SHARPNESS OF THE ASSUMPTIONS FOR THE
REGULARITY OF A HOMEOMORPHISM
STANISLAV HENCL
The recent result shows that a homeomorphism 2
11
(
R ) of nite distortion satises 1 2 loc
( (
) R ).
We show that this result is sharp in a sense that the crutial regularity condition j j 2 1 cannot be replaced by j adj j 2 1
or by a requirement that j j belongs to some bigger Orlicz space.
Abstract.
1;n 1
loc
W
f
n
;
f
Df
L
W
;
f
n
Df
;
n
L
Df
1. Introduction
1;1 (
; Rn )
Let Rn be an open set. We say that a mapping f 2 Wloc
has nite (outer) distortion if Jf (x) 0 almost everywhere and Jf (x) =
1;1 (
; Rn )
0 ) jDf (x)j = 0 a.e. Moreover we say that a mapping f 2 Wloc
has nite inner distortion if Jf (x) 0 almost everywhere and Jf (x) =
0 ) j adj Df j = 0 a.e. (for basic properties, examples and applications
see e.g. [10]). Here adj A means an adjugate matrix - see Preliminaries
for the denition.
Our aim is to show the sharpness of the following recent result from
[1] (see also [6], [7], [14], [11] and [8]):
1;n 1 (
; Rn ) be a
Theorem 1.1. Let Rn be an open set and f 2 Wloc
1;1 (f (
); Rn )
homeomorphism of nite inner distortion. Then f 1 2 Wloc
and f 1 is a mapping of nite outer distortion. Moreover
(1.1)
Z
f
(
)
jDf
1 (y )j dy
=
Z
j adj Df (x)j dx:
This statement is actually claimed in [1] only for mappings of nite
outer distortion. However with a very slight modication of the arguments given there (see Section 3 for details) it is possible to show
the statement also for a wider class of mappings of nite inner distortion (see also [4]). Also formula (1.1) is not shown there, but it was
previously shown under stronger assumptions in [7] and under W 1;n 1
regularity assumption in [16]. Let us also note that the assumption
that f has nite inner distortion is not articial, because it was shown
1;1 , J 0
in [9, Theorem 4] that each homeomorhism such that f 2 Wloc
f
1;1 is necessarily a mapping of nite inner distortion.
a.e. and f 1 2 Wloc
2000 Mathematics Subject Classication. 26B30, 46E35.
Key words and phrases. Sobolev mapping, inverse.
The author was supported by MSM 0021620839.
1
2
STANISLAV HENCL
Our aim is to show that assumptions of Theorem 1.1 are sharp in
n 1
a sense that the crutial regularity condition jDf j 2 Lloc
cannot be
weakened. From the inequality (1.1) one may be tempted to believe
that for a conclusion Df 1 2 L1 is could be enough to assume that
adj Df 2 L1. We show that this is not true:
Example 1.2. Let 0 < " < 1 and n 3. There exist a domain Rn
and homeomorphism f 2 W 1;n 1 " (
; Rn ) such that j adj Df j 2 L1 (
),
pointwise derivative rf 1 exists a.e. in f (
), but jrf 1 j 2= L1 (f (
)).
It is known that for any n 3 and 0 < " < 1 there exists homeo1;1 (see [8, Example 3.1] or
morphism f 2 W 1;n 1 " such that f 1 2= Wloc
Example 1.2) and therefore Theorem 1.1 is sharp on a scale of Sobolev
spaces. Let us note that for many problems connected with the theory
of mapping of nite distortion the optimal regularity of Df is not on
the Lebesgue scale, but on some ner Orlicz scale (see [13] and references given there). We show that this is not the case for Theorem 1.1
and no smaller integrability condition of Df is enough.
Example 1.3. Let n 3 and suppose that g : [0; 1) ! (0; 1) is a
decreasing function such that
lim
g(s) = 0:
!1
s
2 W 1;1(B (0; 1); Rn) such that
jDf (x)jn 1g(jDf (x)j) dx < 1;
Then there is a homeomorphism f
(1.2)
Z
B
(0;1)
pointwise derivative rf 1 exists almost everywhere in f (B (0; 1)), but
jrf 1j 2= L1loc(f (B (0; 1))).
