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In Schalg's Classical multilinear and Harmonic analysis, he presented two methods of the proof of Stein-Tomas theorem, one of which is called the fractional integration method. As a matter of fact, in order to prove \begin{equation} \lVert f * \hat\mu \rVert_{L^{p'}(\mathbb{R}^d)}\le C \lVert f \rVert_{L^p(\mathbb{R}^d)}, \quad \text{for } p=\frac{2d+2}{d+3}, \quad d\ge 3, \end{equation} where $\hat{\mu}\triangleq K$ is the Foureir transform of Lebesgue measure of surface with nonvanishing Gaussian curvature (and we may assume it is just the Fourier tranform of Lebesgue measure of unit sphere $\mathbb{S}^{d-1}$), he tore the coordinates into two pieces $x=(x',t)$, where $x'=(x_1,...,x_{d-1})$, then \begin{equation} f*\hat{\mu} (x) =\int_{\mathbb{R}}\int_{\mathbb{R}^{d-1}} K(x'-y',t-s) f(y',s) dy'ds. \end{equation} Thus we may restrict our attention on the behavior of $K(x',t)$ with respect to $x'$. More precisely, if we assume that $(Ug)(x')= \int_{\mathbb{R}^{d-1}} K(x'-y',t)dt$, then Schlag claimed that $U(t)$ satifies \begin{equation} \lVert U(t) \rVert_{L^1(\mathbb{R}^{d-1}) \to L^\infty(\mathbb{R}^{d-1})} \le C |t|^{d-1}, \quad \lVert U(t) \rVert_{L^2(\mathbb{R}^{d-1}) \to L^2(\mathbb{R}^{d-1})} \le C <\infty, \end{equation} where $C>0$ is independent of $t\in \mathbb{R}$, and then we can use Riesz-Thorin interpolation theorem and then use Hardy-Littlewood-Sobolev inequality to get the desired estimates.

And my question is how to check the second estimate (i.e. the uniform bound of $L^2 \to L^2$), Schlag said it suffices to check that $K(\hat{\cdot},t) \in L_{\xi'}^\infty L_t^\infty ( \mathbb{R}^{d-1} \times \mathbb{R})$, where $K(\hat{\cdot},t)$ means the Fourier transform of $K(x',t)$ w.r.t. $x'$. For example, in $d=3$, then the Fourier transform of unit sphere can be represented by $\hat{\sigma}(x)=\frac{\sin{|x|}}{|x|}$ explicitly, but how can I check that \begin{equation} K(\xi',t)= \int_{\mathbb{R}^2} e^{-2\pi i x' \cdot \xi'}\frac{\sin{|(x',t)|}}{|(x',t)|} dx' \in L_{\xi'}^\infty L_t^\infty ( \mathbb{R}^{2} \times \mathbb{R}) \quad? \end{equation}

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  • $\begingroup$ I think 'of' doesn't belong in "the fractional integration of method". $\endgroup$
    – LSpice
    May 1, 2020 at 3:26

1 Answer 1

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Let's first clarify the definitions (also, there are some typos in your post, perhaps you should consider correcting them). For $\xi\in\mathbb{R}^d$ we shall write $\xi=(\xi',\xi_d)$ with $\xi'\in\mathbb{R}^{d-1}$. For a tempered distribution $T$ we shall denote its distributional Fourier transform by $\widehat{T}$. We will use the same symbol for distributions on $\mathbb{S}(\mathbb{R}^d)$ and $\mathbb{S}(\mathbb{R}^{d-1})$, it will be clear from the context.

We work with a surface of the form $$ M=\{(x', \psi(x')): x'\in U\} $$ for some bounded open set $U\subset\mathbb{R}^{d-1}$ (can think $M=\mathbb{S}^{d-1}$). The surface measure on $M$ is given for $f\in\mathbb{S}(\mathbb{R}^d)$ by $$ \int_{\mathbb{R}^d}f(x)d\mu(x)=\int_{U}f(x', \psi(x'))\sqrt{1+|\nabla \psi(x')|^2}dx'. $$ Note that $\sqrt{1+|\nabla \psi(x')|^2}\simeq 1$, which means that this factor is harmless.

We define $$ K(\xi)=\widehat{\mu}(\xi),\qquad \xi\in\mathbb{R}^d. $$ Next for a fixed $t\in\mathbb{R}$ we consider a locally integrable function $K_t$ on $\mathbb{R}^{d-1}$ given by $$ K_t(\xi'):=K(\xi',t),\qquad \xi'\in\mathbb{R}^{d-1}. $$ We $\textbf{shall show that}$ the distributional Fourier transform of $K_t$ coincides with an $L^\infty$ function on $\mathbb{R}^{d-1}$ which is bounded uniformly in $t\in\mathbb{R}$.

$\textbf{Solution:}$ Using the definition of a Fourier transform of a distribution and then applying Fubini's theorem, we get for $\varphi\in\mathbb{S}(\mathbb{R}^{d-1})$ \begin{align*} \langle \widehat{K_t}, \varphi\rangle&=\langle K_t, \widehat{\varphi}\rangle=\int \widehat{\mu}(\xi',t)\widehat{\varphi}(\xi')d\xi'=\int_{\mathbb{R}^{d-1}}\int_{\mathbb{R}^{d}}e^{-2\pi i(x'\xi'+x_d t)}d\mu(x',x_d) \widehat{\varphi}(\xi')d\xi'\\ &=\int_{\mathbb{R}^{d}}e^{-2\pi i x_d t}\left(\int_{\mathbb{R}^{d-1}}e^{-2\pi i x'\xi'}\widehat{\varphi}(\xi')d\xi'\right)d\mu(x',x_d)\\ & =\int_{\mathbb{R}^{d}}e^{-2\pi i x_d t}\varphi(x')d\mu(x',x_d)=\int_U e^{-2\pi i \psi(x') t}\varphi(x')\sqrt{1+|\nabla \psi(x')|^2}dx'\\ &=: \langle F_t, \varphi\rangle, \end{align*} where $F_t(x')=\chi_U(x')\sqrt{1+|\nabla \psi(x')|^2}e^{-2\pi i\psi(x') t}$. Clearly $F_t(x')\in L_{x'}^\infty L_t^\infty ( \mathbb{R}^{d-1} \times \mathbb{R})$, so the claim is proved.

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  • $\begingroup$ thanks for you solution, but I have a question, why $\sqrt{1+|\nabla \psi (x')|}$ is harmless? For example, as for $M=\mathbb{S}^{d-1}$, then $\psi(x')=\sqrt{1-|x'|^2}$, where $x' \in B(0,1) \triangleq U$. But in this case, we can see that $\sqrt{1+|\nabla \psi(x')|^2}= \frac{1}{ \sqrt{1-|x'|^2} }$, which I think will lead to a singularity in a neighborhood of $|x'|=1$. $\endgroup$
    – Tao
    May 10, 2020 at 13:54
  • $\begingroup$ That's true, but notice that one can cover a sphere with finitely many charts without singularities, so it suffices to restrict attention to the part $\{(x', \sqrt{1-|x'|^2}): x'\in\mathbb{R}^{d-1}$ and $|x'|<1/2 \}$. $\endgroup$
    – Tony419
    May 10, 2020 at 14:06
  • $\begingroup$ I get it, thank you very much. $\endgroup$
    – Tao
    May 10, 2020 at 14:24
  • $\begingroup$ You're welcome :] $\endgroup$
    – Tony419
    May 10, 2020 at 14:24

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