Conditions on the domain for Sobolev embeddings I am reading the proof of the Sobolev embedding theorem presented in the book Sobolev Spaces by Robert A. Adams and John J. F. Fournier. I could not understand the proof for part II of the theorem. The purpose of the proof is to embed $W^{1,p}(\Omega)\hookrightarrow C^{0,\lambda}(\Omega)$ where $n<p\leq\infty$, $0<\lambda\leq1-n/p$, and $\Omega\subset\mathbb{R}^n$ is a connected open subset that satisfies the strong local Lipschitz condition (page 83, paragraph 4.9):

In the actual proof, the authors reduced it to the following lemma:

They proved the case in which $\Omega$ is a unit cube, and then proceeds thus:


My Question
I'm at a total loss. Where is the strong Lipschitz condition used? Why can we take such $P$ and $\delta_0,\delta_1$?
 A: Pick $n$ linearly independent vectors $v_1,\dots,v_n\in\mathbb R^n$ with $v_{i,n}\leq -M|(v_{i,1},\dots,v_{i,n-1})|.$ For example, $v_i=e_i-Me_n$ for $1\leq i\leq n-1$ and $v_n=-e_n.$ Let $P_0$ be the set of points of the form $\sum t_iv_i$ where $t\in[0,1]^n,$ and take $P$ to be $\delta\operatorname{diam}(P_0)^{-1}P_0.$
$P$ has diameter $\delta$ by construction.
Now consider an arbitrary $j.$ Let $\zeta_j$ and $f_j$ be the corresponding Cartesian co-ordinates and Lipschitz function given by the strong local Lipschitz condition. Let $P_j$ be the set of points $p$ with co-ordinates $(\zeta_{j,1}(p),\dots,\zeta_{j,n}(p))\in P$; this is just applying a rotation to $P.$ I'll drop the explicit use of $\zeta$ from now on, so $P=P_j.$
If $x\in V_j\cap\Omega$ then $x+P_j\subset \Omega$:


*

*condition (ii) guarantees that $x$ is at distance greater than $\delta$ from the boundary of $U_j$

*condition (iv) says that for $y\in U_j$ we have $y\in \Omega$ iff $y_n<f_j(y_1,\dots,y_{n-1})$

*in particular, $x_n<f_j(x_1,\dots,x_{n-1})$

*by the Lipschitz property of $f_j$ we get $x_n+p_n<f_j(x_1+p_1,\dots,x_{n-1}+p_{n-1})$ for all $p$ with $p_n\leq-M|(p_1,\dots,p_{n-1})|,$ so in particular for $p\in P$

*hence $x+p\in\Omega$ for $p\in P.$
To get $\delta_0$ and $\delta_1,$ one approach is to observe that $P$ is a linear transformation of the unit cube: $P=L[0,1]^n$ for some invertible linear map $L.$ If $|X-Y|<1$ then the vector $Z$ defined by $Z_i=\max(X_i,Y_i)$ satisfies $Z\in (X+[0,1]^n)\cap (Y+[0,1]^n)$ with $|X-Z|,|Y-Z|\leq|X-Y|,$ so $|X-Z|+|Y-Z|\leq 2|X-Y|.$
Applying $L$ distorts lengths by at most a constant factor: there exists $C$ such that $C^{-1}|X|\leq |LX|\leq C|X|$ for all $X\in\mathbb R^n.$ Take $\delta_0=\min(C^{-1},\delta)$ and $\delta_1=2C^2.$ Given $x=LX$ and $y=LY$ with $|LX-LY|<C^{-1},$ we get $|X-Y|<1,$ so there exists some $z=LZ$ such that $|x-z|+|y-z|\leq C|X-Z|+C|Y-Z|\leq 2C|X-Y|\leq 2C^2|x-y|.$
I don't think condition (i) is used.