Let us point out that the conclusion of our examples that rf 1 exists
and is not integrable imply that f 1 2= W 1;1 and even that f 1 2= BV .
2. Preliminaries
The Lebesgue measure of a set A Rn is denoted by Ln(A).
Given a square matrix B 2 Rnn, we dene the norm jB j as the
supremum of jBxj over all vectors x of unit euclidean norm. The
adjugate adj B of a regular matrix B is dened by the formula
(2.1)
B adj B = I det B;
where det B denotes the determinant of B and I is the identity matrix.
The operator adj is then continuously extended to Rnn.
2.1. Dierentiability of radial functions. By kxk we denote the
norm of x 2 Rn, in fact we use either euclidean norm or maximum
norm kx yk = maxfjxi yij : i = 1; : : : ; ng. The following lemma
can be veried by an elementary calculation for the euclidean norm.
The maximum norm can be obtained from the euclidean norm by the
REGULARITY OF THE INVERSE - SHARPNESS
3
bilipschitz change of variables and therefore it is easy to check that the
formulas hold also for this norm.
Lemma 2.1. Let : (0; 1) ! (0; 1) be a strictly monotone, dieren-
tiable function. Then for the mapping
x
f (x) =
kxk (kxk); x 6= 0
we have for almost every x
n (kxk)
o
(kxk) n 1
0
0
Df (x) max
kxk ; j (kxk)j ; Jf (x) (kxk) kxk
n (kxk)
o (kxk) n 2
and j adj Df (x)j max
; j0 (kxk)j
:
kxk
k xk
2.2. Area formula. We say that a mapping f : ! Rn satises the
Lusin condition (N ) if the implication jS j = 0 =) jf (S )j = 0 holds
for any measurable set S .
1;1 (
; Rn ) be a homeomorphism and let be a nonLet f 2 Wloc
negative Borel-measurable function on Rn. Without any additional
assumption we have
Z
Z
(2.2)
(f (x))jJf (x)jdx (y)dy:
Rn
Moreover there exists a set of full measure such that the area
formula holds for Zf on 0:
Z
(y)dy
(f (x))jJf (x)jdx =
(2.3)
0
0
f
(
0 )
Also, the area formula holds on each set on which the Luzin condition (N ) is satised. This follows from the area formula for Lipschitz
mappings, from the a.e. approximate dierentiability of f [3, Theorem 3.1.4], and a general property of a.e. approximately dierentiable
functions [3, Theorem 3.1.8], namely that can be exhausted up to a
set of measure zero by sets the restriction to which of f is Lipschitz
continuous.
3. Finite inner distortion
The following lemma from [1, Lemma 4.3] contains the main ingredient for the proof of Theorem 1.1.
1;n 1 (
; Rn ) be a homeomorphism. Then
Lemma 3.1. Let f 2 Wloc
(3.1)
Z
B
jf
1 (y )
cj dy Cr0
Z
( )
f 1 B
for each ball B = B (y0 ; r0 ) f (
), where
c=
Z
B
f 1 (y) dy
j adj Df (x)j dx;
4
STANISLAV HENCL
and C = C (n).
The following theorem was shown in [1, Theorem 4.5].
1;n 1 (
; Rn ) be a
Theorem 3.2. Let Rn be an open set and f 2 Wloc
1;1 (f (
); Rn ) and J 0 a.e. Then
homeomorphism such that f 1 2 Wloc
f
f 1 is a mapping of nite outer distortion.
In order to prove the equality (1.1) we will need the following technical lemma from [4, Lemma 2.1].
