I have a function, $\theta : \mathbb{R}^2 \rightarrow \mathbb{R}$, such that $\theta \in L^1(\mathbb{R}^2) \cap L^p(\mathbb{R}^2) $, where $p >2$.
I have been told that, for such a function, we have the following property: $$ \int_{\mathbb{R}^2} \theta(x) R_i \theta(x) \text{d}x = 0, $$ where $R_i \theta$ is the $i^\text{th}$ Riesz-Transform of $\theta$.
If anyone could provide an outline of a proof, or a reference as to where I might find a proof, I would be most grateful. Thank you.
Attempt
I have made some small progress on solving this problem. We use the Fourier transform of the Riesz Transform, and the Parceval Theorem.
The Fourier Transform can be written as:
$ \mathcal{F}[R_i f] = \frac{\xi_i}{|\xi|}\hat{f} $.
Thus, using the Parceval Theroem, we have:
$ \int_{\mathbb{R}^2} f R_i f \text{d}x = \int_{\mathbb{R}^2} \hat{f} \frac{\xi_i}{|\xi|}\hat{f} \text{d}\xi = \int_{\mathbb{R}^2} \hat{f}(\xi)^2 \partial_{\xi_i}|\xi| \text{d}\xi $. We hope to somehow use integration by parts to show this is $0$.
We know that, for $f \in L^1(\mathbb{R})$, $ \int_{\mathbb{R}} (\partial_x f) f \text{d}x = [f^2]^{\infty}_{x = -\infty} - \int_{\mathbb{R}} f (\partial_x f) \text{d}x = 0 - \int_{\mathbb{R}} f (\partial_x f) \text{d}x $.
Then $ \int_{\mathbb{R}} (\partial_x f) f \text{d}x = 0 $.
Similarly, for $ f \in L^1(\mathbb{R}^2) $, $ \int_{\mathbb{R}^2} (\partial_{x_i} f) f \text{d}x = 0$ for each $i = 1,2$.
Thus $ \int_{\mathbb{R}^2} (\partial_{x_i} f) f \text{d}x = \int_{\mathbb{R}^2} \hat{f} \xi_i \hat{f} \text{d} \xi = 0 $, fo reach $ i = 1,2 $.
Is it possible to use the above facts to show that
$ \int_{\mathbb{R}^2} \frac{\hat{f} \xi_i \hat{f}}{|\xi|} \text{d} \xi = 0 $?
This seems false to me, unless the function is even.
Take for example $\varphi≠ 0$ a radial, nonnegative and smooth function compactly supported in the unit ball. Then consider $\hat{f}(\xi) = \varphi(\xi-e_i)$ with $e_i$ the unit vector with the same direction as $\xi_i$. Then $$ ∫_{\mathbb{R}^2} \frac{\hat{f}(\xi)^2\xi_i}{|\xi|}\mathrm{d}\xi = ∫_{\mathbb{R}^2} \frac{\varphi(\xi)^2(\xi_i+1)}{|\xi+e_i|}\mathrm{d}\xi = ∫_{B(0,1)} \frac{\varphi(\xi)^2|\xi_i+1|}{|\xi+e_i|}\mathrm{d}\xi > 0. $$
And of course, $f = \mathcal{F}^{-1}(\varphi(\xi-e_i))$ is in every $L^p$.