José A. CañizoResearch · Publications · Teaching · Other

Projecting on radial functions

A function $g: \R^d \to \R$ is radial or invariant by rotations if it happens that $g(R(x)) = g(x)$ for all $x \in \R^d$ and all rotations $R: \R^d \to \R^d$ (linear orthogonal transformations with determinant 1). The set of real or complex radial functions is clearly a vector space, and we may consider the projection of a function to it. What is its expression?

More precisely, consider $L^2$ the usual real or complex Hilbert space of square-integrable functions on $\R^d$, with the inner product $\langle{ \cdot,\cdot }\rangle$ and norm denoted by $\|\cdot\|$. Call $\mathcal{R}$ the subspace of $L^2$ formed by all functions which are invariant by rotations. What is the expression of the orthogonal projection to $\mathcal{R}$?

Notice that $\mathcal{R}$ is a closed subspace of $L^2$, so we are entitled to consider the orthogonal projection to it: if $R$ is a rotation and $f \in L^2$ we can define the rotation of $f$ by $Rf:\R^d \to \R$, $Rf(x) = f(Rx)$. This $R:L^2 \to L^2$ is an isometry (continuous in particular), and $\mathcal{R}$ is the set of all functions $g$ such that $Rg = g$ for all rotations $R$; that is, $\mathcal{R}$ is the intersection of all sets of the form $(R-I)^{-1}({0})$ for $R$ a rotation, which is an intersection of closed sets and hence closed.

Take $f \in L^2$. Its projection to $\mathcal{R}$, which we call $\pi(f)$, is defined as the element $g \in \mathcal{R}$ which is closest to $f$; that is, such that \begin{equation*} \min_{g \in \mathcal{R}} \|f-g\| = \|f - \pi(f)\|. \end{equation*} One way to find it is the following: for all rotations $R$ and any $g \in \mathcal{R}$ we can write \begin{equation*} \|f-g\|^2 = \ird |f(x) - g(x)|^2 \d x = \ird |f(Rx) - g(Rx)|^2 \d x = \ird |f(Rx) - g(x)|^2 \d x. \end{equation*} Consider normalised Haar measure on $\mathcal{O}$. Averaging by all rotations $R \in \mathcal{O}$ we have then \begin{multline*} \|f-g\|^2 = \int_\mathcal{O} \ird |f(Rx) - g(x)|^2 \d x \d R = \ird \int_\mathcal{O} |f(Rx) - g(x)|^2 \d R \d x \ \geq \ird \Big| \int_\mathcal{O} f(Rx) \d R - g(x) \Big|^2\d x. \end{multline*} This shows then that \begin{equation*} \pi(f) = \int_{\mathcal{O}} f(Rx) \d R, \end{equation*} since we have proved the function defined by this expression is the closest one to $f$ among all functions in $\mathcal{R}$. Of course, it is in fact in $\mathcal{R}$: for any rotation $T$, \begin{equation*} T \int_\mathcal{O} Rf \d R = \int_\mathcal{O} TRf \d R = \int_\mathcal{O} Rf \d R, \end{equation*} since the measure on $\mathcal{O}$ is Haar measure.