Lemma 3.3. Let f : ! Rn be a homeomorphism such that f 2
1;1 (
; Rn ) and f 1 2 W 1;1 (f (
); Rn ). Set
Wloc
loc
E = y 2 f (
) : f 1 is approximatively dierentiable at y
and jJf 1 (y)j > 0 :
Then there exists a Borel set A E such that jE n Aj = 0,
f 1 (A) E~ := fx 2 : f is approximatively dierentiable at x
and jJf (x)j > 0g
(3.2)
and Df 1 (y) = Df (f 1 (y)) 1
Moreover jE~ n f 1 (A)j = 0.
for every y 2 A:
Proof. It is enough to show that jE~ n f 1 (A)j = 0, because everything
else is stated and shown in [4, Lemma 2.1]. Suppose for contradiction
that there is a Borel set G E~ nf 1(A) such that jGj > 0. Without loss
of generality we can also suppose that (2.3) holds for G (i.e. G 0)
and thus
Z
Z
Jf (x) dx =
f (G) (y) dy = jf (G)j:
Rn
G
Since Jf > 0 on G we obtain that jf (G)j > 0. We know that the area
formula holds for f 1 on a Borel subset M f (G) of full measure.
From
Z
Jf (x) dx = jM j > 0
we obtain that jf
f 1 to conclude
1 (M )j
Z
( )\M
f G
( )
f 1 M
> 0. Therefore we can use area formula for
jJf (y)j dy = jG \ f 1(M )j > 0:
1
It follows that Jf 1 > 0 on a subset of f (G) of positive measure. Clearly
f 1 is approximatively dierentiable a.e. on f (G) and therefore f (G) \
A 6= ; gives us a contradiction.
REGULARITY OF THE INVERSE - SHARPNESS
5
Proof of Theorem 1.1. We claim that there is a function g 2 L1loc (f (
))
such that
Z
(3.3)
( )
f 1 B
j adj Df j =
Z
B
g:
This and Lemma 3.1 imply that the pair f; g satises a 1-Poincare
inequality in f (
): From [2, Theorem 9] we then deduce that f 1 2
1;1 (f (
); Rn ).
Wloc
There is a set 0 of full measure such that the area formula
(2.2) holds for f on 0. We dene a function g : f (
) ! R by setting
(
j adj Df (x)j if x 2 0 and J (x) > 0;
f
Jf (x)
g(f (x)) =
0
otherwise.
Since f is a mapping of nite inner distortion, we have
j adj Df (x)j = g(f (x)) Jf (x) a.e. in :
Hence for every A f (
)
Z
(3.4)
( )
f 1 A
j adj Df (x)j dx =
=
Z
( )\
0
f 1 A
Z
A
g(f (x)) Jf (x) dx
g(y) dy:
For A = B this gives (3.3) and for other sets A it also implies g 2 L1loc.
1;1 and from Theorem 3.2 we obtain that f 1 has nite
Hence f 1 2 Wloc
outer distortion.
We will use Lemma 3.3 to prove (1.1). First let us notice that the
Lusin (N ) condition is valid on f 1(A) and therefore we can use (2.3)
there. Indeed, let S f 1(A) be a set of measure zero and let us nd
a Borel measurable set S1 S of measure zero. We can use (2.2) for
f 1 and = S1 and we obtain
Z
( )
f S1
jJf j jS1j = 0:
1
Since Jf 1 > 0 on A it follows that jf (S1)j = 0. Since f 1 is a mapping
of nite distortion and each W 1;1 function is approximatively dierentiable almost everywhere we obtain
Z
f
and analogously
Z
(
)
jDf
1 (y )j dy
=
Z
E
Z
~
E
jDf 1(y)j dy
j adj Df (x)j dx = j adj Df (x)j dx
6
STANISLAV HENCL
since f is a mapping of nite inner distortion. Now we can use jE n Aj =
0, (2.3), (3.2), (2.1) and jE~ n f 1(A)j = 0 to obtain
Z
f
(
)
jDf
=
=
1 (y )j dy
Z
( )
f 1 A
Z
( )
f 1 A
=
jDf
Z
A
jDf 1(y)j dy
1 (f (x))j J
f
(x) dx =
Z
Z
( )
f 1 A
j(Df (x)) 1jJf (x) dx
j adj Df (x)j dx = j adj Df (x)j dx:
4. Construction of examples
In this section we use a notation Q(c; r) for an open cube in Rn 1
centered at c with edge length 2r.
