0

I was reading a lemma from IMPLICIT PARTIAL DIFFERENTIAL EQUATION by Bernard Dacorogna and I am not getting the meaning of this notation:

$W^{1, \infty}(\Omega, \mathbb R^m)$ where $\Omega \subseteq \mathbb R^n, open.$

Mini_me
  • 2,235
  • 3
  • 15
  • 39
  • I don't know this book in particular, but I know another book which says that if $u \in W^{1,\infty}(\Omega,\mathbb{R}^m)$ then $u_i \in W^{1,\infty}(\Omega)$ for all $i=1,...,m$. Sometimes one also sees the notation $(W^{1,\infty}(\Omega))^m$. – Cahn Aug 21 '17 at 20:53

2 Answers2

1

Begin by taking $\alpha$ to be some multi-index with $\lvert \alpha \rvert \leq 1$. Then the space $W^{1,\infty}(\Omega,\mathbb{R}^m)$ is defined to be the space of functions $$ W^{1,\infty}(\Omega,\mathbb{R}^m) := \left\lbrace f \in L^{\infty}(\Omega) \; | \; D^{\alpha}f \in L^{\infty}(\Omega)\right\rbrace $$ where $D^{\alpha}f$ is only asserted to exist in the weak sense. More generally, if $k\in \mathbb{N}$ and if $p \in [1,\infty]$ then for all multi-indexes $\alpha$ with $\lvert \alpha \rvert \leq k$, $$ W^{k,p}(\Omega,\mathbb{R}^m) := \lbrace f \in L^{p}(\Omega) \; | \; D^{\alpha}f \in L^{p}(\Omega) \rbrace $$ where $D^{\alpha}f$ again a weak derivative of $f$.

Geoff
  • 1,697
  • 1
  • 13
  • 20
1

The space $W^{1,\infty}(\Omega)$ consists of functions whose first-order weak derivatives exist and are essentially bounded.

It is closely related to the space $\operatorname{Lip}(\Omega)$ of Lipschitz-continuous functions in $\Omega$. However, they are in general not the same. Indeed, $$\operatorname{Lip}(\Omega) \subset W^{1,\infty}(\Omega)$$ holds in general, because Lipschitz continuity implies absolute continuity on lines, hence the existence and boundedness of first-order weak derivatives. The reverse inclusion holds in some domains but not others: see Relation between Sobolev Space $W^{1,\infty}$ and the Lipschitz class.