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.


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 $L^p$.

  • $\begingroup$ I believe there is a mistake in your integrals here, as the value of the integral should be a scalar, not a vector. We are consider functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ here. The variable $\xi_i$ inside the integral is not a vector. It is the scalar value of the $i^{\text{th}}$ component of $\xi$. $\endgroup$ – David Hughes Mar 6 '20 at 11:01
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    $\begingroup$ Oh yes, I modified it, now it is right. $\endgroup$ – LL 3.14 Mar 6 '20 at 15:23
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    $\begingroup$ Is your last comment about $f$ being in $L^p$ necessarily true? I know there are simple functions whose FTs are not lebesugue integrable. What result have you used to say this? $\endgroup$ – David Hughes Mar 6 '20 at 15:30
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    $\begingroup$ I use the fact that the Fourier transform of a $C^\infty$ compactly supported function is in the Schwartz space, which consists of $C^\infty$ functions decaying faster that any polynomial at infinity ;) $\endgroup$ – LL 3.14 Mar 7 '20 at 16:38
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    $\begingroup$ Any textbook about Fourier transform of distributions should have that, since the fact that the Schwartz space is invariant by Fourier transform is the crucial property to define Fourier transform by duality. Most of the references I personally know are in french, but for example you can find that in math.mit.edu/~rbm/iml/Chapter1.pdf, Equation (1.61). $\endgroup$ – LL 3.14 Mar 10 '20 at 13:53

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