One of the main ingredients of the proof of Lemma 3.1 is the fact
that homeomorphism f 2 W 1;n 1 must satisfy the (n 1)-dimensional
Lusin (N ) condition on almost all hyperplanes. First we construct an
auxiliary mapping that fails the Lusin (N ) condition in Rn 1. For a
construction of a homeomorphism that does not satisfy the Lusin condition (N ) we use Cantor type construction from [12] (see also [15],[5]).
Example 4.1. Let 0 < " < 1 and n 3. There is a homeomorphism
g 2 W 1;n 1 " (( 1; 1)n 1 ; ( 1; 1)n 1 ) such that Jg 2 L1 (( 1; 1)n 1 )
and j adj Dg j 2 L1 (( 1; 1)n 1 ), but g does not satisfy the Lusin condition (N ).
Proof. By V we denote the set of 2n vertices of the cube [ 1; 1]n 1 .
The sets Vk = V : : : V, k 2 N, will serve as the sets of indices for
our construction.
Let us denote
1
1
1
(4.1)
ak = and bk = 1 + n 1 :
k
2
k
Set z0 = z~0 = 0 and let us dene
(4.2)
rk = ak 2 k and r~k = bk 2 k :
It follows that ( 1; 1)n 1 = Q(z0; r0) and further we proceed by induction. For v = [v1; : : : ; vk ] 2 Vk we denote w = [v1; : : : ; vk 1] and we
dene
k
X
1
1
zv = zw + rk 1 vk = z0 +
2
2 j=1 rj 1vj ;
Q0v = Q(zv ; rk2 1 ) and Qv = Q(zv ; rk ):
REGULARITY OF THE INVERSE - SHARPNESS
7
Fig. 1. Cubes Qv and Q0v for v 2 V1 and v 2 V2 .
The number of the cubes fQv : v 2 Vk g is 2(n
to nd out that the resulting Cantor set
1 [
\
k
=1 v2Vk
1)k .
It is not dicult
Qv =: CA = Ca : : : Ca
is a product of n 1 Cantor sets in R. Moreover Ln 1(CA) = 0 since
Ln
[
1
v 2Vk
Qv = 2(n 1)k (2ak 2 k )n 1 k!1
! 0:
Analogously we dene
k
X
1
1
z~v = z~w + r~k 1 vk = z~0 +
2
2 j=1 r~j 1vj ;
Q~ 0v = Q(~zv ; r~k2 1 ) and Q~ v = Q(~zv ; r~k ):
The resulting Cantor set
1 [
\
k
=1 v2Vk
Q~ v =: CB = Cb : : : Cb
satises Ln 1(CB ) > 0 since limk!1 bk > 0. It remains to nd a homeomorphism g which maps CA onto CB and satises our assumptions.
Since Ln 1(CA) = 0 and Ln 1(CB ) > 0 we will obtain that g does not
satisfy the (N ) condition.
P~ 0
P0
@@ P~
@
@P
@@
g-k
@@
Fig. 2. The transformation of Q0 n Q onto Q~ 0 n Q~ Again we will proceed by induction and we will nd a sequence of
homeomorphisms gk : ( 1; 1)n 1 ! ( 1; 1)n 1. We set g0(x) = x and
8
STANISLAV HENCL
for k 2 N8we dene
S
>
for x 2= v2Vk Q0v
<gk 1 (x)
gk (x) = gk 1 (zv ) + (k kx zv k + k ) kxx zzvv k for x 2 Q0v n Qv ; v 2 Vk
>
:
gk 1 (zv ) + rr~kk (x zv )
for x 2 Qv ; v 2 Vk
where the constants k and k are given by
(4.3)
k rk + k = r~k and k rk2 1 + k = r~k2 1 :
It is not dicult to nd out that each gk is a homeomorphism and
maps
[
[
Qv onto
Q~ v :
v 2Vk
v 2Vk
The limit g(x) = limk!1 gk (x) is clearly one to one and continuous
and therefore a homeomorphism. Moreover it is easy to see that g is
dierentiable almost everywhere, absolutely continuous on almost all
lines parallel to coordinate axes and maps CA onto CB .
