A Sobolev embedding theorem for functions on spheres

Question

$L^2(\mathbb{S}^{d-1})$ is embedded in $H^{-s}(\mathbb{R}^d)$ with $s>\frac{1}{2}$, which means for $f\in L^2(\mathbb{S}^{d-1})$, the following holds: $$\DeclareMathOperator{\Dm}{\operatorname{d}\!} \int\limits_{\mathbb{R}^d} \bigg|\int\limits_{\mathbb{S}^{d-1}}f(\omega)e^{-2\pi i x\cdot \omega}\Dm\sigma(\omega)\bigg|^2\frac{\Dm x}{(1+|x|^2)^s}\lesssim \int_{\mathbb{S}^{d-1}}|f(\omega)|^2 \Dm\sigma(\omega) $$

This is a lemma in Luis Vega's article [1] (Page 2). He gives a somewhat roundabout proof. Intuitively, it may help to expand $f$ into spherical harmonics. But I don't know the accurate behavior of spherical harmonics under Fourier transform.

Could you please present a proof or give a explicit expression of spherical harmonics' Fourier transform?

Another idea from Willie Wong is to use Fourier restriction. By Tomas-Stein, we can deduce the conclusion when $s>\frac{d}{d+1}$.

Reference

[1] Luis Vega, "Schrödinger equations: Pointwise convergence to the initial data" (English) Proceeding of the American Mathematical Society 102, No. 4, 874-878 (1988), DOI 10.2307/2047326, MR0934859, Zbl 0654.42014.

Answer 1

As noted in Vega's article, this statement is a special case of the trace theorem, which in particular says that the restriction operator, which restricts a test function to a compact submanifold $\Omega$ of codimension $1$, is bounded from $H^s(\mathbb{R}^d)$ to $L^2(\Omega)$ for $s>1$. The lemma is than a statement about boundedness of the adjoint operator. However, I don't know where to find a proof of the trace theorem in a sufficiently convenient form (with restriction to a submanifold, but with Sobolev spaces described in terms of the Fourier transform), which is probably also why Vega found it easier to give a full proof.

I should also note that the main result from Vega's paper has been superseded by https://annals.math.princeton.edu/2019/189-3/p04

Answer 2

This inequality may be proved by a calm calculation. We can assume that $f$ supports near $(0,\cdots,0,1)\in \mathbb{S}^{d-1}$. Denote the first $d-1$ coordinates of $x\in \mathbb{R}^d$ by $x^{'}$. We have
\begin{equation*} \begin{split} (fd\sigma)^{\vee}(x^{'},t)&=\int_{\mathbb{S}^{d-1}}f(\omega)e^{2\pi i(x^{'},t)\cdot\omega} d\sigma(\omega) \\ &=\int_{\{y^{'}\in \mathbb{R}^{d-1}:|y^{'}|^2\leq 1-c^2\}} f(y^{'},\sqrt{1-|y^{'}|^2})e^{2\pi ix^{'}\cdot y^{'}}e^{2\pi it\sqrt{1-|y^{'}|^2}} \frac{dy^{'}}{\sqrt{1-|y^{'}|^2}} \\ &=\mathscr{F}^{-1} (f(y^{'},\sqrt{1-|y^{'}|^2})\frac{e^{2\pi it\sqrt{1-|y^{'}|^2}}}{\sqrt{1-|y^{'}|^2}})(x^{'}) \end{split} \end{equation*} where $^{\vee}$ is the inverse Fourier transform in $\mathbb{R}^d$ and $\mathscr{F}^{-1}$ in $\mathbb{R}^{d-1}$. Then by Plancherel, \begin{equation} \begin{split} \int_{\mathbb{R}^d} |\int_{\mathbb{S}^{d-1}}f(\omega)e^{2\pi i x\cdot \omega}d\sigma(\omega)|^2\frac{d x}{(1+|x|^2)^s}&\leq \int_{\mathbb{R}^d} |\int_{\mathbb{S}^{d-1}}f(\omega)e^{2\pi i (x^{'},t)\cdot \omega}d\sigma(\omega)|^2 dx^{'}\frac{d t}{(1+|t|^2)^s} \\ &=\int_{\mathbb{R}}\int_{\mathbb{R^{d-1}}}|f(y^{'},\sqrt{1-|y^{'}|^2})\frac{e^{2\pi it\sqrt{1-|y^{'}|^2}}}{\sqrt{1-|y^{'}|^2}}|^2 dy^{'}\frac{d t}{(1+|t|^2)^s} \\ &=C_s\int_{\mathbb{S}^{d-1}}|f(\omega)|^2 d\sigma(\omega) \end{split} \end{equation}