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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) f ∈Wloc1,1(D,Rn);

2) the Jacobian determinant Jf(x) is locally integrable;

3) there exists a finite a.e. measurable functionK:D→[0,∞] such that

|f(x)|n≤K(x)·Jf(x) a.e.

Notice that whenK ∈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 H¨older condition.

If the dilatation mapK Lploc(D), p > n−1, and f Wloc1,n(D,Rn), 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→ Rn of finite distortion for which there exists some smooth, strictly increas- ing A : [0,) [0,) with A(0) = 0, lim

t→∞A(t) = and satisfying the 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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(A2) there exists t0>0 such thatA(t)·tincreases to infinity fort≥t0. As in [9], we say that a mapping of finite distortion satisfies condition (A) iff 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 mappingsf :D→Rnof finite distortion and satisfying condition (A):

(i) M(f(Γ))≤MKn−1(Γ) for every path family Γ from D;

(ii) capf(E) capKn−1(E), where E = (G, C) is a capacitor with G⊂⊂D;

(iii) capKn−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 capKn−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) for which there exists r > 0 such that

B(0,r)

exp(A(K(x)))· |x1|2ndx < . 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 Rn a nonconstant map of finite distortion, satifying condition (A),such that MKn−1(F) = 0and there existsr >0such that

B(0,r)

exp(A(K(x)))·|x1|2ndx <

. Then capf(Rn\F) = 0.

We also give in Theorem 1 a version of Zoric’s theorem with some “sin- gular” 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 ofK(x) dilatation satisfying condition (A), and MKn−1() = 0, then there existsE⊂Rn with capE = 0 such thatf :Rn Rn\E is a homeomorphism. Our condition

B(0,r)

exp(A(K(x)))·|x1|2ndx <

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implies that MKn−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)→Rnis a local homeomorphism of finite distortion ofK(x) dilata- tion satisfying conditon (A) with exp(A ◦K)∈L1(B(0, r)), then there exists δ(n, K, r) such thatf is injective onB(0, δ). The result from [11] is not enough for an arbitrary local homeomorphismf :RnRnof finite distortion ofK(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 the- orem for mappings of finite distortion using [11]. An example from [8] shows that the fundamental condition

B(0,r)

exp(A(K(x)))· |x1|2ndx < used in Theorems 2 and 4 is necessary in order to ensure thatf :RnRnis a global homeomorphism.

We generalize in Theorem 5 a known equicontinuity result for quasireg- ular mappings as follows.

Theorem 5. Let D Rn be a domain, M Rn with capM > 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 Dand the chordal distance on Rn.

We see that even in the case where all the mappings from the familyW areK-quasiregular, we improve the classical result established for quasiregular mappings in [15], Corollary 2.7, page 66. Indeed, the exceptional setM which is avoided by all the mapsf ∈W can be to be at most countable and chosen such that capM >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 caseA(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 path family in Rn, we define F(Γ) = : Rn [0,] is a Borel map

|

γρds≥1 for every locally rectifiable γ Γ}.Let nowD⊂Rn be open and ω :D [0,] measurable and finite a.e. If ω : Rn [0,] is defined by

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ω(x) =ω(x) if x∈D, ω(x) = 0 if x /∈D, then we define the ω-modulus of Γ byMω(Γ) = inf

ρF(Γ)

Rn

ρn(x)·ω(x)dx. For ω 1 we obtain the usual modulus M(Γ). If Γ1, Γ2are path families inRn, we say that Γ1>Γ2if every path from Γ1 has a subpath in Γ2. As in the classical case, we can prove that if Γ1>Γ2, thenMω1)≤Mω2). Also, we can prove thatMω

i=1 Γi

i=1Mωi), 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 inD.

We say that E = (D, C) is a condenser if C is compact, D is open, C D Rn. Let ω :D [0,] be measurable and finite a.e. We define theω-capacity of E by capωE = inf

Rn|∇u|n(x)·ω(x)dx, where u∈C0(D) and u 1 on C. If ω = 1 we obtain the usual capacity cap E of E. We see that ifu is a test function for capωE, then ρ =|∇u| ∈ FE). This implies that MωE) capω(E). Here, ΓE = : [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 capE = 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 capC = 0 if capK= 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 zeroω-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 thatM(Γ)≤Mω(Γ). Hence, ifMω(A) = 0 then capA= 0.

