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Isoperimetric inequalities for Lp-norms of the distance function to the boundary.

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УЧЕНЫЕ ЗАПИСКИ КАЗАНСКОО ОСУДАСТВЕННОО УНИВЕСИТЕТА
Физико-математические науки
2006
Том 148, кн. 2
UDK 517.5
ISOPERIMETRIC INEQUALITIES FOR
Lp -NORMS
OF THE DISTANCE FUNCTION TO THE BOUNDARY
R.G. Salahudinov
Abstrat
The main goal of the paper is to prove that Lp -norms of dist(x, ?G) and dist?1 (x, ?G)
are dereasing funtions of p , where G is a domain in Rn (n ? 2) . We also obtain a sharp estimation of the rate of dereasing for these norms using Lp norms of the distane funtion for a
onsistent ball. We prove a new isoperimetri inequality for Lp norms of dist(x, ?G) , this inequality is analogous to the inequality of Lp norms of the onformal radii (see Avkhadiev F.G.,
Salahudinov R.G. // J. of Inequal.& Appl. 2002. V. 7, No 4. P. 593601).
Note that L2 -norm of dist(x, ?G) plays an important role to investigate the torsional
rigidity in Mathematial Physis (see, for instane, Avkhadiev F.G. // Sbornik: Math. 1998.
V. 189, No 12. P. 17391748; Ba
nuelos R., van den Berg M., Carroll T. // J. London Math.
So. 2002. V. 66, No 2. P. 499512). As a onsequene we get new inequalities in the
torsional rigidity problem.
Also we generalize the n -dimensional isoperimetri inequality.
Introdution
Let
G
be a domain in
n
R (n ? 2) .
tional [1, 2?
I(p, G)
=
Let us onsider the following geometrial fun-
Z
dist
p
(x, ?G) dA.
(1)
G
Here dist(x, ?G) denotes the distane funtion from
x ? G to the boundary ?G , dA
dx1 , . . . , dxn , and p ? ?1 . In [1? it was noted, that it might be
I(p, G) the p -order eulidean moment of G with respet to its boundary.
is the volume element
justied to all
First we remark some appliations of (1), and further we note onnetion of (1) with
a new oneption of
isoperimetrial monotoniity.
0 -order moment is the volume of G . Also,
d(G) = max dist(x, ?G) . We also found that for a
It is lear that
we an get
as a limit ase of (1),
wide lass of domains
x?G
(?1) -order moment is, up to a onstant fator, the surfae area of G .
Further, we remark a new geometrial funtional of
G
that reently appears in the
elasti torsion problem. This is the seond order eulidean moment or, otherwise, the
eulidean moment of inertia with respet to the boundary. Consider the boundary value
problem
?u = ?1
where
??
in
G,
u=0
on
?G,
is the Dirihlet Laplaian, and let
P(G)
:= 4
Z
u(x) dA,
G
P(G) is exatly the torsional rigidity of a simply onneted domain
all P(G) the torsional rigidity of
G,
even if
n>2
and/or
G
G . Just as in [3? we
is not simply onneted.
R.G. SALAHUDINOV
152
We begin from the ase
n = 2.
In 1995 F.G. Avkhadiev [1? proved the two-sided
inequality
I(2, G)
? P(G) ? 64 I(2, G).
(2)
Later in [4? the left-side hand of the inequality was improved to
Let us remark that an
n -dimensional
3I(2, G) < 2 P(G) .
generalization of was reently proved in [5?
2
I(2, G) ? P(G) ? CG I(2, G),
n
under the additional restrition that
onstant (see [6?), where
CG
G
satises a strong Hardy inequality with some
is a funtional on
G.
In partiular, the two-sided inequali-
When is the torsional rigidity of a simply onneted
ties (2), (3) answered the question: Rn
, however for
domain in
(3)
bounded?
n?3
the question is still open.
Another appliation of (1) was disovered by F.G. Avkhadiev in [2?, this is also
n -dimentional
generalize (2) on an
Kp,q (G) =
ase. Consider the funtional
sup
f ?C0? (G)
?
