MAPPINGS OF FINITE DISTORTION:
ZORIC’S THEOREM,
AND EQUICONTINUITY RESULTS
MIHAI CRISTEA
We generalize some known results from the theory of quasiregular mappings, as
Zoric’s theorem and some equicontinuity results. These generalizations hold for
a special class of mappings with finite distortion satisfying condition (A), which
extends the known class of quasiregular mappings, for which some recent modular
and capacity inequalities established in [9] are valid.
AMS 2000 Subject Classification: 30C65.
Key words: Zoric’s theorem, Montel’s theorem, mapping of finite distortion.
1. INTRODUCTION
A mapping f : D → Rn , where D ⊂ Rn is a domain, is said to have
finite distortion if the following conditions are satisfied:
1,1
(D, Rn );
1) f ∈ Wloc
2) the Jacobian determinant Jf (x) is locally integrable;
3) there exists a finite a.e. measurable function K : D → [0, ∞] such that
|f (x)|n ≤ K(x) · Jf (x) a.e.
Notice that when K ∈ L∞ (D) we obtain the known class of quasiregular
mappings, and we refer the reader to [15], [14] for monographs devoted to this
subject. Quasiregular mappings are either constant or open, discrete, satisfy
condition (N) and a Hölder condition.
1,n
(D, Rn ),
If the dilatation map K ∈ Lploc (D), p > n − 1, and f ∈ Wloc
then it is shown in [11] that a map with finite distortion is open, discrete.
Recently, there were considered in [11], [9], [10], [13] mappings f : D →
n
R of finite distortion for which there exists some smooth, strictly increasing A : [0, ∞) → [0, ∞) with A(0) = 0, lim A(t) = ∞ and satisfying the
t→∞
conditions:
(A0 ) exp(A(K)) ∈ L1loc (D);
∞ (A1 ) 1 A t(t) dt = ∞;
REV. ROUMAINE MATH. PURES APPL., 52 (2007), 5, 539–554
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Mihai Cristea
2
(A2 ) there exists t0 > 0 such that A (t) · t increases to infinity for t ≥ t0 .
As in [9], we say that a mapping of finite distortion satisfies condition (A)
if f satisfies condition (A0 ), (A1 ), (A2 ). In [7] is shown that such nonconstant
mappings are either constant or open, discrete. Our proofs are based on the
following basic ingredients, valid for mappings f : D → Rn of finite distortion
and satisfying condition (A):
(i) M (f (Γ)) ≤ MK n−1 (Γ) for every path family Γ from D;
(ii) cap f (E) ≤ capK n−1 (E), where E = (G, C) is a capacitor with
G ⊂⊂ D;
(iii) capK n−1 (B(x, R), B(x, r)) → 0 as r → 0 and R > 0 is kept fixed.
They were established in [9], Corollary 4.2, Corollary 5.2 and Theorem 5.3.
Here, K is the dilatation map of f . We also give in Lemma 1 a condition
able to ensure that capK n−1 (B(x, R), B(x, r)) → 0 as r > 0 is kept fixed
and R → ∞. This last condition is necessary in the proof of Zoric’s theorem
(Theorems 2 and 1).
We generalize Zoric’s theorem for mappings with finite distortion which
satisfy condition (A) as follows.
Theorem 2 (Zoric’s theorem). Let n ≥ 3. Let f : Rn → Rn be a local
homeomorphism and a mapping of finite distortion
satisfying condition (A)
exp(A(K(x))) · |x|12n dx < ∞.
for which there exists r > 0 such that
B(0,r)
Then f is a global homeomorphism.
We give a Picard type theorem in Theorem 4, in connection with a similar
result from [9].
Theorem 4 (Picard’s theorem). Let F ⊂ Rn be closed, f : Rn \F →
that
Rn a nonconstant map of finite distortion, satifying
condition (A), such
exp(A(K(x)))· |x|12n dx <
MK n−1 (F ) = 0 and there exists r > 0 such that
∞. Then cap f (Rn \F ) = 0.
B(0,r)
We also give in Theorem 1 a version of Zoric’s theorem with some “singular” sets K and B, extending some results from [1] and [2]. Our variants
of Zoric’s theorem may be put in connection with some recent result from
[6] and [11]. However, our results ensure that the map f : Rn → Rn is a
homeomorphism, hence we give a complete generalization of Zoric’s theorem.
Indeed, in [6] is proved that if f : Rn → Rn is a local homeomprhism and a
mapping of finite distortion of K(x) dilatation satisfying condition (A), and
MK n−1 (∞) = 0, then there exists E ⊂ Rn with cap E = 0 such that f : Rn →
exp(A(K(x)))· |x|12n dx < ∞
Rn \E is a homeomorphism. Our condition
B(0,r )
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Mappings of finite distortion: Zoric’s theorem and equicontinuity results
541
implies that MK n−1 (∞) = 0, hence our Theorem 2 generalizes the so called
Global Homeomorphism Theorem from [6]. Also, in [11] is proved that if
f : B(0, r) → Rn is a local homeomorphism of finite distortion of K(x) dilatation satisfying conditon (A) with exp(A ◦ K) ∈ L1 (B(0, r)), then there exists
δ(n, K, r) such that f is injective on B(0, δ). The result from [11] is not enough
for an arbitrary local homeomorphism f : Rn → Rn of finite distortion of K(x)
dilatation and satisfying condition (A) to be a global homeomorphism, as in
the classical case when K(x) ≤ K on Rn , i.e., we cannot prove Zoric’s theorem for mappings of finite distortion
from [8] shows
using [11]. An example
exp(A(K(x))) · |x|12n dx < ∞ used in
that the fundamental condition
B(0,r )
Theorems 2 and 4 is necessary in order to ensure that f : Rn → Rn is a global
homeomorphism.
We generalize in Theorem 5 a known equicontinuity result for quasiregular mappings as follows.
