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 $q\geq 1$?

Question

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.

Answer 1

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.