The sharp constant in Hardy's inequality for complement of bounded domain

<

3

fTt

p-I

o

P p-3

((t + a)(t +

b))Go+l)/pOu

Ot ((t

(t + a)(t + b ) d t

0

t h a t c o m p l e t e s p r o o f o f the l e m m a .

0

+ a)(t + b)) lip

Ot

(t + a)(t + b) dt

,

•

LEMMA l b . I f k l , k 2 > 0 a n d p > 1 then I(y) > ((p - 1)/py °.

Proof. Really, in this case if we d e n o t e a = 1 / k 1, b -- 1 / k 2, t h e n I(y) =

Sff IOu/Otf(a

- t)(b - t) dt

~r o lu/tlP(a - t)(b - t) dt

a n d f r o m S t a t e m e n t 3 it follows, t h a t T _< min(a, b). In exactly the s a m e w a y as in L e m m a l a we o b t a i n :

0

p( a -

t)(b

-

t)dt

=

i ip ulul "-2 0

.At

rP

F u r t h e r let us n o t e t h a t

0 _<

i

t(a_

r)(b-

r)

t

T 2.1-p

rl-v [r dr = (a -

r)(b -

Tp

r ) ~ 1--//

tl-P <-- (a - t)(b - t ) - p-l'

It

+ t

- -

Jtp-1

(2r - (a + b)) d r

b e c a u s e for every z ~ [t, T) we have 2 r - (a + b) < 0. T h e n L e m m a l c follows f r o m H61der's inequality. •

LEMMA lC. I f kl > 0, k 2 < 0 a n d p > 2 then I(y) >_ ((p - 2)/p) p.

Proof. Let us d e n o t e a = l / k ~ , b = I1/k2[, then

j~ IOu/atlP(a - t)(t + b) dt I(y) = jr° [u/t[P(a _ t)(t + b) dt a n d f r o m S t a t e m e n t 3 it follows t h a t T_< a.

112

T. MATSKEWlCH and P. E. SOBOLEVSKII

Again it is sufficient to obtain the following estimate: 0 <_

i

ra_

r ) ( r + b) dr rp

t

= 1rl---~p -p (a-

r ) ( r + b)ltr -

I 1

t 1-p

<_--(ap-1

t

p

t)(t + b) +

-

1 (a -- t)(t + b) +

~r I (at

p-1

I-p

l

r)r 1-p dr

-

p-1

p-2

liT ) r 2-p dr p-2 t

1

t l-P

t)(t + b) + ( p _

( a - t) t + p - 2 b

p-2

1)(p- 2)(a-

)

p-1

=

k2

=

tl-P

<--(a-

t)t 2-p

t)(t+ b).

p-2

And so Lemma l c is also established. LEMMA ld. If kl

( r + b))l - p

( (a--r)'c2-P ;

t 1-p <-p--- l ( a -

-

T I - p dr

T

t ((a- r)-

•

0 then for every p > 1 I(y) >_ ((p - 1)/p) p.

This lemma immediately follows from the classical one-dimensional Hardy's inequality. LEMMA le. If kl = 0 and k 2 > 0 then for every p > 1 I(y) >_ ((p - 1)/p) p. Proof. Really, for a -- l / k 2 we obtain: z(y) =

fro IOu/OtlP(a - t) dt lfflu/tl(a

o 7 (a - t) dt =

- t) dt

o pulul p-2 ~

\jr

rp dr

at

and exactly as in the previous lemma it is sufficient to estimate f f ((a - r ) / z p) dr in the following way

f Ta-

3t rP r d r -

./.1-p

p-

1

and then we apply H61der's inequality.

I"T rl-P

r)ll- It p------i dr

< tl-P ( a -- p----i --

t),

•

LEMMA lf. I f k I = 0 and k 2 < 0 then for e v e r y p > 2 I(y) >_ ((p - 2)/p) p.

The sharp constant in Hardy's inequality

113

Proof. For a = I1/k2] it is sufficient to obtain the following estimate: -

Do Irl

(a+

(

r) d z =

-a

p-

1

p-2,]

t

_
tip p-

1

+

,-

-

p-2

t'-P ( P - 2 t) < t'-P_ (a + t). -p~-2 a~---(- 1 + -p2 It means that for every p > 3 I(y) >_ ((p - 3)/1))p. It follows that the same estimate holds for the ratio obtained in Statement 2 and so the following lemma takes place. LEMMA 1. For p > 3 and for arbitrary domain ~ C R 3 such that its boundary 0 ~ is a C 2 smooth two-dimensional compact manifold lp(U, ~) >_(p - 3)/p for arbitrary u e l~l(f2).

