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 $\delta_0,\delta_1$?


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.

  • $\begingroup$ Sorry I didn't know that I can't award the bounty after it has expired... Thanks very much for your detailed answer! It's unbelievable that the author just states "there exist ..." without any comment as to the reason... $\endgroup$ – Colescu Apr 14 '19 at 7:39

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