin{cases} f(x),&|f(x)|\le \lambda/2,\\ \dfrac{\lambda}{2}\dfrac{f(x)}{|f(x)|},&|f(x)|>\lambda/2, \end{cases} \] and put \[ f_{1,\lambda}=f-f_{2,\lambda}. \] The radial truncation map \[ z\longmapsto \begin{cases} z,&|z|\le\lambda/2,\\ (\lambda/2)z/|z|,&|z|>\lambda/2 \end{cases} \] is continuous and sends \(0\) to \(0\). Since \(f\in C_c(\mathbb R)\), both \(f_{1,\lambda}\) and \(f_{2,\lambda}\) therefore belong to \(C_c(\mathbb R)\), so \(T\) is defined on both of them.
Moreover, \[ \|f_{2,\lambda}\|_\infty\le\lambda/2 \] and \[ |f_{1,\lambda}(x)| =(|f(x)|-\lambda/2)_+ \le |f(x)|\mathbf1_{\{|f(x)|>\lambda/2\}}. \]
3. Estimate the distribution function of \(Tf\).
By the strong \((\infty,\infty)\) hypothesis, \[ \|Tf_{2,\lambda}\|_\infty \le\|f_{2,\lambda}\|_\infty \le\lambda/2. \] Since \(Tf=Tf_{1,\lambda}+Tf_{2,\lambda}\), the implication \[ |Tf(x)|>\lambda \quad\Longrightarrow\quad |Tf_{1,\lambda}(x)|>\lambda/2 \] holds. Therefore the weak \((1,1)\) hypothesis gives \[ \begin{aligned} m\{|Tf|>\lambda\} &\le m\{|Tf_{1,\lambda}|>\lambda/2\}\\ &\le \frac{2}{\lambda}\|f_{1,\lambda}\|_1\\ &\le \frac{2}{\lambda} \int_{\{|f(t)|>\lambda/2\}}|f(t)|\,dt. \end{aligned} \]
4. Integrate the distribution estimate.
Substituting Step 3 into Step 1 and applying Tonelli to the nonnegative integrand, \[ \begin{aligned} \|Tf\|_2^2 &\le 4\int_0^\infty \int_{\mathbb R}|f(t)| \mathbf1_{\{\lambda<2|f(t)|\}}\,dt\,d\lambda\\ &=4\int_{\mathbb R}|f(t)| \left(\int_0^{2|f(t)|}d\lambda\right)dt\\ &=8\int_{\mathbb R}|f(t)|^2\,dt. \end{aligned} \] Thus \[ \boxed{\int_{\mathbb R}|Tf|^2\le8\int_{\mathbb R}|f|^2.} \] So the required estimate holds with \(C=8\).