# the fractional integration method of the proof of Stein-Tomas theorem?

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 $$$$\lVert f * \hat\mu \rVert_{L^{p'}(\mathbb{R}^d)}\le C \lVert f \rVert_{L^p(\mathbb{R}^d)}, \quad \text{for } p=\frac{2d+2}{d+3}, \quad d\ge 3,$$$$ where $$\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})$$, then $$$$f*\hat{\mu} (x) =\int_{\mathbb{R}}\int_{\mathbb{R}^{d-1}} K(x'-y',t-s) f(y',s) dy'ds.$$$$ Thus we may restrict our attention on the behavior of $$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)$$ satifies $$$$\lVert U(t) \rVert_{L^1(\mathbb{R}^{d-1}) \to L^\infty(\mathbb{R}^{d-1})} \le C |t|^{d-1}, \quad \lVert U(t) \rVert_{L^2(\mathbb{R}^{d-1}) \to L^2(\mathbb{R}^{d-1})} \le C <\infty,$$$$ where $$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 $$$$K(\xi',t)= \int_{\mathbb{R}^2} e^{-2\pi i x' \cdot \xi'}\frac{\sin{|(x',t)|}}{|(x',t)|} dx' \in L_{\xi'}^\infty L_t^\infty ( \mathbb{R}^{2} \times \mathbb{R}) \quad?$$$$

• I think 'of' doesn't belong in "the fractional integration of method". – LSpice May 1 '20 at 3:26

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.
• 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 $|x'|=1$. – Tao May 10 '20 at 13:54
• 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 $|x'|<1/2 \}$. – Tony419 May 10 '20 at 14:06