Proof. Let us for mapping number i describing Of2 denote: j/(1)

j/(l)

= =

eV/

y e v/

l

Xi(Y) 0 (y)lVulPJ ' dy dt, Zt(Y)

(Y)

J' d y d t ,

then from Statement 2 and estimates obtained above it follows that for every u e C~(f~) j/(1) > I(,~,rt)eV/ N ] ~ j/(2)

I((,~) e v / ~ - - G

(Pp

IOu/OtlP(1 - 2Ht + Ht 2) dt d~ dr/ F ) ( i ( ( P i ( ~ , r / ) ) I T(e'(~'n)) lu/tlP(1 - 2Ht + Kt2) dt d( drl

-- Fxi(tPi(~, r/))ior(*'~'"))

I

-

3) p

And so we can conclude that

0

for every u e Cg'(f~). But Cg'(O) is dense in Wpl(f2) and so Lemma 1 is established. Let us finally note that Theorem 1 directly follows from Lemmas 1 and 2. 3. E S T I M A T E

FROM ABOVE FOR THE COMPLEMENT ARBITRARY BOUNDED DOMAIN

O F AN

LEMMA 2. Let us suppose that f2 is the complement of some bounded domain D with Cl-smooth boundary: f~ = R a " , D . Then for p _> 3

p-3 inf lp(U, t)) <_ - , ~/,¢
114

T. MATSKEWICH and P. E. SOBOLEVSKII

Proof.

The p r o o f based on compression o f the d o m a i n D. W i t h o u t loss of generality let us assume that D contains an origin o f the coordinate s y s t e m - - p o i n t O. D is b o u n d e d and so it belongs to some ball with center in the origin and with sufficiently large radius R > 0. Let us consider the following h o m o t h e t y t r a n s f o r m a t i o n : x ~ 2x, 2 > 0. This t r a n s f o r m a t i o n m a p s d o m a i n D onto d o m a i n D(2) that belongs to the ball with center in the origin and radius 2R. It m a p s f2 onto d o m a i n ~ ( 2 ) = R3kD(o2) and defines bijection between C~(ff2) and C~°(f2(2)). The h o m e o m o r p h i s m Wpl(~) Wpl(~'~(2)) is also established. F r o m the definition o f h o m o t h e t y it evidently follows that distance between any two points is multiplied by ;t and so p0ao,)(2x) = where Poe(x) and are distances f r o m the point to the b o u n d a r y o f f~(2) and f2, respectively. Let us denote t7 = u(Xx), )? = 2x. Then for arbitrary u e Cg'(O)

between

and

2poa(x),

I. Ivul"~

Poe

laa> Iv~al"~-'x-~ ~

la lulpoal" dx = Iaa> lalpoa(,~)l ",~px-3 din'

o1

lp(u, f~) =

Ip(U,

and so infu ~ ( a ) inf. ~wlta0,)) ~(X)). N o w let us pass to the spherical coordinate system and for every e > 0 let us consider the following function ut (r, ~p, 0) = wherea=

1 -

r~(1 0,

I

r),

0 <_r <_1; otherwise,

3/p+e/p;O__O. Then

Ou..__£= O; Otp

~ a r c~-1 -- (a + 1)r ~, O---r = (. O,

OUt

OUt

0--0 = O;

0 _< r _< 1; otherwise.

So we obtain

(IR3 ]•ueIP dx)I/P = (IO

lOj'O OUt

do)

[ (ll r(~-l)prEdr)l/p 0

_< (4~z) I/p c~

+ (c~ + 1)

(ll)l/p] r~Pr2dr ,,gO

=(47t)l/P[(l__3 [_p)(!)l/'...[__(2 3 _q( '7'pl p p/\p + el J --+

(fil~

2~r ~" oo ld P

~Pcbc)lip= (fo

\l/p

fo fo 2~ r2drdOd(o)

>-(4~z)l/P[(t']r(~-l)pr2dr)l/P - (I] p~r2dr)\l/P]]

The sharp constant in H a r d y ' s inequality

115

Then I1~3 IVu~lPdx

< (1

3/p + e/p)(1/e) t/p+ (2 - 3/p + e/p)(1/(p + e)) t/p (1/e) x/p - ( 1 / ( p + e)) 1/p