If A ⊂D is countable and lim

r→0Mω(∆(B(x, r),B(x, R), D)) = 0 for ev- 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 = Kn−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

and

B(x,ρx)

exp(A(K(y)))dy <, whereAsatisfies conditions (A1) and (A2).

Lemma 1 also gives a condition which ensures thatMKn−1() = 0.

IfD⊂Rnis open andb∈∂D, we defineC(f, b) ={w∈Rn|there exists bp∈D,bp→bsuch thatf(bp)→w}. IfB⊂∂D, we letC(f, B) =

bBC(f, b).

If E, F are Hausdorff spaces, f : E F is a map, p : [0,1] F is a path, x∈E is such thatf(x) =p(0),we say thatq: [0, a)→E is a maximal lifting ofpfromxwhen q(0) =x, 0< a≤1, f◦q =p|[0, a) andq is maximal with this property. Ifqis defined on [0,1], we say thatq is a lifting ofp. IfE, F are

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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 the chordal metric inRn given byq(a, b) =|a−b| ·(1 +|a|2)12 ·(1 +|b|2)12 ifa, b Rn, q(a,∞) = (1 +|a|2)12 if a∈ Rn, and by |a−b| the Euclidean distance betweenaand b inRn. We denote by Bq(x, r), respectivelyB(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 setA⊂Rnwe denote byq(A), respectivelyd(A), the diameter ofA. We denote by Wloc1,p(D,Rn) the Sobolev space of all functionsf :D→Rnwhich are locally inLp(D), together with their first order weak partial derivatives.

IfW is a family of mappingsf :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 everyx∈K and everyf ∈W. IfX, Y are metric spaces andW 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 everyf ∈W. We say thatW 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 mapf :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

t→∞A(t) =,satisfying condition (A1) and (A2) for which there exists r > 0 such that

DB(0,r)

exp(A(K(x)))· |x1|2ndx <. ThenMKn−1(∆(B(0, r)∩D,B(0, s)∩D, D)→0as s→ ∞and r >1is 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 MKn−1rs) = MKn−1grs).

Letρ ∈Frs) 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 MKn−1grs)

Rn

ρn(x)· Kn−1(g(x))dx=

D

ρn(g(x))|g(x)|n·Kn−1(g(x))dx=

D

ρn(g(x))·Kn−1(g(x))·

Jg(x)dx =

g(D)

ρn(y)·Kn−1(y)dy

Rn

ρn(y)·Kn−1(y)dy. This implies that MKn−1g(g(Γrs)) =MKn−1grs)≤MKn−1rs). By the same argument, we haveMKn−1rs) =MKn−1gg(g(Λrs))≤MKn−1grs) and

(1) MKn−1rs) =MKn−1grs).

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Next,

g(D)∩B(0,1r)

exp(A(K(g(x))))dx=

g(D)∩B(0,1r)

exp(A(K(g(x))))·Jg−1(g(x))·Jg(x)dx=

=

DB(0,r)

exp(A(K(y)))·Jg−1(y)dy=

DB(0,r)

exp(A(K(y)))· 1

|y|2ndy <∞. 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 x /∈g(D). Since

B(0,1r)

exp(A(Q(x)))dx <∞, by Theorem 5.3, page 24, in [9], capQn−1(B(0,1r), B(0,1s)) 0 as s → ∞ and r > 0 is kept fixed. We have MKn−1grs) ≤MQn−1rs) ≤MQn−1rs) capQn−1(B(0,1r), B(0,1s)) and this implies that

(2) MKn−1grs)0 ass→ ∞ and r >0 is kept fixed.

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 of f. Then K Lploc(D) for every p > 0 and f Wloc1,p(D,Rn) for every 1≤p < n.

Proof. Using property (A2) we see that the function exp(A(t)) grows faster then anytp to infinity and therefore K ∈Lploc(D) for every 0< p <∞. We use now H¨older’s inequality and the integrability of the Jacobian to see that

B|f(x)|pdx <for every ballB withB ⊂Dand every 1≤p < n.

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 intK = ∅, f :Rn\(K ∪B) Rn a map of finite distortion of K(x)dilatation satisfying condition (A)and a local homeomorphism such that MKn−1(B) = 0, intC(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 that there exists δ0 >0such that

B(00)∩(Rn\(KB))

exp(A(K(x)))·|x1|2n <∞.

