If $\|\hat u\|_{W^{1,p}(\mathbb R^d)}\leq C\|u\|_{W^{1,p}(D)}$ does $\|\hat u\|_{L^q(\mathbb R^d)}\leq C\|u\|_{L^q(D)}$ for all $q\geq 1$? Let $\mathcal D=\{x\in \mathbb R^n\mid r<|x|<R\}$ and $u\in \mathcal C^1(\overline{\mathcal D})$. In particular, $u\in W^{1,p}(\mathcal D)$. I can prolonge $u$ on $W^{1,p}(\mathbb R^d)$ by $\hat u\in W^{1,p}(\mathbb R^d)$ s.t. $\hat u|_{\mathcal D}=u$,
$$\|\hat u\|_{L^p(\mathbb R^d)}\leq C\|u\|_{L^p(\mathcal D)}\quad \text{and}\quad \|u\|_{W^{1,p}(\mathbb R^d)}\leq C\|u\|_{L^p(\mathcal D)}.$$
Now my teacher said that we also have 
$$\|\hat u\|_{L^q(\mathbb R^d)}\leq C\|u\|_{L^q(\mathcal D)},$$
and I didn't see any such theorem, so I have doubt about it. So is it really true ? And if yes, were can I find a proof ? And if no, do you have a counter-example ?

Context
The prblem is the following one :
$$\frac{1}{r}=\frac{a}{p}+\frac{1-a}{q},$$
$p>1$, $q\geq 1$ and $a\in [0,1]$.
I know that for $u\in \mathcal C^1(\mathbb R^d)$ there is $C$ independent of $u$ s.t. $$\|u\|_{L^r(\mathbb R^d)}\leq C\|\nabla u\|_{L^p(\mathbb R^d)}^a\|u\|_{L^q(\mathbb R^d)}^{1-a}.$$
Now, I want to prove that if $u\in \mathcal C^1(\overline{\mathcal D})$ there is $C$ independent of $u$ s.t. $$\|u-\bar u\|_{L^r(\mathcal D)}\leq C\|\nabla u\|_{L^p(\mathcal D)}\|u-\bar u\|_{L^q(\mathcal D)}.$$
So since $u-\bar u\in \mathcal C^1(\mathcal D)$, it's also in $W^{1,p}(\mathcal D)$. Let $\hat u\in W^{1,p}(\mathbb R^d)$ the prolongement of $u-\bar u$ s.t. $$\hat u|_{\mathcal D}=u-\bar u,\quad \|\hat u\|_{L^p(\mathbb R^d)}\leq C\|u-\bar u\|_{L^p(\mathcal D)}\quad \text{and}\quad \|\hat u\|_{W^{1,p}(\mathbb R^d)}\leq C\|u-\bar u\|_{W^{1,p}(\mathcal D)}.$$
Now, (using Poincaré's inequality) I have that 
$$\|u-\bar u\|_{L^r(\mathcal D)}\leq \|\hat u\|_{L^r(\mathbb R^d)}\leq C\|\nabla u\|_{L^p(\mathcal D)}\|^a\|\hat u\|_{L^q(\mathbb R^d)}^{1-a}.$$
Now, my teacher say's that $$\|\hat u\|_{L^q(\mathbb R^d)}\leq C\|u-\bar u\|_{L^q(\mathcal D)},$$
hold. But I don't understand why, and I didn't see such a theorem in any books I read or in the internet. So I want to be sure about that it really hold, and if not, to have a counter-example.
 A: Assume without loss of generality that the average of $u$ is zero. To extend a
function in an annulus you do a double reflection. Take $r<s<t<R$ and consider
a smooth function $\varphi$ such that $\varphi=0$ for $|x|<s$ and $\varphi=1$
for $|x|>t$. Define the function
$$
v(x):=\left\{
\begin{array}
[c]{ll}%
u(x) & \text{if }r<|x|<R,\\
(\varphi u)(xR^{2}/|x|^{2}) & \text{if }|x|>R,
\end{array}
\right.
$$
where we extend $\varphi u:=0$ whenever $\varphi=0$. Note that when
$|x|R^{2}/|x|^{2}=R^{2}/|x|<R$ for $|x|>R$ and for $|x|=R$, we have that
$|x|R^{2}/|x|^{2}=R$ so $(\varphi u)(xR^{2}/|x|^{2})=u(x)$ since $\varphi=1$
for $|x|>t$.