Let k 2 N and v 2 Vk . We need to estimate Dg(x), j adj Dgj and
Jg (x) in the interior of the annulus Q0v n Qv . Since
x zv
g(x) = g(zv ) + (k kx zv k + k )
kx z k
v
there, we can use Lemma 2.1, rk rk 1, r~k r~k 1 (4.3), (4.2) and
(4.1) to obtain
o
n
n r~
1 o
Dg(x) max k ; k max k; n 2 k;
rk
k
r~ n 3
n 2
j adj Dg(x)j jDf (x)j k
kn 2 and J (x) r~k
1:
rk
g
It follows that Jg 2 L1((
rk
k
1; 1)
Moreover we can estimate
Ln 1(Q0v n Qv ) = (rk 1)n 1 (2rk )n 1 2 k(n 1) k1n
and we have 2(k 1)n annuli like that. Therefore
Z
jDg(x)j
n
Q0
1
"
dx 1 XZ
X
k
=1 v2Vk
C
1
X
k
and
Z
Q0
1 ).
n
j adj Dg(x)j dx =1
2(k
Q0v Qv
n
jDg(x)jn
1)n 2 k(n 1)
1 XZ
X
k
=1 v2Vk
C
1
X
k
=1
2(k
Q0v Qv
n
1
1 kn
kn
"
1
dx
"
<1
j adj Dg(x)j dx
1)n 2 k(n 1)
1 k n 2 < 1:
kn
REGULARITY OF THE INVERSE - SHARPNESS
9
Proof of Example 1.2. In this example we will use notation and results
from Example 4.1. Set
h
f (x) = g1 ([x1 ; : : : ; xn 1 ]); : : : ; gn 1 ([x1 ; : : : ; xn 1 ]); e
Furher we dene
= CA (0; 1) [
1[
[
k
=1
Q0
i
:
v
v 2Vk
xn
n Qv (0; log(k + 1)):
Clearly f is a homeomorphism and both f and f 1 are dierentiable
almost everywhere. Moreover it is easy to check, that ( 1; 1)n 1 (0; 1) is an open set.
The matrix Df has a special form, because only one term in the
@fn
last column and in the last row is non-zero. This is the term @x
and
n
therefore it is easy to check that
@e
n
x n o
j adj Df (x)j max jJg (~x)j; j adj Dg(~x)j @x
n
;
where x~ = [x1; : : : ; xn 1]. From
XX
Ln(
) =
Ln 1 Q0v n Qv log(k + 1)
k
=
and jJg j 2 L1((
Z
2N v2Vk
X
k
2N
2(k
1)n 2 k(n 1)
1 log(k + 1) < 1
kn
1; 1) 1) we obtain jJg (~x)j 2 L1(
). Further
n
@e
Z
xn Z
1
j adj Dg(~x)j @x dx j adj Dgj
e x dxn < 1
n
( 1;1)
0
1
and hence j adj Df j 2 L (
). Moreover
Df (x) = maxfjDg(~x)j; j @e@x jg jDg(~x)j
n
n 1
xn
n
and therefore
Z
jDf (x)j
n
1
"
dx 1 X Z
X
k
=1 v2Vk
C
1
X
k
=1
2(k
Q0v Qv
n
jDg(~x)j
1)n 2 k(n 1)
n
1 kn
kn
1
1
"
dx~ log(k + 1)
"
log(k + 1) < 1:
Since CA (0; 1) we obtain that
f 1 f[y; t] 2 f (
) : t 2 (0; 1)g = g 1 (y) (0; 1) for every y 2 CB
and thus
Z 1
Z 1
1
@f
1
jrf (y; t)j dt @t (y; t) dt = 1:
0
0
Since Ln 1(CB ) > 0 we obtain that jrf 1j 2= L1(f (
)).
10
STANISLAV HENCL
Remark 4.2. Let us note that the fact that is unbounded is not
essential for our arguments, it only makes them simpler. It would be
possible to twist our and to obtain a bounded domain with the same
properties.
