Differentiation of a product of Sobolev functions

Question

The following is a proposition and proof from Brezis. I have some questions regarding the proof, if someone may be able to help me.

Proposition 9.4 (differentiation of a product). Let $u,v\in W^{1,\,p}(\Omega)\cap L^{\infty}(\Omega)$ with $1\leq p\leq\infty$. Then $uv\in W^{1,\,p}(\Omega)\cap L^{\infty}(\Omega)$ and, \begin{align} \frac{\partial}{\partial x_{i}}(uv)=\frac{\partial u}{\partial x_{i}}v+u\frac{\partial v}{\partial x_{i}},\quad i=1,2,\ldots,N. \end{align}

Proof. There exist sequences $(u_{n}),(v_{n})$ in $C_{c}^{\infty}(\mathbb{R}^{N})$ such that, \begin{align} u_{n|\Omega}&\rightarrow u, &v_{n|\Omega}&\rightarrow v &\text{in }&L^{p}(\Omega)\text{ and a.e. on }\Omega,\\ \nabla u_{n|\omega}&\rightarrow\nabla u_{|\omega}, &\nabla v_{n|\omega}&\rightarrow\nabla v_{|\omega} &\text{in }&L^{p}(\omega)^{N}\text{ for all }\omega\subset\subset\Omega. \end{align} Checking the proof of Theorem 9.2 (density of $C_{c}^{\infty}(\mathbb{R}^{N})$ in $W^{1,\,p}(\omega)$ for all $\omega\subset\subset\Omega$), we easily see that we have further, \begin{align} \|u_{n}\|_{L^{\infty}(\mathbb{R}^{N})}\leq\|u\|_{L^{\infty}(\Omega)}\quad\text{and}\quad\|v_{n}\|_{L^{\infty}(\mathbb{R}^{N})}\leq\|v\|_{L^{\infty}(\Omega)}. \end{align} On the other hand, \begin{align} \int_{\Omega}u_{n}v_{n}\frac{\partial\varphi}{\partial x_{i}}=-\int_{\Omega}\left(\frac{\partial u_{n}}{\partial x_{i}}v_{n}+u_{n}\frac{\partial v_{n}}{\partial x_{i}}\right)\varphi\quad\forall\varphi\in C_{c}^{1}(\Omega). \end{align} Passing to the limit, byt the dominated convergence theorem, this becomes, \begin{align} \int_{\Omega}uv\frac{\partial\varphi}{\partial x_{i}}=-\int_{\Omega}\left(\frac{\partial u}{\partial x_{i}}v+u\frac{\partial v}{\partial x_{i}}\right)\varphi\quad\forall\varphi\in C_{c}^{1}(\Omega). \end{align}

My Questions

1. The inequalities, \begin{align} \|u_{n}\|_{L^{\infty}(\mathbb{R}^{N})}\leq\|u\|_{L^{\infty}(\Omega)}\quad\text{and}\quad\|v_{n}\|_{L^{\infty}(\mathbb{R}^{N})}\leq\|v\|_{L^{\infty}(\Omega)}, \end{align} are given by Young's inequality for convolutions. My question about these inequalities is: Does Brezis use these inequalities to justify using dominated convergence? It seems that, \begin{align} \|u_{n}\|_{L^{\infty}(\Omega)}\leq\|u_{n}\|_{L^{\infty}(\mathbb{R}^{N})}\leq\|u\|_{L^{\infty}(\Omega)}\implies\|u_{n}-u\|_{L^{\infty}(\Omega)}\rightarrow 0, \end{align} and so $u_{n}$ converges to $u$ uniformly a.e. However we do not know if $\Omega$ has finite measure, do we? So then how can we interchange limit and integral in this case? If this is not the approach that Brezis is taking can someone please explain how we get to the dominated convergence conclusion.

2. Is there a particular reason Brezis explicitly mentions, \begin{align} \|u_{n}\|_{L^{\infty}(\mathbb{R}^{N})}\leq\|u\|_{L^{\infty}(\Omega)}\quad\text{and}\quad\|v_{n}\|_{L^{\infty}(\mathbb{R}^{N})}\leq\|v\|_{L^{\infty}(\Omega)} \end{align} rather immediately stating, \begin{align} \|u_{n}\|_{L^{\infty}(\Omega)}\leq\|u\|_{L^{\infty}(\Omega)}\quad\text{and}\quad\|v_{n}\|_{L^{\infty}(\Omega)}\leq\|v\|_{L^{\infty}(\Omega)}. \end{align}

Answer 1

Let $\varphi \in \mathscr{C}^1_c(\Omega)$.

Then for each $n$, $$\int_{\Omega}{u_nv_n\frac{\partial \varphi}{\partial x_i}}=-\int_{\Omega}{\varphi\left(u_n\frac{\partial v_n}{\partial x_i}+v_n\frac{\partial u_n}{\partial x_i}\right)}.$$

Now, let $K$ be the support of $\varphi$, there are constants $U,V > 0$ such that for all $n$, $|u_n| \leq U$ and $|v_n| \leq V$ on $K$. Then the integrand in the first integral converges pointwise to $uv\frac{\partial \varphi}{\partial x_i}$, and is dominated by $UV\left|\frac{\partial \varphi}{\partial x_i}\right|$, which is $L^1(\Omega)$, so the integral converges to $$\int_{\Omega}{uv\frac{\partial \varphi}{\partial x_i}}.$$

For the second integral, let $K \subset \omega \subset \subset \Omega$.

Then $\alpha_n=\frac{\partial u_n}{\partial x_i}$ converges to $\alpha=\frac{\partial u}{\partial x_i}$ in $L^p(\omega)$ thus in $L^1(\omega)$, and $\beta_n=v_n\varphi$ converges pointwise to $\beta=v\varphi$ and is dominated by $V\varphi \in L^{\infty}(\omega)$.

Thus $(\alpha-\alpha_n)\beta_n$ converges to $0$ in $L^1(\omega)$, while $\alpha(\beta_n-\beta)$ is dominated by $2V\varphi|\alpha| \in L^1(\omega)$ and converges pointwise to $0$. Hence $\int_{\omega}{\alpha_n\beta_n} \rightarrow \int_{\omega}{\alpha\beta}$, and we can replace $\omega$ with $\Omega$.

The convergence for the second integral follows.