Restriction to $\mathbb{R}^{d-1}$ as an operator on $L^2(\mathbb{R}^d)$ Identify $\mathbb{R}^{d-1}$ with $\mathbb{R}^{d-1}\times \{0\}\subseteq \mathbb{R}^d$.
Is there a bounded operator $T: L^2(\mathbb{R}^d)\rightarrow L^2(\mathbb{R}^{d-1})$ such that $T(\phi)=\phi \restriction \mathbb{R}^{d-1}$ for every $\phi \in C_c^{\infty}(\mathbb{R}^d)$?
 A: Let $\omega_{a,b}\in C_c^\infty(\mathbb{R})$ be a bump function such that $\operatorname{supp}(\omega_{a,b})=[-a,a]$, $0\leq\omega_{a,b}\leq 1$ and $\omega_{a,b}=1$ on $[-b,b]$. Now fix $\varepsilon>0$ and consider
$$
\varphi(x)=\omega_{\varepsilon,0}(x_d)\cdot\prod\limits_{i=1}^{d-1}\omega_{1+\varepsilon,1}(x_i)
$$
Then
$$
\Vert\varphi\Vert_{L_2(\mathbb{R}^{d})}^2
=\int\limits_{\mathbb{R}}|\omega_{\varepsilon,0}(x_d)|^2dx_d\cdot 
\prod\limits_{i=1}^{d-1}\int\limits_{\mathbb{R}}|\omega_{1+\varepsilon,1}(x_i)|^2dx_i
\leq 2\varepsilon\cdot\prod\limits_{i=1}^{d-1}(2+2\varepsilon)
\leq 2^d\varepsilon(1+\varepsilon)^{d-1}
$$
$$
\Vert T(\varphi)\Vert_{L_2(\mathbb{R}^{d-1})}^2
=\prod\limits_{i=1}^{d-1}\int\limits_{\mathbb{R}}|\omega_{1+\varepsilon,1}(x_i)|^2dx_i
\geq \prod\limits_{i=1}^{d-1}2=2^d
$$
Hence for all $\varepsilon>0$ we have
$$
\Vert T\Vert\geq \Vert T(\varphi)\Vert_{L_2(\mathbb{R}^{d-1})}/\Vert \varphi\Vert_{L_2(\mathbb{R}^{d})}=(1+\varepsilon)^{(1-d)/2}\varepsilon^{-1/2}
$$
After taking the limit $\varepsilon\to0^+$ we get $\Vert T\Vert=+\infty$, so the desired operator is unbounded.
Thanks to davin for this idea!
