SOME MODULAR INEQUALITIES
MIHAI CRISTEA
We study the dilatations of the balls generated by a homeomorphismf between two domains fromRnsatisfying some generalized modular inequalities. We obtain as a particular case some conditions of quasiconformality.
AMS 2010 Subject Classification: 30C65.
Key words: mappings satisfying generalized modular inequalities.
1. INTRODUCTION
If D⊂Rn is a domain, we set A(D) the set of all path families Γ from D and if Γ ∈ A(D) we put F(Γ) = {ρ : Rn → [0,∞] Borel maps|R
γρds ≥ 1 for every γ ∈ Γ locally rectifiable}. We define for p ≥ 1, Γ ∈ A(D) and ω ∈L1loc(D) thepmodulus of weightω,Mωp(Γ) = inf
ρ∈F(Γ)
R
Dω(x)·ρp(x)dxand for ω = 1 we obtain the classical p modulus Mp(Γ) = inf
ρ∈F(Γ)
R
Rnρp(x)dx. If D, D0are domains fromRnandf :D→D0is a homeomorphism so that there exists K ≥1 so that Mn(f(Γ)) ≤K·Mn(Γ) and Mn(Γ)≤K·Mn(f(Γ)) for every Γ∈A(D), we obtain the known class ofK-quasiconformal mappings. An important class of continuous, open discrete mappingsf :D⊂Rn→Rnwhich generalizes quasiconformal mappings is the class of quasiregular mappings (see [22], [17], [18] the mains books concerning the theory of quasiconformal and quasiregular mappings). The important modular inequality of Poleckii also holds for quasiregular mappings and says that if f is K-quasiregular, then Mn(f(Γ)) ≤K·Mn(Γ) for every Γ∈ A(D), and is the key for proving most of the geometric properties of this class of mappings.
More general classes of continuous, open, discrete mappings (so called mappings of finite distortion) for which hold a generalized modular inequality of type “Mn(f(Γ))≤MKn
I,n(f)(Γ) for every Γ∈A(D)” were studied in [2–6], [11–16], [19–21]. In some recent papers [7, 8] we studied continuous, open, dis- crete mappings f :D⊂Rn→Rn satisfying a generalized modular inequality of type Mq(f(Γ)) ≤ γ(Mωp(Γ)) for every Γ ∈ A(D), with q > n−1, p > 1,
REV. ROUMAINE MATH. PURES APPL.,56(2011),4, 275–282
γ : [0,∞) → [0,∞) strictly increasing with lim
t→0γ(t) = 0 and some weight ω ∈L1loc(D). We gave Liouville, Montel, Picard type theorems, equicontinuity results and we gave estimates of the modulus of continuity. We extended in this way partially basic theorems from the theory of quasiregular mappings and from the class of continuous, open, discrete mappings mentioned before.
The basic tools we used before for proving this things was the modular in- equality Mq(f(Γ)) ≤ γ(Mωp(Γ)) for every Γ ∈ A(D) together with the fact that lim
r→0Mωp(Γx,r,R) = 0 for every x ∈ D and every fixed R > 0 (see [7]).
We gave in [7, 8] examples of such mappings which are not quasiconformal showing in this way that our extensions are effective.
IfD, D0are domains fromRn,f :D→D0is a homeomorphism,g=f−1, x∈D,r >0 is so that B(x, r)⊂D, we set
L(x, f, r) = sup
|y−x|=r
|f(y)−f(x)|, l(x, f, r) = inf
|y−x|=r|f(y)−f(x)|, and we set
H(x, f) = lim sup
r→0
L(x, f, r)
l(x, f, r), H∗(x, f) =H(f(x), g).
It is known that if there exists H ≥1 so thatH(x, f) ≤H for every x∈ D, or if H∗(x, f)≤H for everyx∈D, then f is quasiconformal.
