Existence of operator I want to show that for $ s> \frac{1}{2} $ there is a bounded linear operator $ T: H^s(\mathbb{R}^n) \to H^{s-\frac{1}{2}}(\mathbb{R}^{n-1})$ following the below steps: 


*

*Consider that  $ u \in C_0^{\infty}(\mathbb{R}^n) $ and show that  $ \widehat{Tu}(\xi')=\int_{\mathbb{R}} \widehat{u}(\xi', \xi_n) d{\xi_n}$.

*Show that $ (1+|\xi'|^2)^{s-\frac{1}{2}} |\widehat{Tu}(\xi')|^2 \leq \text{ constant } ||u||_{H^s(\mathbb{R}^n)}^2 $.
Why do we show in that way that there is an operator as the desired one?
Let $ u \in C_0^{\infty}(\mathbb{R}^n)$.
Then $ \widehat{Tu}(\xi')=\int_{\mathbb{R}^n} Tu(\xi') e^{-ix \xi' }d{x} $.
Why is this equal to $\int_{\mathbb{R}} \widehat{u}(\xi', \xi_n) d{\xi_n} $ ?
 A: Here the author probably meant that $T\mu$ is the restriction map
$$
T\mu(\xi_1,\cdots, \xi_{n})=\mu(\xi_{1},\cdots, \xi_{n-1},0)
$$
Now writing out explicitly we have
$$
\widehat{T\mu}(\xi')=\int\mu(\xi,0)e^{-i \xi\cdot \xi'}d\xi
$$
But recall that formally we have (for 1D functions)
$$
f(0)=\frac{1}{2\pi}\int \widehat{f(\xi_{n}')}d\xi_{n}'=\frac{1}{2\pi}\int (\int f(x)e^{-x\cdot \xi'_{n}}dx)d\xi_{n}'
$$
Thus substitute $f(x)=\mu(\xi_{1},\cdots, \xi_{n-1},x)$ into above, we can re-write above as 
$$
\widehat{T\mu}(\xi')=\frac{1}{2\pi}\int \mu(\xi_{1},\cdots,x)e^{-i(\xi,x)\cdot (\xi',\xi_{n}')}dx d\xi_{n}'d\xi, \xi=(\xi_1,\cdots ,\xi_{n-1}),\xi'=(\xi_1',\cdots ,\xi_{n-1}')
$$
which can be further simplified to be
$$
\widehat{T\mu}(\xi')=\frac{1}{2\pi}\int (\mu(\xi_{1},\cdots,x)e^{-i(\xi,x)\cdot (\xi',\xi_{n}')}dxd\xi )d\xi_{n}', \xi=(\xi_1,\cdots ,\xi_{n-1}),\xi'=(\xi_1',\cdots ,\xi_{n-1}')
$$
and you get the desired formula this way. There may be something off with the constants. The $\xi_{n}'$ at here corresponds to the $\xi_{n}$ in your notation. It feels somewhat odd to me that $(\xi',\xi_{n})$ should be conflated together. 
The rest of the proof now should follow from Cauchy-Schwarz if you use inequality of the type 
$$
|\widehat{\mu(\xi',\xi'_{n})}d\xi_{n}'|^2\le \int |\widehat{\mu(\xi',\xi'_{n})}|^2\cdot (|\xi'|^2+|\xi'_{n}|^2)^{s}d\xi_{n}'*\int (|\xi'|^2+|\xi'_{n}|^{2})^{-s}d\xi_{n}'
$$
and choosing appropriate cut-off by assuming $|\xi'|>1$. 
