$W_0^{1,p}(\Omega)\neq W^{1,p}(\Omega)$

Question

If $\Omega=B^n(0,1)\subseteq \mathbb{R}^n$ proof that $W_0^{1,p}(\Omega)\neq W^{1,p}(\Omega)$ where for all $p\in[1,\infty)$

By definition we have that $W_0^{1,p}(\Omega)\subseteq W^{1,p}(\Omega)$ so is enough with show a function $u:\Omega\to\mathbb{R}$ such that $u\in W^{1,p}(\Omega)$ and $u\notin W_0^{1,p}(\Omega)$ but I don't know what function could work or how to find someone.

Answer 1

Since $\partial \Omega$ is smooth, $u \in W^{1,p}_0(\Omega)$ if and only if $u \in W^{1,p}(\Omega)$ and $Tu=0$ on $\partial \Omega$ where $T$ is the trace operator. If $u \in W^{1,p}(\Omega) \cap C(\overline{\Omega})$ then $Tu= u \vert_{\partial \Omega}$. Thus, any function $u \in W^{1,p}(\Omega) \cap C(\overline{\Omega})$ that is not identically zero on $\partial \Omega$ is in $W^{1,p}(\Omega)$ but $W^{1,p}_0(\Omega)$.

For example, take $u\equiv 1$ in $\Omega$.