-

~# [u~/r[P dx -

<_ 1 - - + P

+ Mpe t/p,

where Alp > 0 is s o m e c o n s t a n t ( u n d e r a s s u m p t i o n that e _< 1 we can c h o o s e Alp = (3 + 21p)/(p alp - 1)). It m e a n s t h a t f u n c t i o n a l e v a l u a t e d for u~ a p p r o x i m a t e s the desired e s t i m a t e b u t u~ ¢ 1~'1(~(2)) f o r every 2 > 0. N o w let us correct f u n c t i o n u~ a little, " m o v i n g " it f a r t h e r f r o m the origin. Let us c h o o s e s o m e h > 0 a n d b u i l d the following f u n c t i o n q/~,h:

O, O<_r<_h; cgu~ 0 r (r' O' O)' h < r <_ 1;

I

~,h(r, o, O) =

f u~(h, O, 0) = h~(1 - h), k0,

F u r t h e r let us define f u n c t i o n U~,h(r, O, O) T h e f o l l o w i n g estimates can be o b t a i n e d :

(127r lTr l°° Olde OUe,hP 2 o

o

o

Or

~r

r

= ~toIg~,h(S, O, O) ds.

dOdo) 1/p

sin0dr

<(4rO1/p

h OU. Pr2 dO1/P +

-

o

lu.(h, o, O)lPr2 dr) l/p]

Or

1

[(i hrf~-t)p+zd r)

< (4zr) 1/p a

+ ( a + 1)

0

he/p

CI

r~P+Zd

\O0

r;"

- h ) ( t 2 r2dr)\I/Pq]

+ h'~(1

- ~

1 < r _< 2;

r>2.

/

h~,1/p

(4.)t/Ptot + (or + 1)(P + e)i/p +

hl

-s/P(1-

/7\ 1/p"~ he/p

h)t~) ) <- e-'~Np,

w h e r e Np is s o m e positive c o n s t a n t (we can a s s u m e t h a t e < 1, h _< 1). I f we n o w c h o o s e h = h(e) = e 2/e t h e n

30 jo o

Or

-~r

r sin 0 d r dO d o )

-<

Np ,y,IIp

a n d f r o m scalar H a r d y ' s i n e q u a l i t y it follows t h a t 2.

•

o

oo

~

u._

sin 0 d r dO d o

-<

p-1

Npe I/p= l~p~,1/p.

116

T. MATSKEWICH

and P. E. SOBOLEVSKII

Note, that for E < 1

2 (477)‘“(I Then denoting (4n)1’P(1 - (l/~)“~) j? jz j: I&,,,/&l”r” j? j; jr ldu,,JrIpr2

-

(;LJl”> .

by C we obtain:

sin 19dr dt? dq 1’p I (jr jz jt ldu,/drlPr2 sin 0 dr d0 dq)l’P + Npz?‘p sin 0 dr de da, (jr j: j: ldu,/rlpr2 sin 8 dr de dp)l’P - iVp~l’P 5 1 - 3/p + Mp~l’P + (N,/C)E”~ 1 - (Rp/c)&l’P ’ 3 5 1 - - +R,E”~, P

for sufficient small E, here R, is some positive constant. Further, let us note that support of the function u,,h lies outside the ball with radius For every x E supp u,,h we h and SO SUpp I.&h C n(n), for A < h/R u,,h E &j@(n)). have: 1x1 = r > h; +zR

h >

and so

Ip(ue,h W)) =

jr j; jr l&&h/prlPr2 sine dr de dy, SF K jr

b&h/Pan(X)

1’p

lpr2sin8drdedy,

1 - 3 + Rp$‘P P

=

(1+$)(l -; +R,E~/~).

Now it is sufficient to take A = .ee2”.

REFERENCES 1. Matskewich, T. and Sobolevskii, P. E., The best possible constant in generalized Hardy’s inequality for convex bounded domain in R”. Elliptic and Parabolic P.D.E. 3 and Applications, Capri, 19-23 September 1994, Summaries. 2. Hardy, G. H., Littlewood, J. E. and Polya, G., Inequalities. Cambridge University Press, Cambridge, 1952. 3. Stoker, J. J., Differential geometry. Pure and Applied Mathematics. A Series of Texts and Monographs.

New York, 1969.