?
Kp,q (G)
?
?1/q
ZZ
q
?
|f | (x) dA?
G
ZZ
G
?1/p .
|grad f |p (x) dA?
appears in the Poinare Sobolev's inequality. In [2? it was proved the two
sided estimations for
Kp,q (G)
using (1).
The rst property of isoperimetri monotoniity was onjetured by J. Hersh [7?,
and it was proved by M.-Th. Kohler-Jobin [8?. We shall onsider the following boundary
value problem [9?
?v + ?v + 1 = 0
in
G,
v=0
on
?G ( ? ? < ? < ?1 (G))
(4)
and the orresponding funtional
Q(?) :=
Z
v dA.
G
In partiular, we have
A(G)
Q(0) = P(G)/4 , Q(?1 ) = ?j04 (8?21 (G)) ,
and
(??)Q(?) ??
????
(see, for example, [12?).
By
Q(?)
we denote the radius of a ball in
Rn
with same
Q(?) .
Let G be a bounded domain and not a ball in Rn , then Q(?)
is a dereasing funtion on ? . If G is a ball, then Q(?) is a onstant funtion.
Theorem A [8, 10?.
This monotoniity property ontains several well-known plane isoperimetri inequal-
?1 (G) , P(G) , and A(G) . On the other hand,
?1 (G) are expressing in terms of I(?, G) (see [1,
ities (see [12?), whih inlude funtionals
existene theorems for P(G) , and
11?). Further, other monotoniity properties were disovered by C. Bandle [13?, and by
J. Hersh [12?. All of these isoperimetri monotoniity properties were onneted with
the solutions of the dierential equations (4). So, the onjeture on the isoperimetri
monotoniity property of (1) is a geometrial analog of Theorem A. On the other hand,
this work was intended as an attempt to bind ontinuously the well-known geometrial
Rn suh as the surfae area, the volume, the radius of the
quantities of a domain in
largest ball ontained in the domain.
ISOPERIMETRIC INEQUALITIES FOR LP -NORMS OF THE DISTANCE . . .
1.
Let
? > n ? 1.
B1
:= [I(? ? n, B1 )]
Rn ,
is a unit ball in
Theorem 1.
< ?.
Z
Then
Main results and orollaries
Further it would be suitable to use a onstant
?,n
where
153
and
?1
?(·)
?(n /2 )?(? + 1)
,
2? n/2 ?(? ? n + 1)?(n)
=
(5)
is Euler's Gamma funtion.
Let ? ? ? ? n ? 1 , and G is a domain in Rn suh that I(? ? n, G) <
?+??n
dist
(x, ?G) dA
?
G
?,n ?,n
?+?,n
Z
dist
??n
(x, ?G) dAЧ
G
Ч
Z
dist
??n
(x, ?G) dA
(6)
G
The equality holds i G is a ball in Rn .
This kind of inequality was rst proved in [14? for onformal moments of a simply
onneted plane domain. Also, in [14? a hain of plane isoperimetri inequalities was
obtained in order to get sharp lower bound in the torsional problem. This hain is
similar to `
disrete isoperimetri monotoniity.'
three geometrial harateristis of
Note that the inequality (6) ontains
G.
Corollary 1. Let n = 2 and ? = ? = 1 in Theorem 1 , then inequality (6) turns
to the plane lassial isoperimetri inequality
A(G)
?
L2 (G)
.
4?
In the next two assertions we prove isoperimetri monotoniity properties for integral
funtionals whih depend on dist(x, ?G) .
Theorem 2. i)
Let ||dist(x, ?G)||p < ? , then
||dist(x, ?G)||p?
||dist(x, ?D1 )||p?
?
,
||dist(x, ?G)||p??
||dist(x, ?D1 )||p??
(7)
where p? ? p ? p?? ? 0 , p? > p?? , and D1 is a ball suh that ||dist(x, ?D1 )||p :=
||dist(x, ?G)||p . Equality holds only for a ball in Rn .
?1
ii) Let ||dist
(x, ?G)||p < ? , then
||dist?1 (x, ?G)||p?
||dist?1 (x, ?D2 )||p?
