Some intuition on a specific problem on Sobolev's embedding theorem with its relation to Fourier transform of restricted functions I have recently encountered this problem in my studies of Sobolev spaces and generalized functions (distributions), on which I can say I might have some intuition but cannot stumble across a final solution, but first some notations:
$ \mathcal{S} $ is the Schwartz class of functions and $ \mathcal{S}' $ is the class of distributions ("generalized functions") on $ \mathcal{S} $
$ \partial ^ {\alpha} $ is the multi-index distribution derivative on $ L^2 $
$ \Lambda _ s f = { [(1+|\xi|^2)^{\frac{s}{2}} \widetilde{f} ]^{\vee} } $ is a continuous linear operator on the Schwartz distributions. 
The Sobolev space $ H_s = \{ f \in S' | \Lambda_s f \in L^2 \}  $

The Sobolev Embedding Theorem: suppose that $ s > k + \frac{1}{2}n $
a. If $ f \in H_s $ then $ \widehat{\partial ^ \alpha f} \in L^1 $ and $ ||\widehat{\partial ^ \alpha f}||_1 \leq C ||f||_{(s)} $ for multi-index $ |\alpha| < k $ where C depends only on $ k-s $
b. $ H_s \subset C_0 ^ k $ and the inclusion map is continuous.
A reference exercise I managed to do which the problem states as relevant:
If $ f \in L^1(R^{n+m}) $ we define $ P(f) = \int{f(x,y)dy} $ where we have $ x \in R^n $ and $ y \in R^m $ then $ Pf \in L^1(R^n) $ and also we have $ ||Pf||_1 \leq ||f||_1 $ and $ \widehat{ Pf(\xi) } = \widehat{f}(\xi,0) $

And now the actual exercise/problem I am faced with, from Folland's real analysis second edition page $309$ exercise #$35$:

The Sobolev theorem states that if $ s > \frac{1}{2}n $, then it makes sense to evaluate functions in $ H_s $ at a point. For $ 0 \leq s \leq \frac{1}{2}n $ functions in $ H_s $ are defined almost everywhere but if $ s > \frac{1}{2}k $ with $ k < n $ it makes sense to restrict functions in $ H_s $ to subspaces of codimension k. More precisely let us write $ R^n = R^{n-k} \times R^k $, $x=(y,z)$ and dual coordinates $ \xi = (\eta,\zeta) $ and define $ R : \mathcal{S}^n \to \mathcal{S}^{n-k} $ by $ Rf(y) = f(y,0) $
a. We are to show $ \widehat{(Rf)}(\eta) = \int{\widehat{f}(\eta,\zeta)}d\zeta $ via the mentioned exercise stated earlier.
b. We are to show that if $ s > \frac{1}{2}k $ then the following inequality holds:
$ |\widehat{Rf}(\eta)|^{2} \leq C_s (1+|\eta|^2)^{\frac{k}{2}-s} \int{|\widehat{f}(\eta,\zeta)}|^2 (1+|\eta|^2 + |\zeta|^2)^s d\zeta $
c. We are to show that $R$ extends to a bounded map from $ H_s(R^n) $ to $ H_{s-\frac{k}{2}} (R^{n-k}) $ provided that $ s > \frac{k}{2} $

Now I can say I have some intuition for part a via the aforementioned exercise but I still cannot seem to actually find the solutions (rigorous ones) to this problem, so to sum up I really need the help on this seemingly difficult question. I would appreciate any kind of help.
 A: It would help if you actually copied down the problem statement correctly. The inequality in (b) should be
    $$\left|\widehat{Rf}(\eta)\right|^{2}\leq C_{s}(1+\left|\eta\right|^{2})^{\frac{k}{2}-s}\int_{\mathbb{R}^{k}}\left|\widehat{f}(\eta,\zeta)\right|^{2}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{s}d\zeta$$

