# Integral of a Function Times Its Riesz Transform Is $0$

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

$$\int_{\mathbb{R}^2} \frac{\hat{f} \xi_i \hat{f}}{|\xi|} \text{d} \xi = 0$$?

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 $$L^p$$.
• 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 $\xi$. – David Hughes Mar 6 '20 at 11:01
• 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? – David Hughes Mar 6 '20 at 15:30
• 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 ;) – LL 3.14 Mar 7 '20 at 16:38