Theorem 5. Let D ⊂ Rn be a domain, M ⊂ Rn with cap M > 0, W
a family of mappings f : D → Rn \M of finite distortion satisfying condition
(A) and having the same dilatation map K. Then W is equicontinuous if we
consider the Euclidean distance on D and the chordal distance on Rn .
We see that even in the case where all the mappings from the family W
are K-quasiregular, we improve the classical result established for quasiregular
mappings in [15], Corollary 2.7, page 66. Indeed, the exceptional set M which
is avoided by all the maps f ∈ W can be to be at most countable and chosen
such that cap M > 0, hence of zero capacity.
We immediately obtain a Montel type theorem for mappings of finite
distortion satisfying condition (A), extending in this way Theorem 8.14.1, [4],
established in the particular case A(t) = λt.
Theorem 6 (Montel’s theorem). Let D ⊂ Rn be a domain. Let W be
a bounded family of mappings f : D → Rn of finite distortion with the same
dilatation map K and satisfying condition (A). Then W is a normal family.
2. PRELIMINARIES
We call open a path q : [0, 1) → Rn . A point x ∈ Rn will be called
a limit point of q if there exists tp 1 such that q(tp ) → x. If Γ is a
n
n
path
family in R , we define F (Γ) = {ρ : R → [0, ∞] isn a Borel map
| ρds ≥ 1 for every locally rectifiable γ ∈ Γ}. Let now D ⊂ R be open and
γ
ω : D → [0, ∞] measurable and finite a.e. If ω
: Rn → [0, ∞] is defined by
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Mihai Cristea
4
ω
(x) = ω(x) if x ∈ D,
(x) = 0 if x ∈
/ D, then we define the ω-modulus of Γ
ω
ρn (x) · ω
(x)dx. For ω ≡ 1 we obtain the usual modulus
by Mω (Γ) = inf
ρ∈F (Γ) Rn
M (Γ). If Γ1 , Γ2 are path families in Rn , we say that Γ1 > Γ2 if every path from
Γ1 has a subpath in Γ2 . As in the classical case, we can prove that if Γ1 > Γ2 ,
∞ ∞
Mω (Γi ),
then Mω (Γ1 ) ≤ Mω (Γ2 ). Also, we can prove that Mω Γi ≤
i=1
i=1
and if ω1 ≤ ω2 , then Mω1 (Γ) ≤ Mω2 (Γ). For D ⊂ Rn open, E, F ⊂ D, we
denote by ∆(E, F, D) the family of all paths, open or not, which join E with
F in D.
We say that E = (D, C) is a condenser if C is compact, D is open,
C ⊂ D ⊂ Rn . Let ω : D → [0, ∞] be
and finite a.e. We define
measurable
|∇u|n (x) · ω(x)dx, where u ∈ C0∞ (D)
the ω-capacity of E by capω E = inf
Rn
and u ≥ 1 on C. If ω = 1 we obtain the usual capacity cap E of E. We see
that if u is a test function for capω E, then ρ = |∇u| ∈ F (ΓE ). This implies
that Mω (ΓE ) ≤ capω (E). Here, ΓE = {γ : [a, b] → D is a path | γ(a) ∈ C
and γ has a limit point in ∂D} and from Proposition 10.2 in [15, page 54],
we have cap E = M (ΓE ). If C ⊂ Rn is compact, we say that cap C = 0 if
cap(A, C) = 0 for every C ⊂ A ⊂ Rn open. By [15, Lemma 2.2, page 64], the
definition is independent on the open set A such that C ⊂ A. If C ⊂ Rn is
arbitrary, we say that cap C = 0 if cap K = 0 for every compact K ⊂ C.
Let now D ⊂ Rn be open, ω : D → [0, ∞] measurable and finite a.e.,
A ⊂ D a set. We say that A is of zero ω-modulus, and we write Mω (A) = 0,
if the ω-modulus of all paths having some limit point in A is zero. If ω ≥ 1,
we see that M (Γ) ≤ Mω (Γ). Hence, if Mω (A) = 0 then cap A = 0.
If A ⊂ D is countable and lim Mω (∆(B(x, r), B(x, R), D)) = 0 for evr→0
ery x ∈ A and every 0 < R with B(x, R) ⊂ D, we can prove as in the
classical case that Mω (A) = 0. Using Theorem 5.3 in [9], we see that this
holds, for instance, if ω = K n−1 , where K : D → [0, ∞] is measurable and
finite a.e.
and for every x ∈ A there exists ρx > 0 such that B(x, ρx ) ⊂ D
exp(A(K(y)))dy < ∞, where A satisfies conditions (A1 ) and (A2 ).
and
B(x,ρx )
Lemma 1 also gives a condition which ensures that MK n−1 (∞) = 0.
If D ⊂ Rn is open and b ∈ ∂D, we define C(f, b) = {w ∈ Rn | there exists
bp ∈ D, bp → b such that f (bp ) → w}. If B ⊂ ∂D, we let C(f, B) = C(f, b).
b∈B
If E, F are Hausdorff spaces, f : E → F is a map, p : [0, 1] → F is a path,
x ∈ E is such that f (x) = p(0), we say that q : [0, a) → E is a maximal lifting
of p from x when q(0) = x, 0 < a ≤ 1, f ◦ q = p | [0, a) and q is maximal with
this property. If q is defined on [0, 1], we say that q is a lifting of p. If E, F are
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Mappings of finite distortion: Zoric’s theorem and equicontinuity results
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domains in Rn and f is continuous, open, discrete, then there always exists
a maximal lifting. We denote by µn the Lebesgue measure in Rn and by q
1
1
the chordal metric in Rn given by q(a, b) = |a − b| · (1 + |a|2 )− 2 · (1 + |b|2 )− 2
1
if a, b ∈ Rn , q(a, ∞) = (1 + |a|2 )− 2 if a ∈ Rn , and by |a − b| the Euclidean
distance between a and b in Rn . We denote by Bq (x, r), respectively B(x, r),
the ball of center x and radius r if we consider on Rn the chordal metric,
respectively the Euclidean metric. In the same way, for a set A ⊂ Rn we denote
1,p
(D, Rn ) the
by q(A), respectively d(A), the diameter of A. We denote by Wloc
Sobolev space of all functions f : D → Rn which are locally in Lp (D), together
with their first order weak partial derivatives.