We shall show that if f : D → D0 is a homeomorphism between two domains from Rn, p > 1, q > n−1, γ : [0,∞) → [0,∞) is strictly increas- ing with lim
t→0γ(t) = 0 so that Mq(f(Γ)) ≤ γ(Mωp(Γ)) for every Γ ∈ A(D), x ∈ D and g = f−1 and there exists Kx, rx > 0, 0 ≤ αx < p−1 so that R
B(x,δ)ω(z)dz≤Kx·δp ln e·rδxαx
for every 0< δ≤rx, then we can calcu- late a bound for the local dilatation of the map g around the pointy =f(x), i.e. we can findρx>0, a function C: (0, δx)→(0,∞),
C(r) =C(r, p, q, γ, n, Kx, αx)
= exp exp
Kx·ep·
∞
X
k=1
1
kp−αx/γ−1 Cn,q·rn−q 1p!!
e
for every 0< r≤δx so that L(y,g,r)l(y,g,r) ≤C(r) for every 0< r≤ρx.
We say that a mapf :D⊂Rn→ Rn is ACL iff is continuous and for every cube Q⊂⊂ D with the sides parallel to coordinate axes and for every face S of Q it results that f|PS−1(y) ∩Q : PS−1(y)∩Q → Rn is absolutely continuous for a.e.y∈S, wherePS:Rn→S is the projection onS. An ACL map has a.e. partial derivatives and ifp >1 we say thatf is ACLp iff is ACL and the partial derivatives are locally in Lp. We say that f :D⊂Rn → Rn satisfies condition (N) ifµn(f(A)) = 0 for every set A⊂D with µn(A) = 0.
Here µn is the Lebesgue measure from Rn. If A ∈ L(Rn,Rn), detA 6= 0, p > 1, we set K0,p(A) = |det|A|pA|, KI,p(A) = |detl(A)A|p , where |A| = sup
|x|=1
|A(x)|,
l(A) = inf
|x|=1|A(x)| and |x| = n
P
i=1
x2i 12
if x = (x1, . . . , xn) ∈ Rn. If f : D → D0 is a homeomorphism between two domains from Rn and p > 1, we define the map NI,p(f) : D → R+ by NI,p(f)(x) = KI,p(f0(x)) if f is differentiable in x and Jf(x)6= 0, NI,p(f)(x) = 0 otherwise and we define the map N0,p(f) :D→ R+ by N0,p(f)(x) =K0,p(f0(x)) if f is differentiable in x and Jf(x) 6= 0, N0,p(f)(x) = 0 otherwise. A quasiconformal map f :D→D0 is a.e. differentiable, is ACLn and Jf(x) 6= 0 a.e. If D ⊂ Rn is a domain and E, F ⊂ D, we set ∆(E, F, D) = {γ : [0,1]→ Rn path|γ(0) ∈ E, γ(1)∈ F and γ((0,1)) ⊂ D} and if f : D → Rn is continuous, we put L(x, f) = lim sup
h→0
|f(x+h)−f(x)|
|h| . If γ : [a, b] → Rn is a rectifiable path, we set sγ(t) = l(γ|[a, t]) for t ∈ [a, b] and we define a reparametrisation γ0 : [0, l(γ)] → Rn of the path γ given by γ(t) = γ0(sγ(t)) for t ∈ [a, b]. A domain A ⊂ Rn is a ring if CA has exactly two components C0, C1 and we set A = R(C0, C1) and ΓA = ∆(C0, C1,Rn). Given p > 0 and r > 0, we put Φp(r) the set of all rings A = R(C0, C1) so that 0 ∈ C0 and there exists a ∈ C0 so that
|a| = 1 and so that ∞ ∈ C1 and there exists b ∈ C1 so that |b| = r and we denote Hn,p(r) = inf
A∈Φp(r)M(ΓA). We see from Theorem 8 from [2] that if n−1 < p, then Hn,p(1) = Cn,p > 0. If x ∈ Rn and 0 < r < R, we set Γx,r,R= ∆(B(x, r), S(x, R), B(x, R)\B(x, r)).
Using the idea from [15], page 94, Lemma 5.2, we extend Theorem 2 from [5].