5. Sharpness on the Orlicz scale
Lemma 5.1. Let h : (0; 1) ! (0; 1) be an increasing function such
that limt!0+ h(t) = 0. Then there is a function f : (0; 1) ! (0; 1)
such that limt!0+ f (t) = 0,
Z 1
Z 1
f (t)h(t)
f (t)
dt = 1 and
dt < 1:
0
t
0
t
Proof. We can easily nd an increasing dierentiable
function h1 h
th01 (t)
that satises limt!0+ h1(t) = 0 and limt!0+ h1(t) = 0 which is some
sort of strong concavity near 0. Thus we may assume without loss of
generality that h is dierentiable and that the function
th0 (t)
(5.1)
f (t) :=
satises tlim
f (t) = 0:
!0+
h(t)
An elementary computation gives us
Z 1
Z 1 0
h
it=1
f (t)
h (t)
dt =
dt = log h(t)
=1
t=0
0 t
0 h(t)
Z 1
Z 1
h
it=1
f (t)h(t)
0
and
< 1:
dt = h (t) dt = h(t)
0
t
=0
0
t
Proof of Example 1.3. We write ei for the i-th unit vector in Rn , i.e.
the vector with 1 on the i-th place and 0 everywhere else. Given x =
[p
x1 ; : : : ; xn ] 2 Rn we denote x~ = [x1 ; : : : ; xn 1 ] 2 Rn 1 and kx~k =
x21 + : : : + x2n 1 .
From Lemma 5.1 we can nd a function a : (0; 1) ! (0; 1) such
that
lim
a(t) = 0;
!0+
t
(5.2)
(5.3)
Z 1
an 1 (t)
dt = 1
t
0
and
Z 1
an 1 (t) 1 g p dt < 1:
t
t
0
Without loss of generality we may also suppose that
1 a(t) for every t 2 (0; 1 )
(5.4)
2
2
log n 1 1t
REGULARITY OF THE INVERSE - SHARPNESS
11
since the integral in (5.2) is nite for the left hand side. Therefore it
is easy to see that without loss of generality we can also assume that
a is increasing and concave.
Set
n 1
X
xi
a(kx~k) + en xn + kx~k sin a(kkx~x~kk)
kx~k
i=1
if kx~k > 0 and f (x) = enxn if kx~k = 0. Our mapping f is clearly
continuous and it is easy to check that f is a one-to-one map since
z
xi
a(kx~k) = i a(kz~k) for every i 2 f1; : : : ; n 1g )
kx~k
kz~k
) a(kx~k) = a(kz~k) which implies kx~k = kz~k and hence
xi = zi for every i 2 f1; : : : ; n 1g:
Therefore f is a homeomorphism.
By Lemma 2.1 we obtain that the partial derivatives of fi, i 2
f1; : : : ; n 1g, are smaller than
n a(kx
o
~
k
)
a(kx~k)
0
(5.5)
C max
; a (kx~k) C
kx~k
kx~k ;
since a is concave and a(0) = 0. Moreover,
@fn (x)
=
x1 kx~k 1 sin a(kkx~x~kk)
@x1
(5.6)
a0 (kx
~
k
)
x1 a(kx~k)x1 a(kx
~ k) cos
+ kx~k kx~k2
kx~k
kx~k3
f (x) =
ei
can be also bounded by (5.5). Analogously we can bound other derivatives of fn and therefore we can use substitution to spherical coordinates in Rn 1 to obtain
Z
Z
a(kx~k)n 1 a(kx~k) n 1
jDf (x)j g(jDf (x)j) dx C
kx~kn 1 g C kx~k dx
B (0;1)
B (0;1)
Z 1
n 1 C a(t) g C a(t) tn 2 dt:
0
tn 1
t
From (5.4) we can nd " > 0 such that for every t 2 (0; ") we have
C a(tt) p1t and therefore the last integral is nite by (5.3) and the
condition (1.2) follows.
The inverse of f is given by
f
1 (y ) =
n 1
X
yi 1
ei a (ky~k) + en yn
ky~k
i=1
a
1 (ky~k) sin
a
ky~k 1 (ky~k)
if ky~k > 0 and f 1(y) = enyn if ky~k = 0. The dierential of f
clearly continuous outside the segment f[0; ; 0; t] : t 2 Rg.