Then

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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 intE = 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)orf|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 = : [0,1) Rn\K path|∞ ∈Im γ}. By Lemma 1,MKn−1(Γ) = 0.Using (i), we see that M(f(Γ))≤MKn−1(Γ) = 0, henceM(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). LetQbe the component off−1(B(0, r)) which containsxand letU be the component off−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 homeomor- phism 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 thatRn\∂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|U0 :U0 B(0, s) is a homeomorphism, ∂U0 bounds a Jordan domain, and U0 is the unbounded component ofRn\∂U0. Let α >0 be such that ∂U ∪∂U0 B(0, α). Then B(0, α) U ∩U0, hence U = U0 and we obtain thatf−1(B(0, s)) has a single component U on which f is injective, K⊂U and U =f−1B(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 liftp 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 thatf is injective onU, we obtain thatq1(1) =q2(1).We now use the property of the uniqueness of path lifting for local homeomorphisms to conclude thatx1 =x2. We therefore proved that f is injective onRn\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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Indeed, if there existsb=a,b∈Rn such that f(a) =f(b), let U1∈ V(a) and V ∈ V(f(a)) be such that f(U1) =V. Sincef is continuous at b, we can find U2∈ V(b) withU1∩U2=such thatf(U2)⊂V, and since intE =, we can findα ∈U2\E. Then we can findβ ∈U1\E withf(α) =f(β). So, we reached a contradiction, since we proved that f is injective onRn\E. It results now immediately that eitherf|Rn\K:Rn\K Rn\f(K) orf|Rn\K :Rn\K Rn\C(f, K) is a homeomorphism. If we additionally suppose that f also is an open map onK, we find thatf|Rn:RnRn is a homeomorphism.

Proof of Theorem 2. We takeK = in Theorem 1.

The following version of Zoric’s theorem seems to be new even for quasire- gular mappings.

Theorem 3. Let n 3. Let K Rn be compact, f : Rn Rn con- tinuous on Rn and of finite distortion of K(x) dilatation satisfying condition (A) on Rn\K for which there exists r >0 such that f is a local homeomor- phism on B(0, r)and

B(0,r)

exp(A(K(x)))·|x1|2ndx <. If fis unbounded, then there exists δ >0 such that f is injective on B(0, δ) while if f is open discrete on K,then f :RnRn is a homeomorphism.

Proof. Suppose first that f is unbounded and let ρ > 0 be such that f(K)⊂B(0, ρ). Letp: [0,1]→B(0, ρ) be defined byp(t) = (1−t)f(x) +ty fort∈[0,1] and some x∈Rn\K and y ∈CB(0, ρ). As in Theorem 1 in [2], we try to liftpfrom x. Let 0< a≤1 and letq : [0, a)Rn\K be a maximal lifting ofp from x withf ◦q=p|[0, a). As in our Theorem 1 and Theorem 1 in [2] we can find spheresS(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 = s<r(Ds\Dr) and Br = B(p(a), r)∪B(0, ρ) for 0 < r ≤r0. Then Br is connected for 0 < r r0 and let us fix 0 < s < r r0. We show that f((Ds\Dr)\K)⊂Br. Indeed, otherwise we can find a point α (Ds\Dr)\K such that f(α) CBr. Let y Br and let Γ : [0,1] Br 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) Br. On the other hand, we have f(z) = f

plim→∞γ(tp)

= lim

p→∞f(γ(tp)) = lim

p→∞Γ(tp) = Γ(c) Br and we so reach a contradiction.

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We see that we can lift Γ from α, hence we can find a path γ : [0,1] (Ds\Dr)\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 Br, we proved that Br f((Ds\Dr)\K) ⊂f(Ds), so we reached a contradiction so f(Ds) is compact andBr is unbounded.

We therefore proved that f(Ds\Dr) Br for 0 < s < r r0. Let A = lim

r→∞Wr. Then Wr is connected and f(Wr) Br for 0 < r r0, hence C(f, A) =

r>0f(Wr) is connected and C(f, A) Br for 0 < r r0. 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 thatA∩K=and since f is a light map on Rn\K, we have cardA = 1. ThenA={∞} and f maps a neighbourhood of

near p(a) and this contradicts the fact that f is unbounded. We therefore proved that the caseDr⊂Ds for 0< s < r≤r0 cannot hold.