Proposition 1. Let n ≥2, p >1, M > 0, 0 ≤ α < p−1, D ⊂Rn a domain,x∈Dand0< r < Rso thatB(x, r)⊂D,ω :D→[0,∞]measurable and finite a.e. so that R
B(x,δ)ω(z)dz≤M·δp· ln Reδ α
for every 0< δ < R and let C =M·ep·
∞
P
k=1 1
kp−α. Then Mωp(Γx,r,R)≤C/ ln ln Rer p
.
Proof. We can suppose that x= 0 and letρ:Rn→[0,∞] be defined by ρ(z) = 1/
|z| ·ln
Re
|z|
·ln ln Rer
ifz∈B(x, r)\B(x, r),ρ(z) = 0 otherwise.
Let ak =Re−k for k∈ N,Bk =B(0, ak), k∈ N and letAk =Bk\Bk+1 for k ∈N. We see thatρ ∈F(Γx,r,R) and we see that if z∈Ak, then |z|R ≤ek+1
and 1/ln Re
|z|
≤ k+11 for every k∈N. We find that Mωp(Γx,r,R)≤
Z
Rn
ω(z)ρp(z)dz≤
1/
ln ln
Re r
p
· Z
B(x,R)\B(x,r)
ω(z)/
/
|z|p·
ln Re
|z|
p dz≤
1/
ln ln
Re r
p
·
∞
X
k=0
Z
Ak
ω(z)/
/
|z|p
ln Re
|z|
p dz≤
1/
ln ln
Re r
p
·
∞
X
k=0
Z
Bk
ω(z)· e(k+1)p Rp(k+ 1)pdz
≤
M·ep/
ln ln Re
r
p
·
∞
X
k=0
ekp
Rp(k+ 1)p ·(Re−k)p·
ln Re
Re−k α
=
=C/
ln ln
Re r
p
.
Theorem 1. Let n≥2, p >1, n−1 < q, γ : [0,∞) → [0,∞) strictly increasing with lim
t→0γ(t) = 0, D, D0 domains inRn, ω:D→[0,∞]measurable and finite a.e., f :D→ D0 a homeomorphism so that Mq(f(Γ))≤γ(Mωp(Γ)) for every Γ ∈ A(D) and let g = f−1, x ∈ D and y = f(x). Suppose that there exists Kx > 0, rx > 0, 0 ≤ αx < p − 1 so that R
B(x,δ)ω(z)dz ≤ Kx ·δp · ln erδxα
for every 0 < δ ≤ rx and let C(r) = exp(exp((Kx · ep ·
∞
P
k=1 1
kp−αx/γ−1(Cn,q ·rn−q))1p))/e for r > 0. Then there exists ρx > 0 so that L(y,g,r)l(y,g,r) ≤ C(r) for every 0 < r ≤ ρx and if q = n, then H(y, g) ≤ H = H(p, n, γ, Kx, αx). If D = D0 = Rn and there exists Kx > 0 so that R
B(x,δ)ω(z)dz≤Kx·δp for every δ >0, then L(y,g,r)l(y,g,r) ≤C(r) for every r >0.
Proof. Let ρx > 0 be so that L(y, g, r) ≤ rx for 0 < r ≤ ρx. Let 0< r≤ρxandLr=L(y, g, r),lr =l(y, g, r),Ar=R(B(x, lr), CB(x, Lr)) and let f(Ar) =R(Br, Cr). We see that there existsar∈S(x, lr),br ∈S(x, Lr) so that f(ar)∈S(y, r), f(br)∈S(y, r) andy, f(ar)∈Br,y, f(br)∈Cr. Letgr : Rn→Rn,gr(z) = z−yr forz∈Rn and hr=gr◦f. Then hr(Ar) =R(Dr, Er) and we can find points cr ∈ Dr, dr ∈ Er so that cr, dr ∈ S(0,1), 0 ∈ Dr,
∞ ∈ Er and let Cx = Kx·ep·
∞
P
k=1 1
kp−αx. Using Caraman’s result from [1], Theorem 8 and Proposition 1, we find that
0< Cn,q=Hn,q(1)≤Mq(Γhr(Ar)=Mq(Γgr(f(Ar))) = 1
r n−q
·Mq(Γf(Ar)) =
= 1
r n−q
·Mq(f(ΓAr))≤ 1
r n−q
·γ(Mωp(ΓAr)) = 1
r n−q
·γ(Mωp(Γx,lr,Lr))≤
≤ 1
r n−q
·γ
Cx/
ln ln
e·Lr lr
p . We therefore obtain that Llr
r ≤C(r) for every 0< r ≤ρx. IfD=D0 = RnandR
B(x,δ)ω(z)dz≤Kx·δp for everyδ >0, then, takingαx= 0, we obtain that L(y,g,r)l(y,g,r) ≤C(r) = exp exp
Kx·ep·
∞
P
k=1 1
kp/γ−1(Cn,q·rn−q) 1
p
!!
