Skip to main content
fix definition of Hilbert transform
Source Link
George Lowther
  • 17.1k
  • 1
  • 66
  • 98

Given a finite measure $\mu$ on the real line $\mathbb R$, one definition of its Hilbert transform is $(H\mu)(y) =(PV)\int \frac{d\mu(x)}{x-y}$$(H\mu)(y) =\frac{1}{\pi}(PV)\int \frac{d\mu(x)}{x-y}$ which is known to exist almost everywhere on $\mathbb R$. Another way is to define the Borel transform ${\mathcal H}(z) = \frac{1}{\pi}\int\frac{d\mu(x)}{x-z}$ for $z\in \mathbb C\setminus \mathbb R$, and then take the limit of $\Re({\mathcal H}\mu)(x+iy)$ as $y\downarrow 0$, such limit existing almost everywhere. In the case that $\mu$ is absolutely continuous it is stated $explicitly$ in the literature that these agree (almost everywhere), and it is $implicit$ in the use of the term `Hilbert transform' in the literature that the two definitions agree (almost everywhere) for general finite $\mu$. Does anyone know where one can find explicit proof(s)? (Proofs for the $L^p$ case are easy to find.)

Given a finite measure $\mu$ on the real line $\mathbb R$, one definition of its Hilbert transform is $(H\mu)(y) =(PV)\int \frac{d\mu(x)}{x-y}$ which is known to exist almost everywhere on $\mathbb R$. Another way is to define the Borel transform ${\mathcal H}(z) = \frac{1}{\pi}\int\frac{d\mu(x)}{x-z}$ for $z\in \mathbb C\setminus \mathbb R$, and then take the limit of $\Re({\mathcal H}\mu)(x+iy)$ as $y\downarrow 0$, such limit existing almost everywhere. In the case that $\mu$ is absolutely continuous it is stated $explicitly$ in the literature that these agree (almost everywhere), and it is $implicit$ in the use of the term `Hilbert transform' in the literature that the two definitions agree (almost everywhere) for general finite $\mu$. Does anyone know where one can find explicit proof(s)? (Proofs for the $L^p$ case are easy to find.)

Given a finite measure $\mu$ on the real line $\mathbb R$, one definition of its Hilbert transform is $(H\mu)(y) =\frac{1}{\pi}(PV)\int \frac{d\mu(x)}{x-y}$ which is known to exist almost everywhere on $\mathbb R$. Another way is to define the Borel transform ${\mathcal H}(z) = \frac{1}{\pi}\int\frac{d\mu(x)}{x-z}$ for $z\in \mathbb C\setminus \mathbb R$, and then take the limit of $\Re({\mathcal H}\mu)(x+iy)$ as $y\downarrow 0$, such limit existing almost everywhere. In the case that $\mu$ is absolutely continuous it is stated $explicitly$ in the literature that these agree (almost everywhere), and it is $implicit$ in the use of the term `Hilbert transform' in the literature that the two definitions agree (almost everywhere) for general finite $\mu$. Does anyone know where one can find explicit proof(s)? (Proofs for the $L^p$ case are easy to find.)

added a ref-request tag
Link
Yemon Choi
  • 25.8k
  • 9
  • 69
  • 156
Source Link

Hilbert transforms of measures

Given a finite measure $\mu$ on the real line $\mathbb R$, one definition of its Hilbert transform is $(H\mu)(y) =(PV)\int \frac{d\mu(x)}{x-y}$ which is known to exist almost everywhere on $\mathbb R$. Another way is to define the Borel transform ${\mathcal H}(z) = \frac{1}{\pi}\int\frac{d\mu(x)}{x-z}$ for $z\in \mathbb C\setminus \mathbb R$, and then take the limit of $\Re({\mathcal H}\mu)(x+iy)$ as $y\downarrow 0$, such limit existing almost everywhere. In the case that $\mu$ is absolutely continuous it is stated $explicitly$ in the literature that these agree (almost everywhere), and it is $implicit$ in the use of the term `Hilbert transform' in the literature that the two definitions agree (almost everywhere) for general finite $\mu$. Does anyone know where one can find explicit proof(s)? (Proofs for the $L^p$ case are easy to find.)