In the second case, we have Ds Dr for 0 < s < r r0 and let B = r>

0Dr. As before, we see that f(Dr\Ds) Br for 0 < s < r r0, that C(f, B) = r>

0f(Dr) is connected and C(f, B) ⊂Br for 0 < r≤r0 and 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 cardB = 1. We use then the arguments from the proof of Theorem 1 in [2] to see that we can liftp from x.

We therefore proved that we can lift any pathp: [0,1]→B(0, p),p(t) = (1−t)f(x) +tyfort∈(0,1], wherex∈Rn\K andy∈B(0, ρ). We take now V =B(0, ρ), x(Rn\K)∩f−1(V) and letQ 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 thatQis the exterior of a Jordan domainDand the first half of theorem is proved.

Suppose now thatf is open and discrete on K. Let x Rn\K, ρ > 0 such thatf(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 fy) S(f(x), r)[f(x), y] and let Γy : [0,1]→B(f(x), ρ),Γy(t) = (1−t)fy) +ty for t [0,1]. Let Cρ = {y S(f(x), ρ) | Γy cannot be lifted from αy} and Γ ={pathsγy : [0, cy) Rn |y Cρ and γy is a maximal lifting of Γy from αy}. Then∞ ∈Imγfor everyγ∈Γ and by Lemma 1 we haveMKn−1(∞) = 0, henceM(Γ) = 0. Sincef is a map of finite distortion satisfying condition (A) onRn\K, we use (ii) to see thatM(f(Γ)) = 0. This implies thatmn−1(Cρ) = 0, hence we can lift a.e. path Γy from αy. Since ρ > 0 can be taken great enough, it results thatf is unbounded.

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As in the first part of the proof, we can findρ >0 and a Jordan domain Dsuch that f|D:D→B(0, ρ) is a homeomorphism. Then f|∂D:∂D→ S(0, ρ) is a homeomorphism andf is open, discrete onD. 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 thatf :RnRn is a homeomorpism.

Remark 1. Let f : Rn Rn, f(x) = x if x B(0,1), f(x) = |xx|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 mapf from Theorem 3 on the “singular” set K is necessary in order to ensure that f :RnRnis a homeomorphism. We also see that if n≥3, f :RnRnis a nonconstant map of finite distortion ofK(x) dilatation satisfying condition (A) and the condition

B(0,r)

exp(A(K(x)))|x1|2ndx <∞, then the branch set Bf = {x Rn | f is not a local homeomorphism at x} is either empty or unbounded.

Proof of Theorem 4. Since MKn−1(F) = 0, we have cap F = 0, hence that Rn\F is a domain. Since f is nonconstant on Rn\F, the functionf is open, discrete onRn\F. Suppose that capf(Rn\F)>0 and letK⊂Rn\F be compact, connected, with cardK > 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 thatδ < capE. 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 ofK. Then Γ< f(Γ) and ifγ Γ, thenγ has at least a limit point inF∪{∞}. Since

B(0,r)

exp(A(K(x)))·|x1|2ndx <, by Lemma 1 we haveMKn−1() = 0, henceMKn−1(F ∪ {∞}) = 0. This implies that MKn−1(Γ) = 0. We use now (i) and obtain that δ <cap(E) =M)≤M(f(Γ))≤MKn−1(Γ) = 0, so we reach a contradiction. We therefore proved that capf(Rn\F) = 0.

Remark 2. The condition

B(0,r)

exp(A(K(x)))·|x1|2ndx <, which en- sures thatMKn−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

lim sup

p→∞

Kp

lnp =α < n(i.e. it holds for a locally quasiregular mapping having some logarithmic growth of the constant of quasiregularity near). Indeed, takeA(t) =t. If p1≥p0 is such that Kp < α·lnp forp≥p1, then

B(0,p1)

exp(A(K(x)))· 1

|x|2ndx=

pp1

B(0,p+1)\B(0,p)

exp(A(K(x)))· 1

|x|2ndx

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pp1

B(0,p+1)\B(0,p)

exp(Kp+1)· 1

|x|2ndx≤C0·

pp1

(p+ 1)α((p+ 1)n−pn)

p2n

≤C1·

pp1

1

pnα+1 <∞. HereC0 andC1 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

B(b,ρ)

exp(A(K(x)))dx <, which ensures that MKn−1(b) = 0 holds for instance if K(x) Kp on B(b,1p) for p p0 and lim sup

p→∞

Kp

lnp =α < n. This condition holds for locally quasiregular mappings having some logarithmic growth of the constant of quasiregularity near the critical pointb ∂D and it can be used in Theorems 1 and 4. Also, we see that the setB from Theorem 1 can be taken to be countable.