If W is a family of mappings f : D → Rn , we say that W is bounded if
for every compact K ⊂ D there exists M (K) > 0 such that |f (x)| ≤ M (K)
for every x ∈ K and every f ∈ W. If X, Y are metric spaces and W is a family
of mappings f : X → Y, we say that W is equicontinuous at x if for every
ε > 0 there exists δε > 0 such that d(f (x), f (y)) ≤ ε when d(y, x) ≤ δε for
every f ∈ W . We say that W is equicontinuous if it is equicontinuous at every
point x ∈ X. We say that W is a normal family if every sequence in W has
a subsequence which converges uniformly on the compact subsets of X to a
map f : X → Y.
Lemma 1. Let n ≥ 2. Let D ⊂ Rn be an unbounded domain, K :
D → [0, ∞] measurable and finite a.e., A : [0, ∞) → [0, ∞) smooth, strictly
increasing, with A(0) = 0, lim A(t) = ∞, satisfying condition (A1 ) and (A2 )
t→∞
exp(A(K(x))) · |x|12n dx < ∞.
for which there exists r > 0 such that
D∩B(0,r)
Then MK n−1 (∆(B(0, r) ∩ D, B(0, s) ∩ D, D) → 0 as s → ∞ and r > 1 is kept
fixed.
Proof. Let g : Rn → Rn be defined by g(x) =
x
|x|2
if x ∈ Rn and
let Γrs = ∆(B(0, r) ∩ D, B(0, s) ∩ D, D), Λrs = ∆(g(D) ∩ B(0, 1s ), g(D) ∩
B(0, 1r ), g(D)) for 1 < r < s. We show that MK n−1 (Γrs ) = MK n−1 ◦g (Λrs ).
Let ρ ∈ F (Γrs ) and ρ : Rn → [0, ∞] be defined by ρ(x) = ρ (g(x)) ·|g (x)| if
x ∈ D, ρ(x) = 0 if x ∈
/ D. Then ρ ∈ Γ(Λrs ) and MK n−1 ◦g (Λrs ) ≤ ρn (x) ·
Rn
K n−1 (g(x))dx = ρn (g(x))|g (x)|n ·K n−1 (g(x))dx = ρn (g(x))·K n−1 (g(x))·
D
nD
ρ (y) · K n−1 (y)dy ≤ ρn (y) · K n−1 (y)dy. This implies that
Jg (x)dx =
g(D)
Rn
MK n−1 ◦g (g(Γrs )) = MK n−1 ◦g (Λrs ) ≤ MK n−1 (Γrs ). By the same argument, we
have MK n−1 (Γrs ) = MK n−1 ◦g◦g (g(Λrs )) ≤ MK n−1 ◦g (Λrs ) and
(1)
MK n−1 (Γrs ) = MK n−1 ◦g (Λrs ).
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Mihai Cristea
Next,
exp(A(K(g(x))))dx =
g(D)∩B(0, 1r )
exp(A(K(g(x))))·Jg−1 (g(x))·Jg (x)dx =
g(D)∩B(0, 1r )
exp(A(K(y))) · Jg−1 (y)dy =
=
6
D∩B(0,r)
exp(A(K(y))) ·
D∩B(0,r)
1
dy < ∞.
|y|2n
Let now Γrs = ∆(B(0, 1s ), B(0, 1r ), B(0, 1r )\B(0, 1s )) for 1 < r < s and
Q : Rn → [0, ∞] be defined by Q(x) = K(g(x)) if x ∈ g(D), Q(x) = 1 if
exp(A(Q(x)))dx < ∞, by Theorem 5.3, page 24, in [9],
x∈
/ g(D). Since
B(0, 1r )
capQn−1 (B(0, 1r ), B(0, 1s )) → 0 as s → ∞ and r > 0 is kept fixed. We have
MK n−1 ◦g (Λrs ) ≤ MQn−1 (Λrs ) ≤ MQn−1 (Γrs ) ≤ capQn−1 (B(0, 1r ), B(0, 1s )) and
this implies that
MK n−1 ◦g (Λrs ) → 0 as s → ∞ and r > 0 is kept fixed.
(2)
Using (1) and (2), the proof is complete.
Lemma 2. Let D ⊂ Rn be open. Let f : D → Rn be a map of finite
distortion satisfying condition (A) and K : D → [0, ∞] the dilatation map
1,p
(D, Rn ) for every
of f . Then K ∈ Lploc (D) for every p > 0 and f ∈ Wloc
1 ≤ p < n.
Proof. Using property (A2 ) we see that the function exp(A(t)) grows
faster then any tp to infinity and therefore K ∈ Lploc (D) for every 0 < p < ∞.
We use
now Hölder’s inequality and the integrability of the Jacobian to see
that |f (x)|p dx < ∞ for every ball B with B ⊂ D and every 1 ≤ p < n. B
3. PROOFS OF THE MAIN RESULTS
We start with
Theorem 1. Let n ≥ 3. Let B ⊂ Rn be closed, K ⊂ Rn \B closed in
Rn \B with int K = ∅, f : Rn \(K ∪ B) → Rn a map of finite distortion of
K(x) dilatation satisfying condition (A) and a local homeomorphism such that
MK n−1 (B) = 0, int C(f, K) = ∅, C(f, K) closed and Rn \C(f, K) connected.
Let E = K ∪ B ∪ f −1 (C(f, K)). Suppose that exp(A(K)) ∈ L1loc (Rn \K) and
exp(A(K(x))) · |x|12n < ∞.
that there exists δ0 > 0 such that
B(0,δ0 )∩(Rn \(K∪B))
Then
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Mappings of finite distortion: Zoric’s theorem and equicontinuity results
545
1) We can extend f to a local homeomorphism around each point x ∈ B
with possibly f (x) = ∞ for some point x ∈ B. Also, extending if necessary
the map f to a homeomorphism near ∞, we can lift every path: p[0, 1] →
Rn \C(f, K) from every point x ∈ Rn with f (x) = p(0).