?
,
?1
||dist (x, ?G)||p??
||dist?1 (x, ?D2 )||p??
where 0 ? p? ? p ? p?? < 1 , p? < p?? , and D2 is a ball suh that ||dist?1 (x, ?D2 )||p :=
||dist?1 (x, ?G)||p . Equality holds i G a ball in Rn .
In the plane ase for
p = p?? = 0
the inequality (7) is an analogous of famous
p = p?? = 0 , and for p? = 1 the
??
inequality (7) was proved by J. Leavitt and P. Ungar [15?, and in the ase p = p = 0 ,
?
and p ? 0 in [16?.
St Venant's and Polya's inequality for P(G) . In the ase
Using a result from [4? we an get the following lower estimations for the torsional
rigidity of a plane domain.
R.G. SALAHUDINOV
154
Corollary 2. Let G be a simply onneted plane domain, and let P(G) is bounded,
then we have for ? ? 4
?
Z
3?
? ?(? ? 1)
?(? ? 1)
2?
??2
dist
G
?4/?
(x, ?G) dA?
Now we shall onsider a normalized version of I(?
?
I? (G) = ??,n
where
? > n ? 1.
Z
dist
??n
G
< P(G).
? n, G)
?1/?
(x, ?G) dA?
,
(8)
In?1 (G) does not exist, however we give a
? = n ? 1 . In that ase we suppose that G has a
1
boundary ?G , whih belongs to C (G) , and we put
?
?1/(n?1)
Z
?(n
/2
)
In?1 (G) :=
lim
I? (G) = ?
dS ?
,
(9)
??(n?1)+0
2? n/2
In the Lebesgue's sense
meaningful denition for the ase
smooth
?G
where
dS
is the volume element of
?G .
On the other hand, we have
lim I? (G) = d(G).
(10)
???
Therefore, d(G)
< ?
is the neessary ondition to apply our theorems. Further we
prove that the evaluations given in
(9)
and
(10)
are well dened.
Theorem 3. Let G is a domain in R , and suppose that I?0 (G) is bounded for
some ?0 ? n ? 1 . Then
n
i) If G is not a ball in R , then I? (G) is a stritly dereasing funtion of ? for
? ? ?0 and I? (G) = d(G) .
n
ii) If G is a ball in R , then I? (G) is the radius of the ball G for all ? ? n ? 1 .
n
Corollary 3. Let G is a domain in R
suh that I? (G) < ? for ? ? [n ? 1, n) ,
then
In (G) ? I? (G), and I? (G) ? In?1 (G).
In the both equalities the equalities hold only for a ball in Rn .
n
The last inequalities are a generalization of the lassial
n -dimentional isoperimetri
inequality, whih was proved by E. Shmidt [17?, moreover, our assertion is a more
sharper result.
2.
Proofs of theorems
We begin by proving some simple important properties of the funtional I(?, G) for
? > ?1 .
Throughout the proof we will use the following notations
G? (?) := {x ? G |dist? (x, ?G) > ?} ,
?? (?) := {x ? G |dist? (x, ?G) = ?}
a(G)
:=
Z
G
dA,
(11)
ISOPERIMETRIC INEQUALITIES FOR LP -NORMS OF THE DISTANCE . . .
and denote by
[·]S
the Shwarz symmetrization of a domain in
? in
G? := G? (?) .
over the domain. We will usually x a parameter
brevity, we will drop
?,
for example,
Rn ,
155
and of a funtion
our proof, and in those ases, for
Let ? > 0 , and G be an unbounded domain in Rn suh that I(?, G) <
< ? . Then: 1) a(G? ) < ? for 0 < ? ? d? (G) , in partiular, d(G) < ? ;
2) ? a(G? ) ?? 0 .
Lemma 1.
??0
? > 0 , then we have a hain of inequalities
Z
?
?
dist (x, ?G) dA ? a(G? ) inf dist (x, ?G) = ?a(G? ),
I(?, G) ?
Proof. Let
x?G?
G?
so a(G? )
? ??1 I(?, G) < ? .