The sharp constant in Hardy's inequality

117

APPENDIX P r o o f o f Statement 1 Our purpose now is to prove that for the bounded domain ~ c R n such that its boundary 8~ is a C2-smooth two dimensional manifold there exists subset 91: C ~ such that its Lebesgue measure is equal to zero and such that the transformation (y, t) -~ y + n(y)t defines bijection between the set of all pairs (y, t), where y e 0~, t e [0; T(y)) and ff~\OZ. (1) Let us choose a convenient system of maps, describing 0~ as a smooth manifold. Because 0~) is a compact set we can choose such a system of maps [ U i, Vii, ~Pil~=1 that U i is an open bounded domain in R 2, ~J~= 1 Vi = 0~, tp~ defines bijection between U~ and Vii, tpi, ~ t I ~ C 2, and ~0~can he continued as a C2-smooth function to some neighborhood of U~. (2) Let us use the following notations: Sy = [Ry + n(y)t; 0 < t < T(y)}; Ly = [Y + n(y)t; 0 _< t < +~1; OL(O) = [y + T(y)n(y); y e O~}. We will show that so defined 9~ = OZ(~) satisfies Statement I. (3) Let us firstly prove that transformation (y, t) -~ y + n(y)t defines bijection between a set of pairsy ~ 0~, t ~ [0; T(y)) and ~\9Z. In the beginning we will prove that for every x o ~ ~)\9Z(~) there exists such pair (Yo, to) that Yo e 0fl, t o e [0; T(yo) ) and x o = Yo + n(Yo)to. Because ~ is a compact set, there exists at least one point Yo ~ 0~ that is the nearest to x o. In the neighborhood of this point manifold 0 ~ is described by some map tp: U -~ V,, U C R 2, Yo e V and let us suppose that Yo = (o((0), (0 ~ U. Then the function D(() = ]Ixo - (o(()]]2 has a local minimum in the point Go, and after computation of the partial derivatives of D(~) we can conclude, that x o - Yo is perpendicular to the tangent plane to 0~) in the point Yo. So there exists t o _> 0 such that x o = Yo + n(Yo)to. It remains to show that t o < T(yo). Because x o ¢ 9Z(fl) then t o ~ T(Yo). Let us assume by contradiction that t o > T(yo). Then from definition of T(Yo) it follows, that there exists ~, T(yo) _< i < t o, and)7 e 0~, )7 # Y0 such that dist(y o + n(y0)?, )7) _< dist(y o + n(Yo)t, Yo) = t. But then if f i ¢ Ly o then distO7, Xo) < dist(xo, Yo) as it follows from strong triangle inequality. Let now f i e L,0 then because Yo ~ )7 it follows that if 37 lies between Yo and x o then we obtain obvious contradiction with choice of Yo; if x o is between )7 and Yo then it is clear that dist(fi, Xo) < d i s t ( f , y o + n(Yo)i ) < dist(Yo, Yo + n(Yo)t) <- dist(Y0, x0) and it again contradicts to the choice o f y o . I.e. for every x o ~ ~)\9Z(~) there exists point Yo e 0fl such that x o = Yo + n(Yo)to where t o ~ [0; T(Yo) ) and this point is unique because otherwise we obtain contradiction with the definition of T(Yo). Now let us show that for every pair (y, t), y ~ 0~, t ~ [0; T(y)) transformation y + n(y)t defines the unique point x belonging to ~\gZ(~). Uniqueness of such point directly follows from smoothness of the boundary O~. Further, because for every t e [0; T(y)) y is the nearest point of 0~, 0~ does not intersect the segment ( y ; y + n(y)T(y)) and so point y + n(y)t belongs to g$. On the other hand y + n(y)t $ 9~(~). Really, let us suppose that there exists such a point )7 e O~ that y + n(y)t = ~ + n(f)T(2). Then because t < T(y) it is clear that .17 ;~ y and y is the unique nearest for y + n(y)t point of boundary, particularly the following inequality holds: dist(y, y + n(y)t) < distOT, y + n(y)t). And because the last inequality is strict, then if we move points y + n(y)t a little along the straight line between y + n(y)t and )7 in the direction towards )7, then the sign of the inequality remains unchanged. But it means, that there exists a point of the form )7 + nO7)?", ~"e (0; T(~)) such that for this point the nearest point of 0 ~ is y and so we obtain contradiction with the choice of TO;). Hence it is proved that the considered transformation defines bijection. (4) Now let us prove that Lebesgue measure of 9~(~) is equal to zero. Suppose, that we consider the system of maps, described in the beginning of this section. Because 9Z(ga) = loci= ~ [y + n(y)T(y), y e V/} it is sufficient to prove that for every i Lebesgue measure of the set lY + n(y)T(y), y e ~i(/.)i)} is equal to zero (here Ui denotes the closure of domain Ui). Let us consider one such map: ~0: U--+ V. Let us denote by ( = (~, r/) any point in U and let us for more simplicity denote T(tp(()) by T((), n(tp(()) by n(() and so on. The following statement plays the main role in the proof.