Let $f\in\mathcal{S}(\mathbb{R}^{n})$. For (a), observe that by $n$-dimensional Fourier inversion and Fubini's theorem,
    $$f(y,0)=\int_{\mathbb{R}^{n}}\widehat{f}(\xi)e^{2\pi i\xi\cdot(y,0)}d\xi=\int_{\mathbb{R}^{n-k}}e^{2\pi i\eta\cdot y}d\eta\int_{\mathbb{R}^{k}}\widehat{f}(\eta,\zeta)d\zeta, \quad\forall y\in\mathbb{R}^{n-k}$$
Set $g(\eta):=\int_{\mathbb{R}^{k}}\widehat{f}(\eta,\zeta)d\zeta$. By your previous exercise, $g\in L^{1}(\mathbb{R}^{n-k})$ and observe that the RHS above may be written as
    $${g}^{\vee}(y),\quad\forall y\in\mathbb{R}^{n-k}$$
where we take the $(n-k)$-dimensional inverse Fourier transform. Since $\widehat{g}^{\vee}(y)=f(y,0)$ is in $L^{1}$ (actually Schwartz), by $(n-k)$-dimensional Fourier inversion,
$$\int_{\mathbb{R}^{k}}\widehat{f}(\eta,\zeta)d\zeta=g(\eta)=\int_{\mathbb{R}^{n-k}}g^{\vee}(y)e^{-2\pi iy\cdot\eta}dy=\int_{\mathbb{R}^{n-k}}(Rf)(y)e^{-2\pi iy\cdot\eta}y=\widehat{Rf}(\eta),\quad\forall\eta\in\mathbb{R}^{n-k}$$
Note that an easy way to see (a) is also by the density of tensor products of functions $\mathcal{S}(\mathbb{R}^{n-k})$ and $\mathcal{S}(\mathbb{R}^{k})$ in $\mathcal{S}(\mathbb{R}^{n})$.
For (b), observe first that
    \begin{align*}
\left|\int_{\mathbb{R}^{k}}\widehat{f}(\eta,\zeta)d\zeta\right|^{2}&\leq\left(\int_{\mathbb{R}^{k}}\left|\widehat{f}(\eta,\zeta)\right|d\zeta\right)^{2}\\
&=\left(\int_{\mathbb{R}^{k}}\left|\widehat{f}(\eta,\zeta)\right|(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{\frac{s}{2}}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{-\frac{s}{2}}d\zeta\right)^{2}\\
&\leq\left(\int_{\mathbb{R}^{k}}\left|\widehat{f}(\eta,\zeta)\right|^{2}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{s}d\zeta\right)\left(\int_{\mathbb{R}^{k}}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{-s}d\zeta\right),
 \end{align*}
where we use Cauchy-Schwarz to obtain the ultimate inequality. Since $2s>k$, the second factor in the last inequality is finite. By dilation invariance, we see that
    \begin{align*}
\int_{\mathbb{R}^{k}}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{-s}d\zeta&=(1+\left|\eta\right|^{2})^{-s}\int_{\mathbb{R}^{k}}\left(1+\frac{\left|\zeta\right|^{2}}{1+\left|\eta\right|^{2}}\right)^{-s}d\zeta\\
&=(1+\left|\eta\right|^{2})^{\frac{k}{2}-s}\int_{\mathbb{R}^{k}}(1+\left|\zeta\right|^{2})^{-s}d\zeta
 \end{align*}
Using part (a), we obtain the desired conclusion.
For (c), multiply $\left|\widehat{Rf}(\eta)\right|^{2}$ by the factor $(1+\left|\eta\right|^{2})^{s-\frac{k}{2}}$, and use Fubini's theorem to obtain
    \begin{align*}
\int_{\mathbb{R}^{k}}(1+\left|\eta\right|^{2})^{s-\frac{k}{2}}\left|\widehat{Rf}(\eta)\right|^{2}d\eta&\leq\int_{\mathbb{R}^{k}}\int_{\mathbb{R}^{n-k}}\left|\widehat{f}(\eta,\zeta)\right|^{2}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{s}d\zeta\\
&=\left\|\Lambda_{s}f\right\|_{L^{2}}^{2}<\infty
 \end{align*}
By density of Schwartz functions in $H^{s}(\mathbb{R}^{n})$, we obtain the desired conclusion.
