Support of a convolution when extending $u\in W^{1,p}_0$ by zero (Brezis Ch 9)

Question

(iii) $\Rightarrow$ (i). One can always assume that $\Omega$ is bounded (if not, consider the sequence $\left.\left(\zeta_{n} u\right)\right)$. By local charts and partition of unity this is reduced to the following problem. Let $u \in L^{p}\left(Q_{+}\right)$be such that the function $$ \bar{u}(x)= \begin{cases}u(x) & \text { if } x \in Q, x_{N}>0, \\ 0 & \text { if } x \in Q, x_{N}<0\end{cases} $$

belongs to $W^{1, p}(Q)$; prove that $$ \alpha u \in W_{0}^{1, p}\left(Q_{+}\right) \quad \forall \alpha \in C_{c}^{1}(Q) . $$ Let $\left(\rho_{n}\right)$ be a sequence of mollifiers such that $$ \operatorname{supp} \rho_{n} \subset\left\{x \in \mathbb{R}^{N} ; \frac{1}{2 n}<x_{N}<\frac{1}{n}\right\} $$ one may choose, for example, $$ \rho_{n}(x)=n^{N} \rho(n x) \quad \text { and } \quad \operatorname{supp} \rho \subset\left\{x \in \mathbb{R}^{N} ;(1 / 2)<x_{N}<1\right\} $$ Thus $\rho_{n} \star(\alpha \bar{u}) \rightarrow \alpha \bar{u}$ in $W^{1, p}\left(\mathbb{R}^{N}\right)$ (note that $\alpha \bar{u}$ extended by 0 outside $Q$ belongs to $\left.W^{1, p}\left(\mathbb{R}^{N}\right)\right)$. On the other hand, $$ \operatorname{supp}\left(\rho_{n} \star \alpha \bar{u}\right) \subset \operatorname{supp} \rho_{n}+\operatorname{supp}(\alpha \bar{u}) \subset Q_{+} $$ for $n$ large enough. It follows that $$ \rho_{n} \star(\alpha \bar{u}) \in C_{c}^{1}\left(Q_{+}\right) $$ and thus $\alpha u \in W_{0}^{1, p}\left(Q_{+}\right)$. (Screenshot of pages 289-290)

In Brezis, Proposition $9.18$ show that if $u\in W^{1,p}_0(\Omega)$ with $\partial \Omega$ of class $C^1$ then the extention to zero belongs to $W^{1,p}(\Bbb R^n)$.

My question: Why $\operatorname{supp}(\rho_n *\alpha \overline {u}) \subset Q_+$ and not $\overline {Q_+} $?

Notations:

$1)$ $Q= \{ x=(x',x_n) ; |x'|< 1, |x_n|<1\}$

$2)$ $Q_+= \{ x=(x',x_n) ; |x'|< 1, 0<|x_n|<1\}.$

Answer 1

Ignoring the $Q$ vs $\operatorname{supp} \rho_n$ issue I mentioned in the comment (i.e. either changing the definition of one or the other so they are compatible, so that we can only consider $x_N$):

$\operatorname{supp} \alpha\overline u \subset Q_+$ and is compact. Thus there is some minimal distance between it and $x_N = 0$, which is outside $Q_+$. Once $n$ is large enough $\operatorname{supp} \rho_n$, whose maximum thickness in the $x_N$ direction is $< \frac 1n$, is too thin to cover that distance. Thus $\operatorname{supp} \alpha\overline u + \operatorname{supp} \rho_n$ must remain strictly above $x_N = 0$.

Answer 2

I'm going to answer sort of informally.

First thing is that your (2), the definition of $Q_+$ doesn't correspond to Brezis'. For Brezis, $Q_+ = \{(x', x_N) \in {\mathbb R}^N: x' \in {\mathbb R}^{N-1}, x_N \in {\mathbb R}, |x'| < 1, 0 < x_N < 1\}$.

The major difficulty has to do with the analysis of the $N$-th coordinate. I'm only going to focus on that.

The main ideas are:

• $\operatorname{supp} \rho_n \star \alpha \bar u \subset \overline{ \operatorname{supp} \rho_n + \operatorname{supp}\alpha \bar u }$.

• $\operatorname{supp}\alpha \bar u \subset \overline{Q_+} \cap \operatorname{supp}\alpha$ (check the definition of $\bar u$).

• $\operatorname{supp} \rho_n \subset \left \{x \in \mathbb R^N: \frac 1 {2n} < x_N < \frac 1 n\right \}$.

What this all gives you is that, when you look at the set sum $\operatorname{supp} \rho_n + \operatorname{supp}\alpha \bar u$, the $N$-th coordinate gets lifted up by at least $\frac 1 {2n}$, always. More precisely, for every $x \in \operatorname{supp} \rho_n + \operatorname{supp}\alpha \bar u$, then $x = a + b$, for $a \in \operatorname{supp} \rho_n$ and $b \in \operatorname{supp}\alpha \bar u$. This gives $x_N = a_N + b_N > \frac 1 {2n}$, because $a_N > \frac 1 {2n}$ and $b_n \geq 0$.