Further, note that
I(?, G)
=
Z
?
dist (x, ?G) dA
Z
+
G?
dist
?
(x, ?G) dA.
G\G?
Using the denition of integral by Lebesgues, and integration by parts, we obtain
Z
?
a(G? )
dist
?
(x, ?G) dA =
Z
d
?(a) da = ? a(G? ) +
0
G?
Z(G)
a(Gt ) dt.
?
The last integral bounded by I(?, G) . Hene a(G? ) is integrable on
a(G? )
?0
?.
for all admissible
[0, d? (G)]
beause
Therefore
0 ? lim ? a(G? ) ? lim
??0
??0
Z?
a(Gt ) dt
= 0.
0
In the sequel we will use some well-known properties of the level sets (see, for
example, [7?), in partiular, we will frequently make use of the relations
Z
?
d
?
dist
Z(G)
(x, ?G) dA =
a(G? (?))d?
? > 0,
0
G
Z
for
?
dist
?
(x, ?G) dA = a(G)d (G) +
G
da(G? (1))
=?
d?
Z
Z?
a(G? (?)) d?
for
? < 0,
(12)
d? (G)
dS.
?? (1)
Let G be not a ball in Rn , and [·]S is the Shwarz symmetrization of
a domain G? . Then:
Lemma 2.
1)
Z
dist
?
(x, ?G) dA <
G?
2)
Z
G?
Z
?
dist (x, ?[G]S ) dA
for ? > 0;
[G? ]S
dist
?
(x, ?G) dA >
Z
[G? ]S
?
dist (x, ?[G]S ) dA
for ? 1 < ? < 0.
R.G. SALAHUDINOV
156
Proof. Let
x ? G? ,
then it is easy to hek that
dist(x, ?G)
= dist(x, ?G? ) + ?1/? .
(13)
The assertion follows by means of the Shwarz symmetrization of
[dist(·, ?G? )]S (x) <
have
dist(x, ?[G? ]S ) for
x ? [G? ]S ,
G? .
?
and a ([G? ]S )
Indeed, we
a (([G]S )? )
(see [7? ). Using (13) we obtain
[dist(·, ?G)]S (x) =
h
dist(·, ?G? )
+ ?1/?
i
S
(x) = [dist(·, ?G? )]S (x) + ?1/? <
< dist(x, ?[G? ]S ) + ?1/? ? dist(x, ? ([G]S )? ) + ?1/? = dist(x, ?[G]S ).
Using the basi properties of the Shwarz symmetrization, for
Z
?
dist
Z
(x, ?G) dA =
G?
?
([dist(·, ?G)]S (x)) dA <
[G? ]S
In the ase
?1 < ? < 0
Z
? > 0,
dist
?
we get
(x, ?[G]S ) dA.
[G? ]S
we get the same inequality, but with the revers sign.
D(? Rn ), whih orresponds to the domain G . From (5) we
an onlude that I(?, D) , for xed ? , is a stritly inreasing funtion of the radius of
D , and runs from 0 to ? with the radius of D . Therefore there is exatly one ball
D , up to an Eulidean motion, suh that I(?, D) = I(?, G) . Note that D depends of
? and G , and it plays very important role in our proof.
Aording to Lemma 2 we shall distinguish two ases: 1) ? ? 0 , and 2) ?1 < ? < 0 .
We now dene a ball
Proposition 1.
Let ? ? 0 , and D is the ball dened as above, then
1) I(?, G)
? I(?, D)
for ? > ?;
(14)
2) I(?, G)
? I(?, D)
for 0 ? ? < ?.
(15)
In the both ases the equalities hold if and only if G is a ball.
Proof. To prove the assertion we apply the M.-Th. Kohler-Jobin symmetrization
with a slight modiation. This method was introdued in [8, 10?, and it was applied to
study isoperimetri properties of the solutions of (4),
?1 (G) ,
and the rst eigenfuntion
of (4) (see [3, 8, 10, 12?).
? = 0 is the speial ase of
? > 0 . Further note that the ase ? = 0
with ? = 0 . Indeed, we have
First of all we note that the ase
? = 0.