118

T. M A T S K E W I C H and P. E. SOBOLEVSKII

S t a t e m e n t 4. T ( ( ) is a continuous function of (, ( ~ U.

P r o o f . Let us denote train(f) = 1/kmax(() , where kmax(() = max(k I , kz). Then from the definition of manifold it follows, that rmin(() 7~ 0 for every (.

LEMMA 3. If kmax((O) > 0 and 0 < 8 < rmin((0 ) is some fixed number, then there exists R > 0 such that ~O((o) is the unique nearest point of the boundary to the point tp((0) + n((0)(rmin((0) - ~) a m o n g all points ¢(() such that I( - (01 < R.

LEMMA 4. If kmax ~ 0 and t > 0 is some fixed number than also there exists R > 0 such that tp((o) is the unique nearest point of the boundary to the point ~((o) + n(zo) t a m o n g all points tp(() such that 1( - (01 < R. P r o o f o f L e m m a 3. Let us denote x = tp((0) + n(Zo)(rmin((o) - t$) and consider the function Da(h ) = IIx - ~(~o + hd)[] 2, where d is some vector of the unity length, that defines the correspondent direction. Dd(h)

is twice differentiable with respect to h and so (because D~(0) = 0) we have: O d ( h ) - Dd(O) = D~(O)h + 1D,~(h)h2 = ½O~(?t)h 2,

where h is some number between 0 and h. Then in order to prove L e m m a 3 it is sufficient to find R > 0 such that for every Ihl < R and for every d D ~ , ( h ) > O. It can be shown that if d = (dl, d2) then l ~D,~(0) = (d I

d2)

[(E

F

F)_(rmin_O)(L

M

NM)I(dl)

d2

"

Now let us show that ½D~(0) is strictly positive definite. It means that there exists 2 > 0 such that ½D,'~(0) > ;t for every d. We must check that the following inequalities take place: (a)

E - (rmin - ~)L > 0;

(b,

det[( E

F)-(rm,n-~)G~

/

NM)] > 0 .

TO prove (a) let us note that curvature k~ of the normal section in the direction ~ (d = (1; 0)) is equal to L / E . If0rmi .andsoL>0andE-(rmi n-~)L =E-rminL + d ~ L > 0 . If k~ -< 0 then L = k~E E > 0 . To prove (b) we define y > 0 such that kmax + y = 1/rmin - ~. It is known that

and we want to prove that

Really:

= 2ykmax(EG - F 2) + y ( 2 M F - E N - L G ) + y2(EG - F 2)

= 2y(EG - F2)(kmax - H ) + y 2 ( E G - F 2) = (kmax - kmin)(EG - F 2) + y 2 ( g G - F 2) > 0, and inequality (h) is proved.

T h e s h a r p c o n s t a n t in H a r d y ' s inequality

1 19

N o w let us return to the consideration of D'd(h ). Let us denote M 1 = max~-~ o(1~(01, I~dOI), M z = max~-~

o(l~(OI, l~,~(OI, 1¢,,(OI).

Then ½[O~(h) - 95(0)1 < I(tp(( o + h d ) - x , (a~(( o +

hd)d~ + 2~,(¢o + h a ) a , d2 + ~,.(¢o + ha)a~

- (~P(Zo) - x , ~a~(( o + hd)d21 + 2~p~,((o + h d ) d 1 d 2 + ~ , , ( ( o + h d ) d 2 ) l

+ I((a(Zo) - x, (a~(( o + hd)d21 + 2tp~,(( o + h d ) d l d 2 + tp,,(ff o + hd)d22) - (~P(Zo) - x. ~p~((o)dlz + 2~p~,((o)d , d 2 + ¢,,((o)dff)l + 1(~(~'0 + hd)d~ + ~P,((o + h d ) d 2 , tP~((o + h d ) d l + ~%((0 + h d ) d 2 )

