Grafakos $5.2.1$: Show that directional Hilbert Transform maps $L^1(\mathbb{R}^n)$ to $L^{1,\infty}(\mathbb{R}^n)$

Question

Show that directional Hilbert Transform maps $L^1(\mathbb{R}^n)$ to $L^{1,\infty}(\mathbb{R}^n)$

The definition of directional Hilbert Transform is:

$$H_\theta(f)(x)= \frac{1}{\pi}p.v \int_{-\infty}^{\infty}f(x-t \theta) \frac{dt}{t} $$

To prove the statatement, first I need to show that Hilbert Transform is given by convolution with the distribution $\omega_\theta$ in $S'(\mathbb{R}^n)$ defined by

$$\langle \omega_\theta, \phi\rangle = \frac{1}{\pi}p.v \int_{-\infty}^{\infty} \frac{\phi(t\theta)}{t}dt $$

I would like to know how to prove the first step and then how to use it to prove the original statement.Any hint?

Answer 1

Firstly, we prove that $H_\theta f=\omega_\theta*f$ for all $f\in S(\mathbb{R}^n)$. By definition, $$\omega_\theta*f(x)=\langle\omega_\theta, \tau^x\tilde f\rangle=\frac{1}{\pi}\text{p.v. }\int_{-\infty}^{\infty} \frac{(\tau^x\tilde f)(t\theta)}{t}\,dt=\frac{1}{\pi}\text{p.v. }\int_{-\infty}^{\infty} \frac{f(x-t\theta)}{t}\,dt=H_\theta f(x).$$

To prove the weak $(1,1)$ of $H_\theta$, we don't have to use the fact that we just proved. WLOG we assume that $\theta\in S^{n-1}$. Let $L_\theta=\{\lambda\theta: \lambda\in\mathbb R\}$ and let $L_\theta^\perp$ be its orthogonal complement in $\mathbb R^n$. Then given any $x\in\mathbb R^n$, there exists a unique $t\in\mathbb R$ and $\bar x\in L_\theta^\perp$ such that $x=t\theta+\bar x$. For each $\bar x\in L_\theta^\perp$, we define $f_{\bar x}:\mathbb R\to\mathbb R$ by $f_{\bar x}(t)=f(t\theta+\bar x)$ for all $t\in\mathbb R$. Then for any $x=t\theta+\bar x\in\mathbb R^n$, we have \begin{align*} (H_\theta f)(x)&=\frac{1}{\pi}\text{ p.v. }\int_{-\infty}^{\infty} \frac{f(x-s\theta)}{s}\,ds\\ &=\frac{1}{\pi}\text{ p.v. }\int_{-\infty}^{\infty} \frac{f((t-s)\theta+\bar x)}{s}\,ds\\ &=\frac{1}{\pi}\text{ p.v. }\int_{-\infty}^{\infty} \frac{f_{\bar x}(t-s)}{s}\,ds\\ &=(Hf_{\bar x})(t). \end{align*}

Therefore, for all $\lambda>0$, we have \begin{align*} \left|\left\{x\in\mathbb R^n: |H_\theta f(x)|>\lambda\right\}\right|&=\int_{L_\theta^\perp}\left|\left\{t\in\mathbb R: |Hf_{\bar x}(t)|>\lambda\right\}\right|\,d\bar x \tag{1}\\ &\leq \frac C\lambda \int_{L_\theta^\perp}\int_{\mathbb R}|f_{\bar x}(t)|\,dt\,d\bar x \tag{2}\\ &=\frac C\lambda \int_{L_\theta^\perp}\int_{\mathbb R}|f(t\theta+\bar x)|\,dt\,d\bar x\\ &=\frac C\lambda\int_{\mathbb R^n}|f(x)|\,dx. \end{align*}

$\text{Explanations:}$

$(1)$ Fubini's theorem for sets.

$(2)$ Weak $(1,1)$ of the Hilbert transform $H$.