Therefore let us x
interpretation of Lemma 2
I(?, D)
and using (5), we obtain I(0, D)
Thus we an suppose
= I(?, G) ? I(?, [G]S ),
= a(D) ? a([G]S ) = a(G) = I(0, G) .
? > 0.
Set
i(?) =
Z
?
dist
(x, ?G) dA ? ? a(G? ).
G?
The orresponding value for
D
i? (?? ) .
we denote by
From (12) we obtain
?
d
i(?) =
Z(G)
a(Gt ) dt,
?
Lemma 2 with
also is another
ISOPERIMETRIC INEQUALITIES FOR LP -NORMS OF THE DISTANCE . . .
and sine
i(d? (G)) = 0 ,
157
we have
di(?)
= a(G? ).
d?
?
An analogous omputation for
D??
?
(16)
leads to
di? (?? )
= a(D?? ).
d??
Now let us dene a orrespondene between
?
By the denition of D , for ? = ? =
= i? (?? ) .
= i? (0) .
G? and D?? by requiring that i(?) =
0 we have i(0) = I(?, G) = I(?, D) =
G? is bounded, by Lemma 1, applying Lemma 2 to G? and D?? , we obtain
? a(G? ) . Let ?(i) be the inverse of i(?) . Hene by the dened orrespondene
Beause
a(D?? )
we must have the inequality
d?? (i)
d?(i)
??
.
di
di
?? (I(?, D)) = 0 = ?(I(?, G)) , we obtain
?
Again, using
by integration from
i0 (> 0)
to
I(?, G)
?? (i) ? ?(i).
In partiular, we have the inequality d(D)
? d(G) ,
(17)
whih, indeed, follows immediately
D.
form Lemma 2 and the denition of
G? (?) = G??/? , where ? > 0 .
?di(?) = a(G? )d? , by (16), we get
From the denition (11) easily follows
Lemma 1, (11), (12), and the equality
Z
?
a(G)
?
dist (x, ?G)dA
=
Z
0
G
?
d
?(a)da =
d
Z(G)
a(Gt (?)) dt
=
0
?
?
0
0
d
Z(G)
a(Gt?/? ) dt
=
0
dZ(G)
I(?,G)
Z(G)
Z
?
?
?
?/??1
?/??1
t
a(Gt ) dt = ?
t
di(t) =
??/??1 (i) di.
=
?
?
?
In the both ases
? ? ?,
and
0 < ? < ?,
in the seond ase we suppose that I(?, G)
Using
(18)
0
using (17), we obtain (14), and (15), but
< ?.
The ases of equality immediately
follows from Lemma 2.
Proposition 2.
way like above, then
Let ?1 < ? < 0 , ? ? 0 , and D is the ball dened in the same
1) I(?, G) ? I(?, D)
for 0 ? ? > ?;
2) I(?, G) ? I(?, D)
for ? > ?.
Proof. We will use same ideas as in Proposition 1, but in quite dierent situation.
? , then I(?, D) is an inreasing funtion of the radius
D . Moreover, the diret alulation shows that the same assertion remains true for
a(D? ) with xed ? and ? . We will use this remark bellow. In partiular, from Lemma 2,
and the denition of D , follow
First, we note that if we x
of
a(G)
As in Proposition 1 the ase
Therefore let
? < 0.
? = 0
= a([G]S ) ? a(D).
is the partiular ase of Lemma 2 with
(19)
? = 0.
R.G. SALAHUDINOV
158
z
z=?
z=d?(G)
y
G
?1/?
G?
x
Fig. 1
Set
i(?) =
Z?
?
where
b
a(G? ) =
?
?a(G)
?
a(G? )
b
a(Gt ) dt,
for
0 ? ? ? d? (G),
for
? > d? (G).
In partiular, using (12) we get
?
i(0) = a(G)d (G) +
Z?
a(Gt ) dt
= I(?, G).
d? (G)
For the onveniene of the reader we give an illustration for the plane ase (see Fig. 1).
? ?
The orresponding value for D we denote by i (? ) .
