In Schalg's [Classical multilinear and Harmonic analysis](https://www.cambridge.org/core/books/classical-and-multilinear-harmonic-analysis/0F3A1DE477BFF6547CB8A93F6B3588E8?__cf_chl_jschl_tk__=f8145904b8dc6b8f9f7ba2a471ef08ae91a8ccb5-1588300573-0-Af7EgtxN6uQKGgw21kmfN1GlCaZKgJFr1K_Gs5K5CjM-fkaZ0qVCEzNoFyCaNsfI7XN5SD4fRyY2IiEArEScN8jfMo_GRP3q15P1FajpqawrCPbNa9oUmzMeV0YhPHz-74JuHG7T70GBxvvs0X4w5v0nySrnwWdzv1F5a2cpxtZ-oKHjdNMiNLfkpieGnCXofxMrPEbGctpxcXgN7dj8pqUEfU8hA5xeIMihQ7svsYYHIPXnOH6tBslAifJSxovsdWZENGfj_vkjiandl3U9_0jk5hK-m47XhprwNIEYK2M_LqMmtir1bEJ8o177ZtAISDNoivNDySAslBhDMRZzutHaKR7Q5HWO3YpdydCtnfS5l29Q3WuFstwP_gA5p_bcdKmnlCC_aOQX9sQ8vEPGDz59CyMv3Ru9fFKkp6zmQBpK63PpXJ67h2y8zd3m4_DJbg), 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}



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