2) If C(f, K) is compact, then int E = ∅ and f is injective on Rn \E.
If f can be continuously extended to K, then either f |Rn \K : Rn \K →
Rn\C(f, K) or f |Rn \K : Rn \K → Rn \C(f, K) is a homeomorphism. Finally,
if f can be continuously extended to an open map on K, then f : Rn → Rn
is a homeomorphism.
Proof. By Theorem 1.1 [13], we can take B = ∅. Let Γ = {γ : [0, 1) →
Rn \K path |∞ ∈ Im γ}. By Lemma 1, MK n−1 (Γ) = 0. Using (i), we see that
M (f (Γ)) ≤ MK n−1 (Γ) = 0, hence M (f (Γ)) = 0 and by Theorem 1 in [2], 1) is
proved.
Suppose now that C(f, K) is compact and let r > 0 be such that
C(f, K) ⊂ B(0, r), let 0 < r < s and let x ∈ Rn \K be such that f (x) ∈
B(0, s). Let Q be the component of f −1 (B(0, r)) which contains x and let U
be the component of f −1 (B(0, s)) which contains x. Then U ⊂ Q and since
B(0, r) is simply connected and f |Q : Q → B(0, r) is a local homeomorphism which lifts the paths, f |Q : Q → B(0, r) is a homeomorphism, hence
f |U : U → B(0, s) also is a homeomorphism. Since f |∂U : ∂U → S(0, s)
is a homeomorphism, Jordan’s theorem implies that Rn \∂U has exactly two
components, one of them being U . Since U is homeomorphic to B(0, s), U
is unbounded. If U0 is any other component of f −1 (B(0, s)), by the same
argument, f |U 0 : U 0 → B(0, s) is a homeomorphism, ∂U0 bounds a Jordan
domain, and U0 is the unbounded component of Rn \∂U0 . Let α > 0 be such
that ∂U ∪ ∂U0 ⊂ B(0, α). Then B(0, α) ⊂ U ∩ U0 , hence U = U0 and we
obtain that f −1 (B(0, s)) has a single component U on which f is injective,
K ⊂ U and U = f −1 B(0, s).
Since f is a local homeomorphism on Rn \K and int C(f, K) = ∅, we
have int E = ∅. Let x1 , x2 ∈ Rn \E be such that f (x1 ) = f (x2 ) = y. Let
z ∈ B(0, s) and p : [0, 1] → Rn \C(f, K) a path such that p(0) = y and
p(1) = z. We can lift p from x1 and x2 and find some paths qi : [0, 1] → Rn
such that qi (0) = xi , f ◦ qi = p, i = 1, 2. Since qi (1) ∈ f −1 (B(0, s)) =
U, f (qi (1)) = p(1), i = 1, 2 and we proved that f is injective on U , we obtain
that q1 (1) = q2 (1). We now use the property of the uniqueness of path lifting
for local homeomorphisms to conclude that x1 = x2 . We therefore proved that
f is injective on Rn \E.
Suppose now that f can be continuously extended to K and let a ∈ Rn
be a point such that f is open at a. Let us show that {a} = f −1 (f (a)).
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Mihai Cristea
8
Indeed, if there exists b = a, b ∈ Rn such that f (a) = f (b), let U1 ∈ V(a) and
V ∈ V(f (a)) be such that f (U1 ) = V . Since f is continuous at b, we can find
U2 ∈ V(b) with U1 ∩ U2 = ∅ such that f (U2 ) ⊂ V , and since int E = ∅, we can
find α ∈ U2 \E. Then we can find β ∈ U1 \E with f (α) = f (β). So, we reached
a contradiction, since we proved that f is injective on Rn \E. It results now
immediately that either f |Rn \K : Rn \K → Rn \f (K) or f |Rn \K : Rn \K →
Rn \C(f, K) is a homeomorphism. If we additionally suppose that f also is
an open map on K, we find that f |Rn : Rn → Rn is a homeomorphism.
Proof of Theorem 2. We take K = ∅ in Theorem 1.
The following version of Zoric’s theorem seems to be new even for quasiregular mappings.
Theorem 3. Let n ≥ 3. Let K ⊂ Rn be compact, f : Rn → Rn continuous on Rn and of finite distortion of K(x) dilatation satisfying condition
exists r > 0 such that f is a local homeomor(A) on Rn \K for which there
exp(A(K(x)))· |x|12n dx < ∞. If f is unbounded,
phism on B(0, r ) and
B(0,r )
then there exists δ > 0 such that f is injective on B(0, δ) while if f is open
discrete on K, then f : Rn → Rn is a homeomorphism.
Proof. Suppose first that f is unbounded and let ρ > 0 be such that
f (K) ⊂ B(0, ρ). Let p : [0, 1] → B(0, ρ) be defined by p(t) = (1 − t)f (x) + ty
for t ∈ [0, 1] and some x ∈ Rn \K and y ∈ CB(0, ρ). As in Theorem 1 in [2],
we try to lift p from x. Let 0 < a ≤ 1 and let q : [0, a) → Rn \K be a maximal
lifting of p from x with f ◦ q = p|[0, a). As in our Theorem 1 and Theorem 1
in [2] we can find spheres S(p(a), r) and sets Qr which bound some Jordan
domains Dr such that f |Qr : Qr → S(p(a), r) is a homeomorphism for a.e.
r > 0, and we can suppose that B(0, ρ) ∩ B(p(a), r) = ∅ for a.e. 0 < r ≤ r0 .
As in Theorem 1 in [2], we have two cases.