Using the change of variables $y=xR^{2}/|x|^{2}$ or $x=yR^{2}/|y|^{2}$ you can
chech that $v$ is an extension to $\mathbb{R}^{d}\setminus B_{r}$ and that for
every $q$,
\begin{align*}
\int_{\mathbb{R}^{d}\setminus B_{r}}|v|^{q}dx  & =\int_{D}|u|^{q}%
dx+\int_{\mathbb{R}^{d}\setminus B_{r}}|(\varphi u)(xR^{2}/|x|^{2})|^{q}dx\\
& =\int_{D}|u|^{q}dx+\int_{B_{R}}|(\varphi u)(y)|^{q}R^{4}/|y|^{4}dy\\
& \leq\int_{D}|u|^{q}dx+\int_{B_{R}\setminus B_{s}}|u(y)|^{q}R^{4}%
/|y|^{4}dy\\
& \leq\int_{D}|u|^{q}dx+R^{4}/s^{4}\int_{B_{R}\setminus B_{s}}|u(y)|^{q}dy.
\end{align*}
while by the chain rule
\begin{align*}
\frac{\partial v}{\partial x_{i}}(x)  & =\sum_{j=1}^{d}[\varphi(xR^{2}%
/|x|^{2})\frac{\partial u}{\partial x_{j}}(xR^{2}/|x|^{2})+u(xR^{2}%
/|x|^{2})\frac{\partial\varphi}{\partial x_{j}}(xR^{2}/|x|^{2})]\frac
{\partial}{\partial x_{i}}(x_{j}R^{2}/|x|^{2})\\
& =\sum_{j=1}^{d}[\varphi(xR^{2}/|x|^{2})\frac{\partial u}{\partial x_{j}%
}(xR^{2}/|x|^{2})\\&\quad+u(xR^{2}/|x|^{2})\frac{\partial\varphi}{\partial x_{j}%
}(xR^{2}/|x|^{2})](\delta_{ij}R^{2}/|x|^{2}-2x_{i}x_{j}R^{2}/|x|^{4}).
\end{align*}
and so
\begin{align*}
|\nabla v(x)|  & \leq c[\varphi(xR^{2}/|x|^{2})|\nabla u(xR^{2}/|x|^{2}%
))|+|u(xR^{2}/|x|^{2})|\nabla\varphi(xR^{2}/|x|^{2}))|]R^{2}/|x|^{2}\\
& \leq c[\varphi(xR^{2}/|x|^{2})|\nabla u(xR^{2}/|x|^{2}))|+|u(xR^{2}%
/|x|^{2})|\nabla\varphi(xR^{2}/|x|^{2}))|]
\end{align*}
since $R^{2}/|x|^{2}\leq1$. Hence, changing variables as before
\begin{align*}
\int_{\mathbb{R}^{d}\setminus B_{r}}|\nabla v|^{p}dx  & \leq\int_{D}|\nabla
u|^{p}dx+c\int_{\mathbb{R}^{d}\setminus B_{r}}|\varphi(xR^{2}/|x|^{2})\nabla
u(xR^{2}/|x|^{2}))|^{p}dx\\
& \quad+c\int_{\mathbb{R}^{d}\setminus B_{r}}|u(xR^{2}/|x|^{2})\nabla
\varphi(xR^{2}/|x|^{2}))|^{p}dx\\
& \leq\int_{D}|\nabla u|^{p}dx+cR^{4}/s^{4}\int_{B_{R}\setminus B_{s}}|\nabla
u(y)|^{p}dy+R^{4}/s^{4}\int_{B_{t}\setminus B_{s}}|u(y)|^{p}dy.
\end{align*}
Finally to extend to $B_{r}$ you reflect again and define
$$
w(x):=\left\{
\begin{array}
[c]{ll}%
v(x) & \text{if }r<|x|,\\
v(xr^{2}/|x|^{2}) & \text{if }|x|<r.
\end{array}
\right.
$$
Since $\varphi=0$ for $|x|<s$ you have that $v(x)=0$ for $R^{2}/|x|<s$ that is
$|x|>R^{2}/s$ and so you can use the same type of calculations to conclude
that
\begin{align*}
\int_{B_{r}}|w|^{q}dx  & =\int_{\mathbb{R}^{d}\setminus B_{r}}|v(y)|^{q}%
r^{4}/|y|^{4}dy\leq\int_{\mathbb{R}^{d}\setminus B_{r}}|v(y)|^{q}dy\\
& \leq\int_{D}|u|^{q}dx+R^{4}/s^{4}\int_{B_{R}\setminus B_{s}}|u(y)|^{q}dy
\end{align*}
and similar estimates for the derivative, which I skip.