/e for every r >0.
Corollary 1. Let n ≥ 2, p > 1, D, D0 domains from Rn, ω : D → [0,∞] measurable and finite a.e., γ : [0,∞) → [0,∞) strictly increasing with
t→0limγ(t) = 0, and f : D → D0 a homeomorphism so that Mn(f(Γ)) ≤ γ(Mωp(Γ)) for every Γ ∈ A(D). Suppose that there exists K > 0, a > 0 so that for every x ∈ D there exists rx > 0 and 0 ≤ αx < p−1−a so that R
B(x,δ)ω(z)dz≤K·δp· ln e·rδxα
for every 0< δ≤rx. Thenf is quasicon- formal and
H∗(x, f)≤H=H(p, n, γ, K, a) =
= exp exp
K·ep·
∞
X
k=1
1
k1+a/γ−1 C(n) 1
p
!!
e
for every x∈D.
Proof. We take q = n and K = Kx in Theorem 1 and we see that
∞
P
k=1 1 kp−αx ≤
∞
P
k=1 1
k1+a and the theorem is proved.
Corollary 2. Let n ≥ 2, p > 1, n−1 < q, ω :Rn → [0,∞] measu- rable and finite a.e., γ : [0,∞) →[0,∞) strictly increasing with lim
t→0γ(t) = 0, f : Rn → Rn a homeomorphism, let g = f−1 and suppose that Mq(f(Γ))
≤ γ(Mωp(Γ)) for every Γ ∈ A(Rn) and that there exists K > 0 so that R
B(x,δ)ω(z)dz≤K·δp for every x∈Rn and every δ >0. Then L(y, g, r)
l(y, g, r) ≤C(r) = exp exp
K·ep·
∞
X
k=1
1
kp/γ−1(Cn,q·rn−q) p1!!
e
for every y∈Rn and every r >0.
Remark1. We see that the homeomorphismg:Rn→Rnfrom Corollary 2 satisfies a kind of quasisimmetry relation, i.e., if r > 0 there exists C(r) >0
so that L(y,g,r)l(y,g,r) ≤ C(r) for every y ∈ Rn and every δ ≥ r, hence for enough great balls B(y, r) their dilatation L(y,g,δ)l(y,g,δ) is uniformly bounded on Rn.
Theorem 2. Let n ≥ 2, n−1 < p ≤ n, D, D0 domains from Rn, f :D→D0 a homeomorphism so thatg=f−1is ACLp, a.e. differentiable and Jg(y)6= 0a.e., letx∈D,y=f(x)and suppose that there exists Kx>0, rx>
0,0≤αx < p−1so thatR
B(x,δ)NI,p(f)(z)dz≤Kx·δp· ln e·rδxαx
for every 0< δ≤rxand letC(r) = exp
exp
Kx·ep·
∞
P
k=1 1
kp−αx/(Cn,prn−p)1p /e for r >0. Then there existsρx>0so that L(y,g,r)l(y,g,r) ≤C(r)for every0< r≤ρx
and if p = n, then H(y, g) ≤ H(p, n, γ, Kx, αx). If D = D0 = Rn and there exists Kx > 0 so that R
B(x,δ)NI,p(f)(z)dz ≤ Kx·δp for every δ > 0, then
L(y,g,r)
l(y,g,r) ≤C(r) for everyr >0.
