= Solution
For $s>\tfrac12$, the <Sobolev trace theorem> states that the restriction map initially defined on smooth functions extends uniquely to a bounded linear operator
$$
\gamma:H^s(\mathbb R^n)
\longrightarrow H^{s-1/2}(\mathbb R^{n-1}),
\qquad
(\gamma u)(x')=u(x',0).
$$
Take first $u$ in the <Schwartz space>. Up to the harmless constant determined by the <Fourier transform> convention,
$$
\widehat{\gamma u}(\xi')
=\int_{\mathbb R}\widehat u(\xi',\xi_n)\,d\xi_n.
$$
The <Cauchy-Schwarz inequality>, with weights $1+|\xi'|^2+\xi_n^2$, gives
$$
\begin{aligned}
|\widehat{\gamma u}(\xi')|^2
&\leq
\left(\int_{\mathbb R}
(1+|\xi'|^2+\xi_n^2)^s
|\widehat u(\xi',\xi_n)|^2\,d\xi_n\right)\\
&\quad\times
\left(\int_{\mathbb R}
(1+|\xi'|^2+\xi_n^2)^{-s}\,d\xi_n\right).
\end{aligned}
$$
After the substitution $\xi_n=(1+|\xi'|^2)^{1/2}t$, the second factor is
$$
C_s(1+|\xi'|^2)^{1/2-s},
\qquad
C_s=\int_{\mathbb R}(1+t^2)^{-s}\,dt<\infty,
$$
where finiteness is exactly the condition $s>1/2$. Multiplying by $(1+|\xi'|^2)^{s-1/2}$ and integrating in $\xi'$ yields
$$
\lVert\gamma u\rVert_{H^{s-1/2}(\mathbb R^{n-1})}
\leq C_s^{1/2}\lVert u\rVert_{H^s(\mathbb R^n)}.
$$
Density of the Schwartz space in the <Sobolev space> completes the unique bounded extension.
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