In the first case, suppose that Dr ⊂ Ds for 0 < s < r ≤ r0 . Let
Wr = (Ds \D r ) and Br = B(p(a), r) ∪ B(0, ρ) for 0 < r ≤ r0 . Then B r
s<r
is connected for 0 < r ≤ r0 and let us fix 0 < s < r ≤ r0 . We show that
f ((Ds \Dr )\K) ⊂ B r . Indeed, otherwise we can find a point α ∈ (Ds \Dr )\K
such that f (α) ∈ CB r . Let y ∈ B r and let Γ : [0, 1] → B r be a path
such that Γ(0) = f (α), Γ(1) = y. If we cannot lift Γ from α, we can find
0 < c ≤ 1 and a maximal lifting γ : [0, c) → Rn \K of Γ from α. Let tp c
be such that γ(tp ) → z. Then z ∈ Ds , hence z = ∞ and z ∈ Qs ∪ Qr ∪ K,
hence
f (z) ∈ f (Qs ∪ Qr ∪ K) ⊂ B r . On the other hand, we have f (z) =
f lim γ(tp ) = lim f (γ(tp )) = lim Γ(tp ) = Γ(c) ∈ B r and we so reach a
p→∞
contradiction.
p→∞
p→∞
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Mappings of finite distortion: Zoric’s theorem and equicontinuity results
547
We see that we can lift Γ from α, hence we can find a path γ : [0, 1] →
(Ds \D r )\K with γ(0) = α and Γ = f ◦ γ, hence y = Γ(1) = f (γ(1)) ∈
f ((Ds \Dr )\K). Since y was arbitrary chosen in B r , we proved that B r ⊂
f ((Ds \Dr )\K) ⊂ f (Ds ), so we reached a contradiction so f (Ds ) is compact
and B r is unbounded.
We therefore proved that f (Ds \Dr ) ⊂ B r for 0 < s < r ≤ r0 . Let
A = lim Wr . Then Wr is connected and f (Wr ) ⊂ B r for 0 < r ≤ r0 ,
r→∞
hence C(f, A) = ∩ f (Wr ) is connected and C(f, A) ⊂ B r for 0 < r ≤ r0 .
r>0
Since p(a) ∈ C(f, A), we have C(f, A) ⊂ B(p(a), r) for 0 < r ≤ r0 , hence
C(f, A) = {p(a)}. This implies that A ∩ K = ∅ and since f is a light map on
Rn \K, we have card A = 1. Then A = {∞} and f maps a neighbourhood of
∞ near p(a) and this contradicts the fact that f is unbounded. We therefore
proved that the case Dr ⊂ Ds for 0 < s < r ≤ r0 cannot hold.
In the second case, we have Ds ⊂ Dr for 0 < s < r ≤ r0 and let
B = ∩ Dr . As before, we see that f (Dr \Ds ) ⊂ B r for 0 < s < r ≤ r0 ,
r>0
that C(f, B) = ∩ f (Dr ) is connected and C(f, B) ⊂ B r for 0 < r ≤ r0 and
r>0
p(a) ∈ C(f, B). This implies that C(f, B) ⊂ B(p(a), r) for 0 < r ≤ r0 , hence
C(f, B) = {p(a)} , B ∩ K = ∅. Since f is a light map on Rn \K, we have
card B = 1. We use then the arguments from the proof of Theorem 1 in [2] to
see that we can lift p from x.
We therefore proved that we can lift any path p : [0, 1] → B(0, p), p(t) =
(1 − t)f (x) + ty for t ∈ (0, 1], where x ∈ Rn \K and y ∈ B(0, ρ). We take now
V = B(0, ρ), x ∈ (Rn \K) ∩ f −1 (V ) and let Q be the component of f −1 (V )
containing x. Then the map g = f |Q : Q → V is a local homeomorphism
which lifts the paths and V is simply connected. This implies that g is a
homeomorphism. We see that Q is the exterior of a Jordan domain D and the
first half of theorem is proved.
Suppose now that f is open and discrete on K. Let x ∈ Rn \K, ρ > 0
such that f (K) ⊂ B(f (x), ρ) and let U ∈ V(x) and 0 < r < ρ be fixed such
that B(f (x), r) ⊂ f (U ). If y ∈ S(f (x), ρ), let αy ∈ U such that f (αy ) ∈
S(f (x), r) ∩ [f (x), y] and let Γy : [0, 1] → B(f (x), ρ), Γy (t) = (1 − t)f (αy ) + ty
for t ∈ [0, 1]. Let Cρ = {y ∈ S(f (x), ρ) | Γy cannot be lifted from αy } and
Γ = {paths γy : [0, cy ) → Rn | y ∈ Cρ and γy is a maximal lifting of Γy from
αy }. Then ∞ ∈ Im γ for every γ ∈ Γ and by Lemma 1 we have MK n−1 (∞) = 0,
hence M (Γ) = 0. Since f is a map of finite distortion satisfying condition (A)
on Rn \K, we use (ii) to see that M (f (Γ)) = 0. This implies that mn−1 (Cρ ) =
0, hence we can lift a.e. path Γy from αy . Since ρ > 0 can be taken great
enough, it results that f is unbounded.
548
Mihai Cristea
10
As in the first part of the proof, we can find ρ > 0 and a Jordan domain
D such that f |D : D → B(0, ρ) is a homeomorphism. Then f |∂D : ∂D →
S(0, ρ) is a homeomorphism and f is open, discrete on D. Using the theorem
of univalence on the border [3], we see that f is injective on D and f (D) =
B(0, ρ). It results that f : Rn → Rn is a homeomorpism. Remark 1. Let f : Rn → Rn , f (x) = x if x ∈ B(0, 1), f (x) = |x|x 2
if x ∈ B(0, 1). Then f is conformal on Rn \S(0, 1), it is not an open map
on S(0, 1) and it is not injective. This shows that the openess of the map f
from Theorem 3 on the “singular” set K is necessary in order to ensure that
f : Rn → Rn is a homeomorphism. We also see that if n ≥ 3, f : Rn → Rn is
a nonconstant map of finite
distortion of K(x)1 dilatation satisfying condition
exp(A(K(x))) |x|2n dx < ∞, then the branch set
(A) and the condition
B(0,r )
Bf = {x ∈ Rn | f is not a local homeomorphism at x} is either empty or
unbounded.