1
is
12
STANISLAV HENCL
Analogously to (5.6) we obtain
@ (f 1 )n (y)
=(a 1)0(ky~k)y ky~k 1 sin
1
@y1
ky~k (ky~k)
a 1
1 0
+ a 1(ky~k) ky~ka y11(ky~k) y1a(a1(k)y~(kk)y~2k) cos
It follows that we can nd > 0 such that
1
(a 1)0(ky~k)
@ (f )n (y ) (5.7)
C
k
y
~
k
@y
a 1 (ky~k)
ky~k :
(ky~k)
a 1
1
for every
y 2 S := y 2 B (0; ) : y1 > 12 jjy~jj; j cos
ky~k j p2 :
(ky~k)
2
a 1
Here we have also used the fact that (5.4) gives us
a 1 (y) exp n12 1 for small enough y:
y
Clearly Ln(S ) = C Ln(G) for
G := y 2 B (0; ) : cos
and thus we can use (5.7) to obtain
Z
(5.8)
( (0;1))
f B
jDf
1 (y )j dy
ky~k p2 (ky~k)
2
a 1
Z 1
@ (f )n (y ) @y1
Z
C ky~k (aa
G
S
Now let us consider a mappings
n 1
X
dy
1 )0 (ky~k)
1 (ky~k)
dy:
xi
a(kx~k) + en xn + kx~k cos a(kkx~x~kk)
kx~k
i=1
if kx~k > 0 and h(x) = enxn if kx~k = 0. Analougously as above we
obtain that h is a homeomorphism that satises (1.2) and that
Z
Z
1 0
1
(5.9)
jDh (y)j dy C ~ ky~k (aa 1)(k(ky~yk~k) ) dy
G
h(B (0;1))
h(x) =
where
ei
p G~ = y 2 B (0; ) : sin a 1k(y~kky~k) 22
for some possibly smaller . By the formula of change of variables and
(5.2) we obtain that
Z Z
1 0
(
a 1 )0 (ky~k)
ky~k a 1(ky~k) dy C s (aa 1)(s()s) sn 2 ds
0
(5.10) B(0;)
Z a 1 ( )
1
C
a(t) a(t)n 2 dt = 1:
t
0
REGULARITY OF THE INVERSE - SHARPNESS
13
From (5.8), (5.9), G [ G~ = B (0; ) and (5.10) we obtain that either
rf 2= L1 or rh 2= L1 which is the desired conclusion.
References
[1] M. Csornyei, S. Hencl and J. Maly, Homeomorphisms in the Sobolev space
1
1 , preprint 2007.
[2] B. Franchi, P. Hajlasz and P. Koskela, Denitions of Sobolev classes on metric
spaces, Ann. Inst. Fourier 49 no. 6 (1999), 1903{1924.
[3] H. Federer, Geometric measure theory, Die Grundlehren der mathematischen
Wissenschaften, Band 153 Springer-Verlag, New York, 1969 (Second edition
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[4] N. Fusco,G. Moscariello and C. Sbordone, The limit of 1 1 homeomorphisms
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1267{1285.
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22{32.
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[11] T. Iwaniec and J. Onninen, Deformations of nite conformal energy: existence,
and removability of singularities, preprint 2008.
[12] J. Kauhanen, P. Koskela and J. Maly, Mappings of nite distortion: Condition
N, Michigan Math. J. 49 (2001), 169{181.
[13] J. Kauhanen, P. Koskela, J. Maly, J. Onninen and X. Zhong, Mappings of nite
distortion: Sharp Orlicz-conditions, Rev. Mat. Iberoamericana 19 (2003), 857{
872.
[14] J. Onninen, Regularity of the inverse of spatial mapping with nite distortion,
Calc. Var. Partial Dierential Equations 26 no. 3 (2006), 331{341.
[15] S. Ponomarev, Examples of homeomorphisms in the class
which do
not satisfy the absolute continuity condition of Banach (Russian), Dokl. Akad.
Nauk USSR 201 (1971), 1053-1054.
[16] K. Quittnerova, Funkce vce promennych s konecnou variac, diploma thesis.
W
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AC T L
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Department of Mathematical Analysis, Charles University, Soko 83, 186 00 Prague 8, Czech Republic
lovska
E-mail address : [email protected]