# How can we extend $\operatorname{div}:C_c^\infty(\Omega)^d\to L^p(\Omega)$ to $W_0^{1,\:p}(\Omega)^d$?

Let

• $d\in\mathbb N$ and $p\ge 1$
• $\lambda$ be the Lebesgue measure on $\mathbb R^d$
• $\Omega\subseteq\mathbb R^d$ be open and $$W_0^{1,\:p}(\Omega):=\overline{C_c^\infty(\Omega)}^{\left\|\;\cdot\;\right\|_{W^{1,\:p}(\Omega)}}$$ with $$\left\|\phi\right\|_{W^{1,\:p}(\Omega)}^p:=\left\|\phi\right\|_{L^p(\Omega)}^p+\left\|\nabla\phi\right\|_{L^p(\Omega,\:\mathbb R^d)}^p\;\;\;\text{for }\phi\in C_c^\infty(\Omega)$$

I want to extend the linear operator $$\operatorname{div}:C_c^\infty(\Omega)^d\to L^p(\Omega)\;,\;\;\;\phi\mapsto\nabla\cdot\phi$$ to $W_0^{1,\:p}(\Omega)^d$.

I guess this is a very simple task, but I'm not very familiar with operator theory and Sobolev spaces.

Let $(\phi_n)_{n\in\mathbb N}\subseteq C_c^\infty(\Omega)$ be a $\left\|\;\cdot\;\right\|_{W^{1,\:p}(\Omega)}$-cauchy sequence $\Rightarrow$

• $\exists u\in L^p(\Omega)$ with $$\left\|\phi_n-u\right\|_{L^p(\Omega)}\stackrel{n\to\infty}\to 0$$ (since $L^p(\Omega)$ is complete)
• $\exists v\in L^p(\Omega,\mathbb R^d)$ with $$\left\|\nabla\phi_n-v\right\|_{L^p(\Omega,\:\mathbb R^d)}\stackrel{n\to\infty}\to 0$$ (since $L^p(\Omega,\mathbb R^d)$ is complete)

Thus, by definition of (weak) derivativeness, $$-\int_\Omega u\frac{\partial\phi}{\partial x_i}\;{\rm d}\lambda=-\lim_{n\to\infty}\int_\Omega u_n\frac{\partial\phi}{\partial x_i}\;{\rm d}\lambda=\lim_{n\to\infty}\int_\Omega\phi\frac{\partial\phi_n}{\partial x_i}\;{\rm d}\lambda=\int_\Omega v_i\phi\;{\rm d}\lambda$$ for all $i\in\left\{1,\ldots,d\right\}$ and $\phi\in C_c^\infty(\Omega)$, i.e. $$v=\nabla u\;\;\;\lambda\text{-almost everywhere}\tag 1$$

$(1)$ is the reason why we can extend $$\nabla:C_0^\infty(\Omega)\to L^p(\Omega,\mathbb R^d)$$ to $W_0^{1,\:p}(\Omega)$ via $$\nabla u:=v\;.$$

How can we do the same for $\operatorname{div}$?

• By what you wrote you can extend $C_c\infty(\Omega)^d \to L^p(\Omega,\mathbb R^d)^d =L^p(\Omega)^{d\times d}$, $(f_1,\ldots,f_d)\mapsto (\nabla f_1,\ldots, \nabla f_d)$ and then you can compose with the linear continuous map $L^p(\Omega)^{d\times d} \to L^p(\Omega)$, $(g_{i,j})_{i,j} \mapsto \sum_i g_{i,i}$. – Jochen Jun 27 '16 at 10:14
• @Jochen Could you explain this in more detail? Isn't there any more direct way to extend $\operatorname{div}$? – 0xbadf00d Jun 28 '16 at 11:34