Proof of Theorem 4. Since MK n−1 (F ) = 0, we have cap F = 0, hence
that Rn \F is a domain. Since f is nonconstant on Rn \F , the function f is
open, discrete on Rn \F . Suppose that cap f (Rn \F ) > 0 and let K ⊂ Rn \F
be compact, connected, with card K > 1. Then E = (f (Rn \F ), f (K)) is a
capacitor and q(f (K)) > 0, and by Lemma 2.6, page 65, in [5], there exists
δ > 0 such that δ < cap E. Let Γ = ∆(f (K), f (Rn \F ), Rn ) and let Γ be the
family of all maximal liftings of some paths from Γ starting from some points
of K. Then
Γ < f (Γ) and if γ1 ∈ Γ, then γ has at least a limit point in F ∪{∞}.
exp(A(K(x)))· |x|2n dx < ∞, by Lemma 1 we have MK n−1 (∞) = 0,
Since
B(0,r)
hence MK n−1 (F ∪ {∞}) = 0. This implies that MK n−1 (Γ) = 0. We use now
(i) and obtain that δ < cap(E) = M (Γ ) ≤ M (f (Γ)) ≤ MK n−1 (Γ) = 0, so we
reach a contradiction. We therefore proved that cap f (Rn \F ) = 0. exp(A(K(x))) · |x|12n dx < ∞, which enRemark 2. The condition
B(0,r)
sures that MK n−1 (∞) = 0, used in our generalizations of Zoric’s and Picard’s
theorems, holds for instance if K(x) ≤ Kp for x ∈ B(0, p), p ≥ p0 and
K
lim sup ln pp = α < n (i.e. it holds for a locally quasiregular mapping having
p→∞
some logarithmic growth of the constant of quasiregularity near ∞). Indeed,
take A(t) = t. If p1 ≥ p0 is such that Kp < α · ln p for p ≥ p1 , then
B(0,p1 )
1
exp(A(K(x))) · 2n dx =
|x|
p≥p1
B(0,p+1)\B(0,p)
exp(A(K(x))) ·
1
dx ≤
|x|2n
11
Mappings of finite distortion: Zoric’s theorem and equicontinuity results
exp(Kp+1 ) ·
p≥p1
B(0,p+1)\B(0,p)
549
(p + 1)α ((p + 1)n − pn )
1
≤
2n dx ≤ C0 ·
p2n
|x|
p≥p1
≤ C1 ·
p≥p1
1
pn−α+1
< ∞.
Here C0 and C1 are suitable constants.
We can generalize in this way Theorems 8 and 9 from [2], obtaining a
generalization of Zoric’s theorem for locally quasiregular mappings having a
logarithmic growth of the constant
of quasiregularity near ∞.
Similarly, the condition
exp(A(K(x)))dx < ∞, which ensures that
B(b,ρ)
MK n−1 (b) = 0 holds for instance if K(x) ≤ Kp on B(b, 1p ) for p ≥ p0 and
lim sup
p→∞
Kp
ln p
= α < n. This condition holds for locally quasiregular mappings
having some logarithmic growth of the constant of quasiregularity near the
critical point b ∈ ∂D and it can be used in Theorems 1 and 4. Also, we see
that the set B from Theorem 1 can be taken to be countable.
Proof of Theorem 5. Let x ∈ D be fixed and let ε > 0 with B(x, ε) ⊂ D.
We fix ε > 0 and suppose that there exist ρ > 0, rp → 0 and fp ∈ W such
that q(fp (B(x, rp ))) ≥ ρ for every p ∈ N. Then the fp are open, discrete
mappings for every p ∈ N and fp (U ) ∩ M = ∅ for every open, nonempty
U ⊂ D and every p ∈ N. Then fp (B(x, rp )) ∩ M = ∅ for every p ∈ N and
we can find δ > 0 such that δ ≤ cap (M , fp (B(x, rp ))) for every p ∈ N (see
Lemma 2.6, page 65, in [15]). By (ii) we have δ ≤ cap(M , fp (B(x, rp ))) ≤
cap(fp (B(x, ε)), fp (B(x, rp ))) ≤ capK n−1 (B(x, ε), B(x, rp )) → 0 if rp → 0 and
ε > 0 is kept fixed, so we reach a contradiction. We therefore proved that for
every ε > 0 there exists δε > 0 such that f (B(x, δε )) ⊂ Bq (f (x), ε) for every
f ∈ W , i.e., W is equicontinuous at x. Proof of Theorem 6. Let x ∈ D, δ > 0 be such that B(x, δ) ⊂ D and
M > 0
such that |f (y)|≤ M for every y ∈ B(x, δ) and every f ∈ W . Let
W0 = f |B(x, δ)|f ∈ W . By Theorem 5, W0 is equicontinuous at x, hence
W is equicontinuous at x. Take on D the Euclidean metric and on Rn the
chordal metric. By Theorem 20.4, page 68 in [16], if (fp )p∈N is a sequence of
mappings from W , we can find a subsequence (fpk )k∈N and a map f : D → Rn
such that fpk → f uniformly on the compact subsets of D. If x ∈ D is fixed
and Mx > 0 is such that |fpk (x)| ≤ Mx for every k ∈ N, then |f (x)| ≤ Mx ,
hence f takes finite values and it results that W is a normal family. 550
Mihai Cristea
12
Theorem 7. Let D ⊂ Rn be a domain, W a family of mappings f : D →
of finite distortion, with the same distortion map K satisfying condition
(A), and let x ∈ D be such that there exists r, δ > 0 such that B(x, δ) ⊂ D
and f (B(x, δ)) ⊂ B(f (x), r) for every f ∈ W . Then W is equicontinuous at
x, if we take on D and on Rn the Euclidean distance.
