2. MAIN RESULT 7
Then a straightforward calculation gives for any ρ 0:
(2.10) W Opα(b)W
a −1
= Opβ(b,ρ),
a
b,ρ(x, y, ξ) = b( x, y,ρξ), β = αρ.
In particular,
(2.11) W χΛW
−1
= χΛ , Λ =
−1Λ,
It is important that the norm (2.5) is invariant under certain linear transformations.
Note first of all that
(2.12)
N(n1,n2,m)(p;
, ρ) =
N(n1,n2,m)(p
1,ρ1
;
−1,ρρ1
1
−1),
for arbitrary positive ,
1
, ρ, ρ1. Moreover, the norm is also invariant under Eu-
clidean isometries:
(2.13)
N(n1,n2,m)(p;
, ρ) =
N(n1,n2,m)(pO,k,k1
; , ρ).
Sometimes we refer to and ρ as scaling parameters.
Now we can specify the classes of domains which we study. We always assume
that Λ and Ω are domains with smooth boundaries in the standard sense. However,
for the reference convenience and to specify the precise conditions on the objects
involved, we state our assumptions explicitly.
Definition 2.1. We say that a domain Γ
Rd,
d 2, is a
Cm-graph-type
domain, with some m 1, if one can find a real-valued function Φ
Cm(Rd−1),
with the properties
(2.14)

⎪Φ(ˆ)



0 = 0,
∇Φ is uniformly bounded on
Rd−1,
∇Φ is uniformly continuous on
Rd−1,
and some transformation E = (O, k) E(d) such that
E−1Γ
= {x : xd Φ(ˆ)}, x ˆ x = (x1,x2,...,xd−1).
In this case we write Γ = Γ(Φ; O, k) or Γ = Γ(Φ), if the omission of the dependence
on E does not lead to confusion.
We often use the notation
(2.15) = ∇Φ
L∞
.
It is clear that Γ(Φ; O, k) = EΓ(Φ; I, 0) with E = (O, k). The point k is on the
boundary of the domain Γ(Φ; O, k).
If Λ = Γ(Φ; O, k), then the domain Λ (see definition (2.11)) has the form
(2.16) Λ = Γ(Φ ; O,
−1k),
Φ (ˆ) x =
−1Φ(
ˆ). x
Note that the value (2.15) is invariant under scaling:
(2.17) ∇Φ
L∞
= ∇Φ
L∞
.
In what follows we extensively use the relations (2.6), (2.7) and (2.8) in order to
reduce the domains or symbols to a more convenient form. For these purposes, let
us make a note of the following elementary property for the domain Λ = Γ(Φ; I, 0).
According to (2.7), for any k ∂Λ we have
(2.18) ΛI,k = Γ(Φk; I, 0), where Λ = Γ(Φ; I, 0), Φk(ˆ) x = Φ(ˆ x +
ˆ)
k kd.
Clearly,
Φk(ˆ)
0 = 0 and ∇Φ
L∞
= ∇Φk
L∞
.
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