Proof. Let Γ ∈A(D), let ρ ∈ F(Γ) and letρ0 :Rn → [0,∞] be defined by ρ0(z) =ρ(g(z))·L(z, g) if z∈ D0, ρ0(z) = 0 otherwise. Let A ={z ∈D | f is differentiable inz and Jf(z) 6= 0} and B = {z ∈ D0 | g is differentiable in z and Jg(z) 6= 0}. Then CB = 0 and A = g(B), hence A 6= φ and let Γ0 = {β ∈ f(Γ) |β is locally rectifiable and g◦β0 is absolutely continuous}.
We see from Theorem 5.3, page 12 from [22] thatρ0∈F(Γ0). Using Fuglede’s theorem (see Theorem 28.2, page 95, from [22]) and the change of variable formulae (3) from [10] we find that
Mp(f(Γ)) =Mp(Γ0)≤ Z
Rn
ρ0p(z)dz= Z
D0
ρp(g(z))·L(z, g)pdz=
= Z
D0
ρp(g(z))· |g0(z)|pdz= Z
B
ρp(g(z))·N0,p(g)(z)· |Jg(z)|dz=
= Z
B
ρp(g(z))·NI,p(f)(g(z))· |Jg(z)|dz≤ Z
A
ρp(u)·NI,p(f)(u)du≤
≤ Z
Rn
ρp(u)·NI,p(f)(u)du.
It results thatMp(f(Γ))≤MNp
I,p(f)(Γ) for every Γ∈A(D) and we apply now Theorem 1 with ω=NI,pp (f) andγ(t) =tfor every y≥0.
Corollary 3. Let n ≥ 2, D, D0 domains from Rn, f : D → D0 a homeomorphism so that g =f−1 is ACLn, a.e. differentiable and Jg(y) 6= 0 a.e. and suppose that there exists K > 0, a > 0 so that for every x ∈ D there exists rx > 0 and 0 ≤ αx < n−1−a, so that R
B(x,δ)NI,n(f)(z)dz ≤
Kx ·δn· ln erδxαx
for every 0 < δ ≤ rx. Then f is quasiconformal and H∗(x, f)≤exp
exp
K·en·
∞
P
k=1 1
k1+a/C(n) 1n
/e for everyx∈D.
Corollary 4. Letn≥2,n−1< p≤n,f :Rn→Rna homeomorphism so that g = f−1 is ACLp, a.e. differentiable and Jg(y) 6= 0 a.e. and so that there exists K >0 so thatR
B(x,δ)NI,p(f)(z)dz≤K·δp for everyx∈Rn and every δ > 0 and let C(r) = exp
exp
K·ep·
∞
P
k=1 1
kp/(Cn,prn−p)1p /e.
Then L(y,g,r)l(y,g,r) ≤C(r) for everyy∈Rn and every r >0.
Corollary 5. Let n ≥ 2, D, D0 domains from Rn, f : D → D0 a homeomorphism satisfying condition(N)and having an ACLn inverse so that there exists Q > 0 so that lim sup
r→0
R−
B(x,r)
NI,n(f)(z)dz ≤ Q for every x ∈ D.
Then f isK(Q, n)-quasiconformal.
Proof. Let g=f−1. We see from Lemma 6.7, page 190 from [17] that g is a.e. differentiable and satisfies condition (N). Let Dg ={y ∈D0 |g is not differentiable in y} andZg ={y∈D0 |g is differentiable iny and Jg(y) = 0}.
Using Sard’s lemma (see [2]) and the fact that g satisfies condition (N), we see that µn(g(Dg∪Zg)) = 0 and since f also satisfies condition (N), we see that µn(Dg∪Zg) = 0. We proved that g is a.e. differentiable and Jg(y) 6= 0 a.e. We apply now Corollary 3.
A related result with Corollary 1 may be found in [9].
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Received 25 February 2012 University of Bucharest
Faculty of Mathematics and Computer Sciences Str. Academiei 14
010014 Bucharest, Romania [email protected]