Rn
Proof. Let f ∈ W. If f is constant on B(x, δ), then f (B(x, ρ)) = {f (x)} ⊂
B(f (x), ε) for every 0 < ρ < δ, 0 < ε < r. Suppose that f is open, discrete
on B(x, δ) and let 0 < ε < r, 0 < ρ < δ, and E = (B(x, δ), B(x, ρ)). Then
f (E) = (f (B(x, δ), f (B(x, ρ)) is a capacitor and let νn be the function defined
in [15], page 60. Keep δ >0 fixed andlet ρ → 0 such that capK n−1 (E) ≤ νn ( rε )
(x)|
≤ cap f (E) ≤ capK n−1 (E) ≤ νn ( rε ) for
for 0 < ρ < δε . Then νn |f (y)−f
r
every y ∈ B(x, ρ). Since νn is strictly increasing, we have |f (y) − f (x)| ≤ ε
for every y ∈ B(x, δε ) and every f ∈ W , i.e., W is equicontinuous at x. Remark 3. Theorem 7 shows the interesting thing that if W is a family
of mappings f : D → Rn of finite distortion with the same distortion mapping
K and satisfying condition (A), and x is a point from D, then it is sufficient
to exist a single r > 0 and a single δ > 0 such that f (B(x, δ)) ⊂ B(f (x), r)
for every f ∈ W to obtain the equicontinuity of the family W at the point
x. Taking a mapping f : D → Rn of finite distortion and satisfying condition
(A), fλ = f +λ, W = (fλ )λ∈Rn , we see immediately that W is equicontinuous.
Since g(x) = Rn for every x ∈ D, we cannot apply Theorem 5 to deduce
g∈W
the equicontinuity of the family W , but we can use instead Theorem 7.
Theorem 8. Let D ⊂ Rn be a domain, W a family of homeomorphisms
f : D → f (D) of finite distortion with the same distortion map K satisfying
condition (A) for which there exists r > 0 such that for every f ∈ W there
/ Im f with q(af , bf ) ≥ r. Then W is equicontinuous if we take
exist af , bf ∈
the Euclidean distance on D and the chordal distance on Rn .
Proof. We follow the proof of Theorem 19.2, page 65 in [16]. Let λn be
the function defined in [16], 12.4, page 38 and let x0 ∈ D and 0 < ε < r. Let
Q0 = B(x0 , α), Q1 = B(x0 , β) be such that 0 < α < β and B(x0 , β) ⊂ D,
and let A = R(Q0 , Q1 ). Then f (A) = R(f (Q0 ), f (Q1 )) and q(f (Q1 )) ≥
q(af , bf ) ≥ r, q(f (Q0 )) ≥ q(f (x), f (x0 )) for every x ∈ Q0 . Let x ∈ Q0 be
fixed and t = min{q(f (x0 ), f (x)), r}. We keep β > 0 fixed and choose α
small enough such that MK n−1 (ΓA ) ≤ λn (ε). Then λn (t) ≤ M (f (ΓA )) ≤
MK n−1 (ΓA ) ≤ λn (ε) and since λn is increasing, we have t ≤ ε. Since ε < r, we
get t = q(f (x), f (x0 )). We therefore proved that q(f (x), f (x0 )) ≤ ε for every
x ∈ Q0 and every f ∈ W.
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Mappings of finite distortion: Zoric’s theorem and equicontinuity results
551
Theorem 9. Let D ⊂ Rn be a domain, W a family of homeomorphisms
f : D → f (D) of finite distortion with the same dilatation map K satisfying
condition (A) and such that one of the following condition is satisfied:
(i) there exist x1 , x2 ∈ D and r > 0 such that each f ∈ W omits a point
af with q(af , f (xi )) ≥ r for i = 1, 2;
(ii) there exist xi ∈ D and r > 0 such that q(f (xi ), f (xj )) ≥ r for i = j,
i, j = 1, 2, 3 and every f ∈ W.
Then W is equicontinuous.
Proof. Suppose that condition (i) is satisfied and let Dk = D\ {xk } for
k = 1, 2. By Theorem 8, the families Wk = {f |Dk | f ∈ W } are equicontinuous
on Dk for k = 1, 2, hence W is equicontinuous. Suppose now that condition
(ii) is satisfied and let Dij = D\ {xi , xj } and Wij = {f |Dij | f ∈ W } for i, j =
1, 2, 3. By Theorem 8, the families Wij are equicontinuous on Dij for i, j =
1, 2, 3, hence W is equicontinuous. Corollary 1. Let D ⊂ Rn be a domain, W a family of homeomorphisms f : D → f (D) of finite distortion with the same dilatation mapK,
satisfying condition (A) such that f (ai ) = bi , i = 1, 2, 3 for every f ∈ W ,
where a1 , a2 , a3 are three different points in D and b1 , b2 , b3 are three different
points in Rn . Then W is equicontinuous.
Theorem 10. Let D, Dj be domains in Rn , fj : D → Dj homeomorphisms of finite distortion with the same dilatation map K satisfying condition
(A) and such that fj → f . If card(Im f ) ≥ 3, then f : D → D is a homeomorphism onto a domain D from Rn while if fj → f uniformly on the
compact subsets from D, then f is either constant or a homeomorphism onto
a domain from Rn .
Proof. We follow the proof of Theorem 21.1 in [16] page 9. Let b1 , b2 , b3
be three different points in Im f , ak ∈ D such that f (ak ) = bk for k = 1, 2, 3,
and let r > 0 be such that q(f (ai ), f (aj )) ≥ r for i = j, i, j = 1, 2, 3. Then
there exists j0 ∈ N such that q(fj (ai ), fj (ak )) ≥ 2r for i = k, i, k = 1, 2, 3 and
j ≥ j0 . It follows from Theorem 9, condition 2), that the family W = (fj )j≥j0
is equicontinuous. By Theorem 20.3 in [16], page 68, we have fj → f uniformly
on the compact subsets of D, hence f is continuous on D. By Brouwer’s
theorem, it is enough to prove that f is injective on D.