Further, almost everywhere we have
?
An analogous omputation for
D??
?
di(?)
=b
a(G? ).
d?
leads to
di? (?? )
=b
a(D?? ).
d??
G? and D?? by requiring that i(?) =
? = ?? = 0 we have i(0) = I(?, G) = I(?, D) =
Let us dene a orrespondene between
= i? (?? ) .
= i? (0) .
By the denition of
D,
(20)
for
ISOPERIMETRIC INEQUALITIES FOR LP -NORMS OF THE DISTANCE . . .
159
G? and D?? , the note at the beginning of the proof, and
?b
a(G? ) . Let ?(i) be the inverse of i(?) . Hene by the dened
Applying Lemma 2 to
(19), we obtain b
a(D?? )
orrespondene we must have the inequality
?
Using
d?? (i)
d?(i)
??
.
di
di
?? (I(?, D)) = 0 = ?(I(?, G)) ,
we obtain by integration from
i0 (> 0)
to I(?, G)
?? (i) ? ?(i).
(21)
G? (?) = G??/? , where ? < 0 .
?di(?) = b
a(G? )d? , by (20), we get
From the denition (11) easily follows
Lemma 1, (11), (12), and the equality
Z
?
dist (x, ?G) dA
= a(G)d? (G) +
a(Gt (?)) dt
= a(G) (d? (G))
?/?
+
d? (G)
G
+
Z?
?
d
Z?
Z(G)
?
?/??1
a(G) dt +
t?/??1 a(Gt ) dt =
t
?
?
a(Gt?/? ) dt =
?
d? (G)
?
=
?
d? (G)
0
Z?
t
?/??1
0
Let
Z?
Using
? < ? < 0,
then
?
a(Gt ) dt = ?
b
?
?/? ? 1 < 0 .
Z?
I(?,G)
t
?/??1
0
Z
Z
(?(i))?/??1 di.
(22)
0
From (21) and (22) we easy obtain
I(?,G)
?
I(?, G) =
?
?
di(t) =
?
I(?,D)
?/??1
(?(i))
?
di ?
?
0
Z
(?? (i))?/??1 di = I(?, D),
0
whih is the rst inequality of our proposition. If we interhange
?
and
?,
then the
seond inequality follows from the rst.
This ompletes the proof of Proposition 2.
Proof of Theorem
I(?, D)
Proof of Theorem
I(p, D)
= I(p, D1 ) .
= I(p, D2 ) .
We just apply Proposition 1 and Proposition 2 for the ase
2.
To prove the rst part we apply Proposition 1 for the ase
The seond part follows from Proposition 2 for the ase I(p, D)
Proof of Theorem
for
1.
= I(?, G) .
? ? n ? ?
3.
First note that the isoperimetri inequality
=
I? (G) ? I? (G)
follows easily from Lemma 2, and moreover, we have the same lower
and the upper bound
In (G)
for
I? (G)
and
I? (G)
respetively. Now we prove that
the denition (9) is well-dened for domains with smooth boundaries. For the brevity
we will use the denotation
lim :=
?
lim
??(n?1)+0
. Indeed, using (5), (11), and (12), and
R.G. SALAHUDINOV
160
applying a integration by parts, we obtain
Z
lim(I? (G))? = lim ?,n
?
?
??n
(x, ?G) dA
dist
=
G
a(G)
= lim ?,n
?
Z
d(G)
??n
? (a(G? (1)))
da(G? (1)) = lim
?
?,n
??n+1
0
0
= lim
?
?
d(G)
dS
Z
Note that, from (5) and (8) we an see
G
??n+1
?
0
?d(G) (1)
the radius of the ball
Z
dSd???n+1 =
?? (1)
?(? + 1)?(n/2)
Ч
2? n/2 ?(? ? n + 2)?(n)
Z
?
Ч ?d(G)??n+1
Let
Z
?
?
d?
??
Z
?? (1)
I? (D) = R
?? ?(n /2 )
dS ?? =
2? n/2
for all
? ? n ? 1,
Z
dS.