- (~(~o)d, + ¢.((o)dz, ~(~o + hd)d, + ¢.(~o + ha)a2)l + (~P~(~o)dl + ~P,(~o)dz, (P~(~o + h a ) d , + ~P,(~o + h d ) d 2 ) - (~p,(~'o)dl + ~p,(~'o)dz, ~P~((o)d, + ~p,((o)d2)l <_ 4U21l~(z o + h a ) -

~(~o)11 + 4/l[HO(ffo +

hd) -

~(~o)ll + I1~,(~o +

hd) -

~,(¢o)111

+ rmi.(Co)[ll~&e(C o + h d ) - ~ee(Co)ll + II~e,(ffo + h d ) - ~e,(~o)ll + I1~,,(~o + h d ) - ~,,(Co)lll. Because ¢p ~ C ~ then we can choose R > 0 such that for every Ih] < R and for every d the expression obtained a b o v e will be smaller t h a n 2 . T h e n D ~ ( h ) > 0 for every d and for every Ihl < R. • Proof of Lemma

4. W e can prove this l e m m a in the s a m e m a n n e r as the previous l e m m a was proved. In this

case we obtain:

Because km~x -< 0 then k~ = I / E < 0 and so E - t L >_ E > O. F u r t h e r , let us note that 1 / t > 0, kma x -< 0 and so we can choose y > 0 such, that 1 / t = kmax + y. T h e n det[( E

F)-t(L

~)]

=t2detI(kmax+)')(E

GF)-(~

NM)I > 0 '

as it was s h o w n in the p r o o f of the previous l e m m a . It can be s h o w n , that ½lO,~(h) - D,~(0)I <-

4mzll(o(Zo +

hd)

+/[11~e¢(¢o + h d )

(o(~o)11 + 4M,[ll(o,((o

-

-

+

hd)

~¢¢(¢o)11 + I1~,(~o + h d )

-

-

~,~(¢o)11 + I1(o,((o +

hd)

~,(~o)11 + II~,,(~o + h d )

(~,((o)111

-

-

~,,(go)ll].

F r o m this inequality an appropriate R > 0 can be chosen. N o w let us pass to the p r o o f of Statement 4. W e a s s u m e by contradiction, that T(() has at least one point Co of discontinuity. T h e n there exists such e > 0 and such sequence of points [~}~ t h a t ~ ~ Co and It((o) - T((,)[ > e for every n. [T~,)}7= ~ is a b o u n d e d sequence (for every n, T(~,) is less that diameter of ~2), so we can choose convergent s u b s e q u e n c e and for simplicity we will a s s u m e that [T(~,)}~= t itself converges: T ( ( , ) ~ T. F r o m the definition of T ( ( , ) it follows, that for every y e 0f~ ]ly - ~P(C.) - T(C.)h(~DI[ -> T ( ( . ) . So passing to the limit we obtain:

Ily - c0(Go) - Tn((o)ll -> T. Hence T _ T((o). Really, if T > T((o) then there exist t such that T((o) _< t < T and )7 e Of~, )7 # y and such that 11)7- tP((o) - tn(Co)ll <- t. T h e n f r o m triangle inequality it follows that 11)7- ~(~o) - tn(~o)l[ > 11)7 - ~((o) - ~n((o )11 - (Y - t) ___ t and so the contradiction is obtained.

120

T. M A T S K E W I C H and P. E. S O B O L E V S K I I

So, it is proved that T _< T((o) and it means that starting f r o m a certain index T((n) < T((0) - e. Let us suppose that there exists such sequence that (n -~ (o, and T((.) < T((o) - e. If kmax((o) > 0 then 0 < rmin < +co and so in the same n e i g h b o r h o o d of the point Go, rmin(() is positive and finite. Further two opportunities exist: (a) Or starting f r o m a certain index IT(Go) - rmin((n)[ < e/2; (b) Or there exists subsequence [(nk] such that ( . . ~ Go and IT((.k) rmin((nk) [ ----.•. If o p p o r t u n i t y (a) takes place, then because 0 ~ is C2-smooth, so rmi.((.) -~ rmi.((o). Further, because 0 < rmin((0) < q-cO f r o m Statement 3 it follows that T((o) _< train(50 ). On the other hand, we can a s s u m e that T((.) -~ T, and so passing to limit in the inequality IT((.) - rmi.(5.)[ < e / 2 we obtain that IT - rmin(5o)[ -< e/2. But it was supposed that T(5.) < T((o) - e and so T _< T(5o) - e. Hence we obtain the contradiction: -

8 8 T(5o) - 2 -< rmin(50) - - <2 -- T <

T((0)-

e.

