# Exercise about the trace of sobolev functions

Let $$Q_1$$ and $$Q_2$$ be two open squares in $$\mathbb{R}^2$$ whose closures have an edge - say L - in common. Let be $$u_i\in W^{1,p}(Q_i)$$ for $$i=1,2$$ and for such $$p\in[1,+\infty]$$.

Suppose that $$Tr(u_1)=Tr(u_2)$$ on $$L$$

Given $$\Omega:=Q_1\cup Q_2\cup L$$, show that the function $$u:\Omega\to\mathbb{R}$$ which coincides with $$u_i$$ on $$Q_i$$ belongs to $$W^{1,p}(\Omega)$$.

My book gives the following theorem about Tr (without proof):

Let $$\Omega\subset\mathbb{R}^N$$ be a bounded open set with $$C^1$$ boundary and $$p\in[1,+\infty]$$.

There exists a linear and continuous operator $$Tr:W^{1,p}(\Omega)\to L^p(\delta\Omega,\lambda^{N-1})$$ such that:

1. $$Tr(u)=u_{|\delta\Omega}$$ if $$u\in C^0(\overline\Omega)$$

2. Exists $$C=C(\Omega,p)$$ such that $$||Tr(u)||_{L^p(\delta\Omega)}\le C||u||_{W^{1,p}(\Omega)} \quad \forall u\in W^{1,p}(\Omega)$$

So, I have two questions:

The subsets of $$\mathbb{R}^2$$ in the exercise don't have $$C^1$$ boundary. So I assume the theorem above holds even if $$\delta\Omega$$ is piecewise $$C^1$$.

Anyway I don't see the point of the exercise since the definition of $$u$$ on $$L$$ does not affect the $$W^{1,p}$$ norm value of $$u$$ due to the fact that $$\lambda^2(L)=0$$. What is wrong with this argument?

• The problem is to show that $u$ has weak derivatives on the combined domain. – daw Feb 5 '19 at 15:34
• Thanks daw, your hint helped me :) – framago Feb 5 '19 at 17:51

My last argument was wrong, it is true that the definition of $$u$$ on $$L$$ does not affect its $$L^p(\Omega)$$ norm value though. The point, as suggested in the comment by daw, is to show that $$u$$ has weak derivative in $$L^p(\Omega)$$.
Pick $$\phi\in C^\infty_c\left(\Omega;\mathbb{R}^2\right)$$
$$\int_\Omega\nabla u \cdot\phi\,dx=-\sum_i\int_{Q_i} u_i \,div(\phi)\,dx=\sum_i\left[-\int_{\delta Q_i} Tr(u_i\phi)\cdot\nu_i\,dS+\int_{Q_i}\nabla u_i\cdot\phi\,dx\right]=\sum_i\left[-\int_{\delta Q_i} Tr(u_i)\,\phi\cdot\nu_i\,dS+\int_{Q_i}\nabla u_i\cdot\phi\,dx\right]=\sum_i \left[-\int_{\delta Q_i\setminus L} Tr(u_i)\,\phi\cdot\nu_i\,dS+\int_{Q_i}\nabla u_i\cdot\phi\,dx\right]=\\ 0+\sum_i\int_{Q_i}\nabla u_i\cdot\phi\,dx=\int_\Omega \left(\chi_{Q_1}\nabla u_1+\chi_{Q_2}\nabla u_2\right)\cdot\phi\,dx$$
The fact that $$Tr(u_i\phi)=Tr(u_i)\,\phi$$ follows by the extension theorem, the density of $$C^\infty_c\left(\mathbb{R}^2\right)$$ in $$W^{1,p}\left(\mathbb{R}^2\right)$$, the regularity of $$\phi$$ and the continuity of $$Tr$$.
The integral on the boundary is $$0$$ because $$\nu_1+\nu_2=0$$ on $$L$$ and because $$\phi=0$$ on $$\delta Q_i\setminus L$$.
So we find out that $$\nabla u=\chi_{Q_1}\nabla u_1+\chi_{Q_2}\nabla u_2$$ which is in $$L^p\left(\Omega;\mathbb{R}^2\right)$$.