?G
where
R
is
D.
is not a ball in
Rn , and x ?(? ?0 ) , then I? (G)
is bounded, by Proposition 1
D(?) in Rn ,
and Proposition 2. Like above we an show that there is exatly one ball
I? (G) = I? (D(?)) . Further we onlude from (5)
? and ? . Therefore for a small
? > 0 , from Proposition 1 and Proposition 2 follow again that I?+? (G) is bounded, and
the ball D(?) gives a larger I?+? (·) than the domain G , that is
up to an Eulidean motion, suh that
and (6) that
I? (D(?)) = I? (D(?))
for all admissible
I?+? (G) < I?+? (D(?)) = I? (D(?)) = I? (G),
(23)
whih is the desired onlusion.
To omplete the proof we have to prove (10). From the monotoni property of
I? (G)
with respet to the domain we get the inequality
be enough for us to establish the reverse inequality. Let
Then, applying (18) for
? > ?0 ,
1/?
I? (G) = (?,n I(? ? n))
?
?
= d(G) ?
where
?
we an write the equality
?
Z
?
?(???0 )/(?0 ?n) (i) di?
0
I(?0 ?n,G)
(? ? n)?,n
(?0 ? n) [d(G)]
?1/?
I(?0 ?n,G)
??n
?
= ??,n
?0 ? n
??0
Z
0
?
?0 ?n
d
(G)
(???0 )/(?0 ?n)
=
?1/?
?
(i) di?
,
?0 ?n
orresponds to the level sets of dist
(x, ?G) . Thus we easily get
I? (G) ? d(G)
Tending
I? (G) ? d(G) . Thus it would
I?0 (G) < ? for some ?0 ? n .
?
1/?
(? ? n) ?,n
I(?0 ? n, G)
.
(?0 ? n) [d(G)]??0
to the innity, we get
I? (G) ? d(G) .
This nishes the proof of the theorem.
In the onlusion we have to make a remark about reverse inequalities in the theorems. Note that, using Proposition 1 we also an 'go bak' from
I? (G)
to
I??? (G) ,
but
ISOPERIMETRIC INEQUALITIES FOR LP -NORMS OF THE DISTANCE . . .
we need to know that
G
I??? (G)
161
is bounded. We now give a simple example of domain
on the plane, suh that it has
I? (G) = ?
and
I? (G)
G,
? > ? . In parI? (G) > cI? (G) for ? > ?
is bounded for
tiular, that means, we annot prove the reverse inequality
c > 0 is a universal onstant.
? > 0 , and onsider the domain plaed
?1/?
between urves x = 1 , y = 0 , and y = x
. There exists X ? R suh that
dist(z, ?G) = y/2 , where z = x + iy , and x > X . Denote by GX the subdomain
of G between x = 1 and x = X , we obtain
without additional restritions on
here
Without loss of generality we suppose
Z
dist
?
(x, ?G) dA ?
G
Z
?
dist
1
(x, ?G) dA +
2
GX
ZX0
dx
?? ?,
x X0 ??
X
nevertheless
Z
?+?
dist
G
where
? > 0.
(x, ?G) dA ?
Z
?+?
dist
(x, ?G) dA +
GX
Z?
dx
< ?,
x1+?/?
X
The generalization to an
n -dimensional
ase is obvious.
The author is thankful to Professor F.G. Avkhadiev for interesting disussions on
the mathematial physis and useful advies.
The work was also supported by Russian Foundation of Basi Researh (grant
No. 05-01-00523).
езюме
Изопериметрические неравенства для Lp -норм ункции расстояния до границы области.
.. Салахудинов.
Доказана изопериметрическая монотонность евклидовых степенных моментов области относительно своей границы. Доказанное свойство, эквивалентно изопериметрическим неравенствам для Lp -норм ункции расстояния до границы области для различных значений p .
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Поступила в редакцию
21.03.06
Салахудинов устем умерович
кандидат изико-математических наук, доцент, старший научный сотрудник НИИ математики и механики им. Н.. Чеботарева
Казанского государственного университета.
E-mail: rsalahudinovksu.ru
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