# Fourier Transform of an even function

Let $$S^n$$ be an $$n$$-dimentional unit sphere.

Consider $$f: S^n \longrightarrow R_+$$, where $$f$$ is an even continuous function.

Denote $$F(f):=\int_0^{\infty}\int_{S^n}f(y)g\left(\frac{|xy|}{t}\right)dy\frac{dt}{t^{n+1}},$$ where $$x \in S^n, \, t>0$$, and function $$g$$ is such that $$\int_{0}^{\infty}s^jg(s)ds=0, \quad j=0,2,4,\ldots, 2\left[(n-1)/2\right]$$ $$\int_1^{\infty}s^{\alpha}|g(s)|ds< \infty, \quad \alpha>n-1.$$

Find the Fourier Transform of $$F$$.

• What does it mean for a function to be "even" on the unit sphere for $n>1$? – Michael Engelhardt Jun 13 '20 at 13:41
• I guess "even" just means $f(x)=f(-x)$ all $x$. – YCor Jun 13 '20 at 13:45
• Ah - parity-even as it's called in some quarters. Yes, that must be it. – Michael Engelhardt Jun 13 '20 at 14:45

This is not a full answer, just an outline. It may require additional regularity assumptions on $$f$$ and $$g$$. $$\newcommand{\bR}{\mathbb{R}}$$ $$\DeclareMathOperator{\SO}{SO}$$ For $$x\in\bR^{n+1}\setminus 0$$ we set $$\bar{x}:=\frac{1}{|x|}x.$$ I assume that $$xy$$ denotes the inner product. Note that $$F[f](x)=\int_0^\infty\left(\int_{S^n} f(y)g(|xy|/t)dy\right) t^{-n-1}dt$$ $$= \int_0^\infty\left(\int_{S^n} f(y)g(|x| |\bar{x} y|/t)dy\right) t^{-n-1}dt$$ ($$t=s|x|$$) $$= |x|^{-n}\int_0^\infty\left(\int_{S^n} f(y)g(|\bar{x} y|/s)dy\right) s^{-n-1}ds=|x|^{-n}F(\bar{x}).$$ For $$s>0$$ define $$\newcommand{\eT}{\mathscr{T}}$$ $$\eT_s:L^2(S^n)\to L^2(S^n),\;\;\eT_s[f](x)=\int_{S^n}f(x)g(|xy|/s) dy,\;\;\forall x\in S^{n}.$$ (This requires some assumption on $$g$$.) Observe next that we have a right action of $$\SO(n+1)$$ on $$L^2(S^n)$$. For $$A\in\SO(n+1)$$ define $$L^2(S^n)\ni f\mapsto A^*f\in L^2(S^n),\;\;A^*f(x)=f(Ax).$$ Note that $$\eT_s[A^*f](x) = \int_{S^n}f(Ax)g(|Axy|/s) dy= \int_{S^n}f(Ax)g(|AxAy|/s) dy$$ $$= \int_{S^n}f(Ax)g(|xy|/s) dy=\eT_s[f](Ax)$$ so that $$\eT_s[A^*f]=A^*\eT_s[f].$$ In other words, the transformation $$\eT_s$$ is equivariant with respect to the action of $$\SO(n+1)$$ and thus, according to Schur's Lemma, it acts as multiplication by constants on the irreducible components of this $$\SO(n+1)$$ representation on $$L^2(S^n)$$.

These are the spaces of homogeneous harmonic polynomials or, equivalently, the eigenspaces of the Laplacian on the round $$n$$-dimensional sphere. As such they coincides with the restrictions to the sphere of homogeneous harmonic polynomilas.

Denote by $$\newcommand{\bH}{\mathbb{H}}$$ $$\bH_d$$ space of (restrictions to $$S^n$$) of homogeneous polynomials of degree $$d$$ on $$\bR^{n+1}$$. Thus, $$\forall s>0$$, $$d>0$$ there exists a constant $$c_d(s)$$ such that $$\eT_s[P]=c_d(s)P,\;\;\forall P\in \bH_d.$$ Let me explain how to find this constant. Denote by $$\newcommand{\bx}{\boldsymbol{x}}$$ $$\bx^+=(1,0,0,\dotsc,0)\in\bR^{n+1}$$ the North Pole in $$S^n$$ and choose $$P\in\bH_d$$ such that $$P(\bx^+)=1$$. Then $$\eT_s[P](\bx^+)=c_d(s)P(\bx^+)=c_d(s).$$ Hence $$c_d(s)=\int_{S^n}P(y)g(|\bx^+y|/s) dy.$$ Fortunately the spaces $$\bH_d$$ are well understood and the above integral can be explicitly described as a $$1$$-dimensional integral involving $$g$$ and Legendre polynomials. This is the so called Funk-Hecke formula ; see Sec. 1.4 of

C. Muller: Analysis of Spherical Symmetries in Euclidean Spaces, Springer Verlag, 1998.

Now observe that if $$f\in\bH_d$$ and $$x\in S^n$$ then $$F[f](x)=\int_0^\infty \eT_s[P] s^{-n-1} ds=\left(\int_0^\infty c_d(s) s^{-n-1} ds\right)P.$$

Thus, everything boils down to computing Fourier transforms of homogeneous functions of the form,

$$\frac{1}{|x|^{n+d}}P_d(x),$$

where $$P_d$$ is a homogenous harmonic polynomial of even degree $$d$$ in $$n+1$$ variables.