For which $s \in \mathbb{R}$ is the surface measure an element of $H^{s}(\mathbb{R}^{m})$ I'm working on Q17 of sheet 2 of the course Distribution Theory and applications of part iii of the mathematical Tripos. I've left the link for the question out due to copyright reasons but you may find it via google.
The question reads: Let $\Gamma=\{x\in \mathbb{R}^{m}:x \cdot n=0 \}$ be a hyperplane with surface element $d \sigma$ and normal $n$. Let $\chi \in \mathscr S(\mathbb{R}^{m})$ be a fixed Schwartz function and set $d \mu = \chi d \sigma$. This defines a tempered distribution $\mu_{\Gamma}$ by
$$
\langle\mu_{\Gamma},\psi\rangle =\int_{\Gamma}\psi d\mu
$$
for each $\psi \in \mathscr S(\mathbb{R}^{m})$. The Fourier transform of $\mu_{\Gamma}$ is defined as
$$
\hat{\mu_{\Gamma}}=\int_{\Gamma}e^{-i\lambda \cdot x} d\mu(x).
$$
We are asked to determine for which $s \in \mathbb{R}$ we have $\mu_{\Gamma} \in H^{s}(\mathbb{R}^{m})$.
To recap,
$$
H^{s}(\mathbb{R}^{m})=\{f:\mathbb{R}^{m} \to \mathbb{R} \mid f\ \text{measurable},\ \|f\|_{H^{s}} < +\infty \}
$$
where
$$
\|f\|_{H^{s}}^{2}= \int_{\mathbb{R}^{m}}(1+|\lambda|^{2})^{s}|\hat{f}(\lambda)|^{2} d\lambda.
$$
In my attempt, I considered the case where $\Gamma$ is the plane $x_{m}=0$. In this case
$$
\hat{\mu_{\Gamma}}=\int_{\mathbb{R}} \dotsb \int_{\mathbb{R}} e^{-i (\lambda_{1}x_{1}+\dots+ \lambda_{m-1}x_{m-1})} \psi(x_{1},\dots,x_{m-1},0) \,dx_{1} \dots dx_{m-1}.
$$
Let $\psi_{0} := \psi(x_{1},\dots,x_{m-1},0)$. Notice that $\psi_{0} \in \mathscr S(\mathbb{R}^{m-1})$. Thus, $\hat{\mu_{\Gamma}}= \hat{\psi_{0}} \in \mathscr S(\mathbb{R}^{m-1}) \subset H^{s}(\mathbb{R}^{m-1})$  for each $s \in \mathbb{R}$.
I'm not sure how to "upgrade" to $H^{s}(\mathbb{R}^{m})$. Any help would be greatly appreciated.
 A: $\newcommand{\R}{\mathbf R}$The space $H^s(\R^m)$ is invariant under rotations, so without loss of generality, $\Gamma = \{x_m = 0\}$. Thus, if we write $\mathbf{R}^m=\Gamma\times\R$, with coordinates $(x,t)$, then $$d\mu(x,t)=\chi(x,t)\,d\sigma = \chi(x,t)\,dx\otimes\delta_0(dt).$$ ($\delta_0(dt)$ is the point mass at the origin of $\R$.) Taking the Fourier transform gives $\newcommand{\F}{\mathscr F}$
\begin{align*}
\widehat{d\mu}(\xi,\tau) &= \int_{\R^m}e^{-i(x\cdot \xi + t\tau)}\,d\mu(x,t)\\
&= \int_{\R^{m-1}}e^{-ix\cdot\xi}(\int_\R\chi(x,t)e^{-it\tau}\,\delta_0(dt))\,dx\\
&= \int_{\R^{m-1}}e^{-ix\cdot\xi}\chi(x,0)\,dx = (\F_1\chi)(\xi,0),
\end{align*}
where $(\F_1\chi)(\xi,t)$ denotes the function we get by taking the Fourier transform of $\chi(x,t)$ in the $x$ variable alone. Put $f(\xi) = (\F_1\chi)(\xi,0)$, so that $f\in\mathscr S(\R^{m-1})$ and $\widehat{d\mu}(\xi,\tau) = f(\xi)$. In other words, $\widehat{d\mu}$ is a function of $(\xi,\tau)$ that is actually independent of $\tau$.
Now we ask for which $s$ we have
$$
\iint_{\R^{m-1}\times\R} |f(\xi)|^2(1+|(\xi,\tau)|^2)^s\,d\xi\,d\tau<\infty.
$$
We get all the decay in $\xi$ we want because $f(\xi)$ is Schwartz, so we only need to choose $s$ to make sure that our integral in $\tau$ is convergent. You can check that $s < -\frac12$ is necessary and sufficient for this integral to be finite.
