37.2 An Extension Theorem
Definition 37.2.1 An open subset, U, of ℝn has a Lipschitz boundary if it satisfies the
following conditions. For each p ∈ ∂U ≡U∖U, there exists an open set, Q, containing p,
an open interval
, a bounded open box B ⊆ ℝn−1, and an orthogonal transformation
R such that
where g is Lipschitz continuous on B,a < min
Letting W = Q ∩ U the following picture describes the situation.
The following lemma is important.
Lemma 37.2.2 If U is an open subset of ℝn which has a Lipschitz boundary, then
it satisfies the segment condition and so Xm,p
Proof: For x ∈ ∂U, simply look at a single open set, Qx described in the above
which contains x. Then consider an open set whose intersection with U is of the form
and a vector of the form
chosen smaller than min
There is nothing to prove for points of
One way to extend many of the above theorems to more general open sets than ℝn is
through the use of an appropriate extension theorem. In this section, a fairly general one
will be presented.
Lemma 37.2.3 Let B ×
be as described in Definition 37.2.1 and let
for g a Lipschitz function of the sort described in this definition. Suppose u+ and u−
are Lipschitz functions defined on V + and V − respectively and suppose that
∈ B. Let
and suppose spt
⊆ B ×
. Then extending u to be 0 off of B ×
, u is
continuous and the weak partial derivatives, u,i, are all in L∞
1 and u,i
,i on V + and u,i
,i on V −.
Proof: Consider the following picture which is descriptive of the situation.
Note first that u is Lipschitz continuous. To see this, consider
There are various cases to consider depending on whether
is above g
Suppose yn1 < g
yn2 > g
The other cases are similar. Thus u
is a Lipschitz continuous function which has compact
support. By Corollary 34.5.4
on Page 4316
it follows that u,i ∈ L∞
It remains to verify u,i
on V +
on V −
. The last claim is
obvious from the definition of weak derivatives.
Lemma 37.2.4 In the situation of Lemma 37.2.3 let
u ∈ C1
Then w ∈ W1,p
and there exists a constant, C depending only on Lip
dimension such that
Denote w by E0u. Thus E0
for all y ∈ V − but E0u
= w is defined on all
of ℝn. Also, E0 is a linear mapping.
Proof: As in the previous lemma, w is Lipschitz continuous and has compact
support so it is clear w ∈ W1,p
The main task is to find w,i
yn > g
and then to extract an estimate of the right sort. Denote by
set of points of ℝn
with the property that
if and only if
and yn > g
Then letting ϕ ∈ Cc∞
suppose first that i < n.
where ein−1 is the unit vector in ℝn−1 having all zeros except for a 1 in the ith position.
Now by Rademacher’s theorem, Dg
exists for a.e.
and so except for a set of measure
zero, the expression,
and also for
not in the
Therefore, since the integrand in 37.2.28 has compact support and because of
the Lipschitz continuity of all the functions, the dominated convergence theorem may be
applied to obtain
whenever i < n which is what you would expect from a formal application of the chain
rule. Next suppose i = n.
which is also expected.
From the definnition, for y ∈ ℝn ∖U ≡
is given by 37.2.29
for i < n.