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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 $F$.

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

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