If o p p o r t u n i t y (b) takes place then let us denote for m o r e simplicity ~.. by ~n and assume that there exists limit l i m . ~ T((.) = 7~. Further, because 0 < rmin(50) < +CO it can be supposed that for all elements of sequence 0 < rmi.(~) < +oo and hence T((.) _< rmi.((.) - e/2. F r o m the definition of T(Z.) it follows that for every n we can find t. such that T(Sn) < t. < T(~.) + e/4 and y . e OO such that y . ~ ~0((.) and II~off.) + t.n((.) - y.l[ -< t.. Because t. is a b o u n d e d sequence and y . is a sequence of the points o n a c o m p a c t set O~ so we can assume that t. --, ~ and y . ~ 37 e 0 ~ (otherwise we can choose such subsequencies). T h e n H¢(5o) + /h((o) - fi[[ -< L But ~ _< l i m . ~ = ( T ( ( . ) + e/4) < T((o) - 3/4e < T(5o) that contradicts to the definition of T(5o) i f y ;~ ~O((o). So, it is only necessary to s h o w that y . does not converge to (O(5o). Let us recall that in L e m m a 3 function Dd(h ) was considered. Besides d and h, this function also depends on the choice of 5o and g. N o w we will denote this function by D~o,d,~(h ). Let us fix 0 = e/4. Because kma~((o ) > 0 then there exists such ;t o that D~'o,d,./4(O) > ;to for every d. But DLd,e/4 is a c o n t i n u o u s function of 5 and so we can choose R 1 > 0 such that D~-,d,./4(O) > ;to/2 for every 5 such that 15 - 5o[ < R~. Let us also recall that the value of D'~',d,e/4(O) was used to choose f r o m the inequality 4M211~0(5 + hd) -

~(z)ll

+ 4M~([1~o~(5 + h d ) -

+ rmin(()(ll~P~e(( + h d ) -

~P~(()[] +

~,~(oll + ll~.(5

II~,,((

+ hd)-

+ hd) - ~p~.(()ll +

~,<5)ll) I1~,,((

+ hd)-

¢o,,(()l[)

< D;,d,~/4(O) such R~ that q~(() is the nearest point to (0(5) + (rmln(5) -- e/4)n(() a m o n g all points of the f o r m ~o((') if

15 - 5'1 < g~. But if I(o - 51 < R1 then Dz,d,e/4(O ) > ;t0/2 and in this n e i g h b o r h o o d o f ( we can estimate rmi~(() -< r < +co for sufficiently large r. T h e n f r o m the inequality above it follows that R( can be chosen uniformly for I~ - ~ol < R , . So it was s h o w n that there exists R and R~ such that if l( - 5ol < R~ and l( - ('l < R, 5' ;e ( t h e n point ~0(O is closer to r/f(5) + (rmi.(5) - e/4)n(() t h a n every point ~0(5'). It was supposed that for every n there exists tn such that T((.) _< t. <_ T(5.) + e/4 and y . ~ 0 ~ such that y . ¢ ~o((.) and II~offD + t.n(5.) - y.ll <- t.. We can a s s u m e that I(o - (.I < R~ for every n because ( . -~ 5o. T h e n ~0(5.) is the unique nearest b o u n d a r y point for ~o(~'.) + (rmi.((.) - e/4)n(5.) in the R - n e i g h b o r h o o d o f ( . . But then it is also the nearest b o u n d a r y point for ~0(5.) + t.n(5.) in the same n e i g h b o r h o o d . So if we suppose by contradiction that y . ~ ~0((o) then for sufficiently large n, y . belongs to the m a p which contains ~O((o)and we can assume that y . = ~0((.). Hence it follows that inequality I(. - (.[ > R m u s t be fulfilled. But 5. ~ 5o and starting f r o m a certain index [5. - 5o[ < R / 2 and 150 - (o1 > [(. - (.1 - [5o - 5.1 > R / 2 . So we obtain that ( . does not converge to 50 and then y . = tp((.) does not converge to SoStatement 4 is established. F r o m this statement there immediately follows that Lebesgue measure of [~0(5) + T(On(O; ( e U] is equal to zero as measure of the continuous image of c o m p a c t plane d o m a i n U. So Statement 1 is also proved.