Proof of Theorem 5. Letx∈Dbe fixed and letε >0 withB(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 findδ >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))) capKn−1(B(x, ε), B(x, rp)) 0 ifrp 0 and ε >0 is kept fixed, so we reach a contradiction. We therefore proved that for everyε >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 fromW, we can find a subsequence (fpk)k∈Nand a mapf :D→Rn such that fpk f uniformly on the compact subsets of D. Ifx ∈D is fixed and Mx >0 is such that |fpk(x)| ≤ Mx for every k N, then |f(x)| ≤ Mx, hencef takes finite values and it results thatW is a normal family.

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Theorem 7. Let D⊂Rnbe a domain,W a family of mappings f :D→ Rn 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 Dand on Rn the Euclidean distance.

Proof. Letf ∈W.Iff is constant onB(x, δ), thenf(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 letνnbe the function defined in [15], page 60. Keepδ >0 fixed and letρ→0 such that capKn−1(E)≤νn(εr) for 0< ρ < δε. Then νn

|f(y)−f(x)|

r

capf(E) capKn−1(E) νn(rε) for every y B(x, ρ). Since νn is strictly increasing, we have |f(y)−f(x)| ≤ ε for everyy∈B(x, δε) and everyf ∈W, i.e., W is equicontinuous at x.

Remark 3. Theorem 7 shows the interesting thing that if W is a family of mappingsf :D→Rnof 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 mappingf :D→Rnof finite distortion and satisfying condition (A), fλ=f+λ, W = (fλ)λ∈Rn, we see immediately thatW is equicontinuous.

Since

gWg(x) =Rn for everyx ∈D, we cannot apply Theorem 5 to deduce the equicontinuity of the familyW, but we can use instead Theorem 7.

Theorem 8. Let D⊂Rnbe 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 exist af, bf ∈/ Im f with q(af, bf)≥r. Then W is equicontinuous if we take 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 letx0 ∈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 MKn−1A) λn(ε). Then λn(t) M(f(ΓA)) MKn−1A)≤λn(ε) and sinceλnis increasing, we havet≤ε. Sinceε < r, we gett=q(f(x), f(x0)). We therefore proved thatq(f(x), f(x0))≤εfor every x∈Q0 and every f ∈W.

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Theorem 9. Let D⊂Rnbe 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∈Dand 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 >0such 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 familiesWk={f|Dk|f ∈W}are equicontinuous on Dk for k = 1,2, hence W is equicontinuous. Suppose now that condition (ii) is satisfied and letDij =D\ {xi, xj} andWij ={f|Dij|f ∈W}fori, j = 1,2,3. By Theorem 8, the families Wij are equicontinuous on Dij for i, j = 1,2,3, henceW is equicontinuous.

Corollary 1. Let D Rn be a domain, W a family of homeomor- phisms 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 Dand 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 homeomor- phisms of finite distortion with the same dilatationmapKsatisfying condition (A) and such that fj →f. If card(Imf) 3, then f :D→ D is a home- omorphism 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. Letb1, b2, b3 be three different points in Imf,ak∈D such that f(ak) =bk fork= 1,2,3, and let r > 0 be such thatq(f(ai), f(aj)) r for i= j, i, j = 1,2,3. Then there exists j0 N such thatq(fj(ai), fj(ak)) r2 fori=k,i, k = 1,2,3 and j≥j0. It follows from Theorem 9, condition 2), that the familyW = (fj)jj0 is equicontinuous. By Theorem 20.3 in [16], page 68, we havefj →f uniformly on the compact subsets of D, hence f is continuous on D. By Brouwer’s theorem, it is enough to prove thatf is injective onD.

Suppose that there existsz1, z2 ∈D,z1 =z2such thatf(z1) =f(z2) and let r > 0 be such that z2 ∈/ B(z1, r). Then fj(S(z1, r)) separates the points fj(z1) and fj(z2), hence we can findxj ∈S(z1, r) such that

(1) q(fj(xj), fj(z1))≤q(fj(z1), fj(z2)) for every j∈N.

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