# Riesz transform

In the mathematical theory of harmonic analysis, the Riesz transforms are a family of generalizations of the Hilbert transform to Euclidean spaces of dimension d > 1. They are a type of singular integral operator, meaning that they are given by a convolution of one function with another function having a singularity at the origin. Specifically, the Riesz transforms of a complex-valued function ƒ on R$^d$ are defined by
| $R_{j}f(x)=c_{d}\lim _{\epsilon \to 0}\int _{\mathbf {R} ^{d}\backslash B_{\epsilon }(x)}{\frac {(x_{j}-t_{j})f(t)}{|x-t|^{d+1}}}\,dt$ | (1) |
for j = 1,2,...,d. The constant c$_d$ is a dimensional normalization given by
    $c_{d}={\frac {1}{\pi \omega _{d-1}}}={\frac {\Gamma [(d+1)/2]}{\pi ^{(d+1)/2}}}.$
where ω$_{d−1}$ is the volume of the unit (d − 1)-ball. The limit is written in various ways, often as a principal value, or as a convolution with the tempered distribution
    $K(x)={\frac {1}{\pi \omega _{d-1}}}\,p.v.{\frac {x_{j}}{|x|^{d+1}}}.$
The Riesz transforms arises in the study of differentiability properties of harmonic potentials in potential theory and harmonic analysis. In particular, they arise in the proof of the Calderón-Zygmund inequality (Gilbarg & Trudinger 1983, §9.4).

## Multiplier properties

The Riesz transforms are given by a Fourier multiplier. Indeed, the Fourier transform of R$_j$ƒ is given by
    ${\mathcal {F}}(R_{j}f)(x)=-i{\frac {x_{j}}{|x|}}({\mathcal {F}}f)(x).$
In this form, the Riesz transforms are seen to be generalizations of the Hilbert transform. The kernel is a distribution which is homogeneous of degree zero. A particular consequence of this last observation is that the Riesz transform defines a bounded linear operator from L$^2$(R$^d$) to itself.
This homogeneity property can also be stated more directly without the aid of the Fourier transform. If σ$_s$ is the dilation on R$^d$ by the scalar s, that is σ$_s$x = sx, then σ$_s$ defines an action on functions via pullback:
    $\sigma _{s}^{*}f=f\circ \sigma _{s}.$
The Riesz transforms commute with σ$_s$:
    $\sigma _{s}^{*}(R_{j}f)=R_{j}(\sigma _{x}^{*}f).$
Similarly, the Riesz transforms commute with translations. Let τ$_a$ be the translation on R$^d$ along the vector a; that is, τ$_a$(x) = x + a. Then
    $\tau _{a}^{*}(R_{j}f)=R_{j}(\tau _{a}^{*}f).$
For the final property, it is convenient to regard the Riesz transforms as a single vectorial entity Rƒ = (R$_1$ƒ,...,R$_d$ƒ). Consider a rotation ρ in R$^d$. The rotation acts on spatial variables, and thus on functions via pullback. But it also can act on the spatial vector Rƒ. The final transformation property asserts that the Riesz transform is equivariant with respect to these two actions; that is,
    $\rho ^{*}R_{j}[(\rho ^{-1})^{*}f]=\sum _{k=1}^{d}\rho _{jk}R_{k}f.$
These three properties in fact characterize the Riesz transform in the following sense. Let T=(T$_1$,...,T$_d$) be a d-tuple of bounded linear operators from L$^2$(R$^d$) to L$^2$(R$^d$) such that
* T commutes with all dilations and translations.
* T is equivariant with respect to rotations.
Then, for some constant c, T = cR.

## Relationship with the Laplacian

Somewhat imprecisely, the Riesz transforms of $f$ give the first partial derivatives of a solution of the equation
    ${(-\Delta )^{\frac {1}{2}}u=f},$
where Δ is the Laplacian. Thus the Riesz transform of $f$ can be written as:
    ${Rf=\nabla (-\Delta )^{-{\frac {1}{2}}}f}$
In particular, one should also have
    $R_{i}R_{j}\Delta u=-{\frac {\partial ^{2}u}{\partial x_{i}\partial x_{j}}},$
so that the Riesz transforms give a way of recovering information about the entire Hessian of a function from knowledge of only its Laplacian.
This is now made more precise. Suppose that $u$ is a Schwartz function. Then indeed by the explicit form of the Fourier multiplier, one has
    $R_{i}R_{j}(\Delta u)=-{\frac {\partial ^{2}u}{\partial x_{i}\partial x_{j}}}.$
The identity is not generally true in the sense of distributions. For instance, if $u$ is a tempered distribution such that $\Delta u\in L^{2}(\mathbb {R} ^{d})$, then one can only conclude that
    ${\frac {\partial ^{2}u}{\partial x_{i}\partial x_{j}}}=-R_{i}R_{j}\Delta u+P_{ij}(x)$
for some polynomial $P_{ij}$.