Suppose that there exists z1 , z2 ∈ D, z1 = z2 such that f (z1 ) = f (z2 ) and
/ B(z1 , r). Then fj (S(z1 , r)) separates the points
let r > 0 be such that z2 ∈
fj (z1 ) and fj (z2 ), hence we can find xj ∈ S(z1 , r) such that
(1)
q(fj (xj ), fj (z1 )) ≤ q(fj (z1 ), fj (z2 )) for every j ∈ N.
552
Mihai Cristea
14
Taking a subsequence, we can suppose that xj → x ∈ S(z1 , r) and
since W is equicontinuous at x, we have q(fj (xj ), f (x)) ≤ q(fj (xj ), fj (x)) +
q(fj (x), f (x)) → 0. Letting j → ∞ in (1), we find that q(f (x), f (z1 )) ≤
q(f (z1 ), f (z2 )) = 0, hence f (z1 ) = f (x). We proved that f is not injective in
any neighbourhood of z1 .
We now prove that every point x ∈ D has a neighbourhood U such that
f is either injective on U or constant on U . Indeed, suppose that this does not
hold. Then we can find a ball U centered at x and distinct points u1 , u2 , u3 in U
with f (u1 ) = f (u2 ), f (u2 ) = f (u3 ). Since W is equicontinuous at x, we can take
U such that qC(fj (U )) ≥ 1 for every j ∈ N, and such that K n−1 (z)dz < ∞.
U
We join u1 and u2 by an arc J0 from U and choose an arc J1 joining u3
with a point u4 ∈ ∂U in U . Then A = R(Im J0 , C(U \ Im J1 )) is a ring and
Aj = fj (A) is a ring R(C◦j , C1j ), where C◦j = fj (J0 ), C1j = fj (U \ Im J1 )
for j ∈ N. Let λn (r, t) be the function defined in [16], 12.6, page 39, and
let rj = q(fj (u1 ), fj (u2 )), tj = q(fj (u2 ), fj (u3 )) for j ∈ N. Then q(C◦j ) ≥
rj , q(C◦j , C1j ) ≤ tj for j ∈ N, tj → 0, and taking a subsequence, we can
suppose that rj ≥ r for every j ∈ N. It results that M (ΓAj ) ≥ λn (rj , tj ) → ∞
for j → ∞. Let now δ = d(Im J0 , C(U \ Im J1 )) > 0 and let ρ : Rn → [0, ∞] be
/ U . Then ρ ∈ F (ΓA ). Hence, by
defined by ρ(z) = 1δ if z ∈ U , ρ(z) = 0 if z ∈
Lemma 2 we have MK n−1 (ΓA ) ≤ ρn (z)·K n−1 (z)dz = δ1n · K n−1 (z)dz < ∞.
Rn
U
Using (i) from Section 1, we obtain that ∞ > Mkn−1 (ΓA ) ≥ M (ΓAj ) → ∞,
and we reach a contradiction.
Let now Q1 = {z ∈ D | there exists V ∈ V(z) such that f is injective
on V }, and Q2 = {z ∈ D | there exists V ∈ V(z) such that f is constant on
/ Q1 , we have z1 ∈ Q2 , hence Q2 = ∅.
V }. Then D = Q1 ∪ Q2 and since z1 ∈
Since D is connected, we have D = Q2 , i.e., f is locally constant on D, hence
f is constant on D. We reached a contradiction, because card Im f ≥ 3. We
therefore proved that f : D → D is a homeomorphism onto a domain D from
Rn . Theorem 11. Let D, D be domains in Rn with card ∂D ≥ 2. Let
F ⊂ D be compact and let W be a family of homeomorphisms f : D → D of
finite distortion with the same dilatation map K and satisfying condition (A).
For every ε > 0 there exists δ > 0 such that if f ∈ W and q(f (F ), ∂D ) < δ,
then q(f (F )) < ε.
Proof. Suppose that the theorem is not true. Then there exist ε > 0 and
a sequence (fj )j∈N from W such that q(fj (F ), ∂D ) < 1j and q(fj (F )) ≥ ε for
/ Im fj for
every j ∈ N. Since card ∂D ≥ 2, there exist at least two points yk ∈
every j ∈ N and k = 1, 2. By Theorem 8 the family (fj )j∈N is equicontinuous.
15
Mappings of finite distortion: Zoric’s theorem and equicontinuity results
553
Taking a subsequence, we can find a map f : D → Rn such that fk → f
uniformly on the compact subsets of D. By Theorem 10, either f is a constant
map on D or f : D → G is a homeomorphism onto a domain G from Rn . Since
q(fj (F )) ≥ ε > 0 for j ∈ N, the function f cannot be constant on D, hence
f : D → G is a homeomorphism.
Let us show that G ⊂ D . Indeed, let y ∈ G and x ∈ D be such
/ f (∂U ). Set
that y = f (x) and let U ∈ V(x) be such that U ⊂ D, f (x) ∈
r = q(f (x), f (∂U )). Since fj → f uniformly on U , there exists j0 ∈ N such
that q(fj (z), f (z)) < 3r for every z ∈ U and every j ≥ j0 . If V = Bq (y, r3 ) then
V ∩ fj (U ) = ∅, ∂fj (U ) = fj (∂U ) and fj (∂U ) ∩ V = ∅ for j ≥ j0 . We have
V = (V ∩ fj (U )) ∪ (V ∩ ∂fj (U )) ∪ (V ∩ fj (U )) = (V ∩ fj (U )) ∪ (V ∩ fj (U ))
for j ≥ j0 , and since V is connected, we have V = V ∩ fj (U ), hence V ⊂ fj (U )
for j ≥ j0 . It results that y ∈ V ⊂ fj (U ) ⊂ D , We therefore proved that
G ⊂ D , hence δ = q(f (F ), ∂D ) > 0. Since fj → f uniformly on F and
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Received 2 February 2006
University of Bucharest
Faculty of Mathematics and Computer Science
Str. Academiei 14
010014 Bucharest, Romania
[email protected]