What is the definition of $W^{1,\infty}_0(\Omega)$?

Question

I can't find a definition of the space $W_0^{1,\infty}(\Omega)$, where $\Omega\subset\mathbb{R}^n$ open bounded of class $C^1$.

Should I take $C_c^{\infty}(\Omega)$ and consider the closure in $W^{1,\infty}$ norm, as in the case $p\in[1, \infty)$?

What is the definition of this space?

Thank you!

Maybe relevant: http://math.stackexchange.com/questions/874982/relation-between-sobolev-space-w1-infty-and-the-lipschitz-class. — Martín-Blas Pérez Pinilla, Sep 07 '16 at 07:48

score 3 · Accepted Answer · answered Sep 07 '16 at 08:49

The space $W^{1,\infty}(\Omega)$ is usually defined as all functions $v \in L^\infty(\Omega)$ with weak derivatives of first order and these derivatives shall belong to $L^\infty(\Omega)$.

Then, you can define $W_0^{1,\infty}(\Omega) := H_0^1(\Omega) \cap W^{1,\infty}(\Omega)$, i.e., all functions in $W^{1,\infty}(\Omega)$ whose trace is zero.

Note that $C_c^\infty(\Omega)$ is not dense in $W^{1,\infty}(\Omega)$. In fact, if $\{v_n\}\subset C_c^\infty(\Omega)$ converges towards $v$ in $W^{1,\infty}(\Omega)$, the (continuous) derivatives of $\{v_n\}$ converge uniformly towards the derivatives of $v$. Hence, $v$ has continuous derivatives. But not all functions in $W^{1,\infty}(\Omega)$ possess continuous derivatives.

What is the definition of $W^{1,\infty}_0(\Omega)$?

1 Answers1