Weighted Hardy's inequalities for negative indices

Publ. Ma . 48 (2004), 423–443
WEIGHTED HARDY’S INEQUALITIES FOR NEGATIVE
INDICES
Dmi y V. P okho o
Abs ac
In he pape we ob ain a p ecise cha ac e iza ion o Ha dy ype
inequali ies wi h weigh s o he nega i e indices and he indices
be ween 0 and 1 and es ablish a duali y be ween hese cases.
1. In oduc ion
The Ha dy inequali y wi h weigh s u, ≥0
Z+∞
0
u(x)Zx
0
(y) (y)dyq
dx1
q
≤CZ+∞
0
(x)pdx1
p
, ≥0
has been comple ely cha ac e ized o p, q > 0 by G. Talen i [11], B. Mu-
ckenhoup [7], J. S. B adley [3], V. G. Maz’ja and A. Rozin [6,§1.3],
G. Sinnamon [9], G. Sinnamon and V. D. S epano [10] and some o he
au ho s (see he monog aphs [8] and [5] o de ails). An analogous p ob-
lem o p, q < 1 was s udied by P. R. Beesack and H. P. Heinig [1], whe e
su icien condi ions and necessa y condi ions unde some es ic ions on
he weigh unc ions we e gi en o he inequali y
(1) Z+∞
0
(x)pdx1
p
≤CZ+∞
0
u(x)
Zx
0
(y) (y)dyq
dx1
q
, ≥0.
In he p esen pape we ob ain a p ecise cha ac e iza ion o (1) o p, q <0
and p, q ∈(0,1). Mo eo e , we es ablish a duali y be ween hese cases.
The pape is o ganized as ollows: Sec ion 2 con ains an explici c i-
e ion o (1) o he cases −∞ < q ≤p < 0 (Theo em 1) and −∞ < p <
q < 0 (Theo em 2) and simila esul s wi h he dual ope a o . Theo em 3
o Sec ion 3 p o es ha he Ha dy ype inequali y wi h nega i e indices
2000 Ma hema ics Subjec Classi ica ion. 26D15, 26D10.
Key wo ds. Ha dy’s inequali y, weigh unc ion, nega i e indices, duali y.
The esea ch was pa ially suppo ed by he Russian Founda ion o Basic Resea ch
(P ojec 03–01–00017) and by he Russian Academy o Science (P ojec 04–3–Γ–01–
049).
424 D. V. P okho o
o in eg a ion is equi alen o he same inequali y wi h he dual ope a o
o conjuga e indices. As a consequence we a e able o cha ac e ize (1)
o p, q ∈(0,1) in he co olla y o Sec ion 4. O he p oposi ions o Sec-
ion 4 supplemen he main esul s by a numbe o cha ac e iza ions o
simila inequali ies.
Th oughou his pape A.Band B&Ameans ha A≤cB, whe e
he cons an cdepends only on p,qand may be di e en in di e en
places. I bo h A.Band A&B, hen we w i e A≈B.Ns ands o
he se o posi i e in ege s, Zis he se o all in ege s and (a, b) is a non-
emp y in e al o he eal line (−∞,+∞). The symbol p0:= p
p−1deno es
he conjuga e numbe o p,q0:= q
q−1and he symbol ma ks he end
o p oo o any s a emen . As usual we abb e ia e “almos e e ywhe e”
by a.e. and “i and only i ” by i . Some imes o simplici y we also use
he no a ion Rb
a := Rb
a (x)dx.
Le us men ion one mo e hing be o e we s a . A peculia i y o he
inequali y (1) caused by he nega i i y o he indices o ce us o wo k
wi h measu able unc ions ha ing hei alues in he ex ended semi-
axis [0,+∞] equipped by a i hme ic:
(2)
0 + (+∞) = a+ (+∞) = a·(+∞) = +∞, a ∈(0,+∞];
0·(+∞) = 0;
(+∞)α= 0−α= +∞,(+∞)−α= 0α= 0, α ∈(0,+∞).
No e also simple co olla ies o he axioms (2) which we o en use:
Le α, β ∈(−∞,+∞) {0}and a, b ∈[0,+∞]. Then
1. (aα)β=aαβ.
2. aαaβ≤aα+β; i (a < +∞and α+β > 0) o (a > 0and α+β < 0),
hen aαaβ=aα+β.
3. aαbα≤(ab)αwi h equali y unless α < 0and {a, b}={0,+∞}.
2. The main esul s
Deno e by M+(a, b) he class o all measu able unc ions : (a, b)→
[0,+∞] and pu
(I× )(x) := Zx
a
(y)dy,
(I× )(x) := Zb
x
(y)dy,
x∈(a, b).
Ha dy’s Inequali ies o Nega i e Indices 425
Theo em 1. Le −∞ < q ≤p < 0,u, ∈M+(a, b)and I=I×o
I=I×. Then he ollowing s a emen
(a)The e exis s a cons an C > 0such ha
(3) Zb
a
(x)pdx!1
p
≤C Zb
a
u(x) [(I( ))(x)]qdx!1
q
o all ∈M+(a, b)
holds
is equi alen o A<+∞, whe e
A:= sup
a< <b
A( ) := sup
a< <b
[(Iu)( )]−1
qh(I p0)( )i−1
p0
.
Mo eo e , A ≈ C o he leas possible cons an Cin (3).
P oo : Le I=I×( he case I=I×can be p o ed analogously). A i s
we no e ha ini eness o he cons an Ais equi alen o he condi ion
(a1)Ra∗
au= 0, R
a∗u < +∞ o all ∈(a∗, b∗), A < +∞and Rb∗
a∗u=
+∞implies Rb∗
a∗ p0= +∞, whe e a∗:= sup{ ∈[a, b)|R
a p0= 0},
b∗:= sup{ ∈[a, b)|R
a p0<+∞} and
A:= sup
a∗< <b∗
A( ) := sup
a∗< <b∗Z
a∗
u−1
qZ
a∗
p0−1
p0
.
Mo eo e , A=A. I (a1) holds hen o ∈(a, a∗] we ha e
R
au−1
q= 0 and o ∈(b∗, b) we ha e R
a p0−1
p0
= 0. Fo
∈(a∗, b∗] we ha e A( ) = A( )≤A. Pu ing hese oge he shows
ha A ≤ A < +∞. Con e sely, i A<+∞ hen A(a∗)<+∞
and Ra∗
a p0= 0 so Ra∗
au= 0. The condi ion ha R
a∗u < +∞
o ∈(a∗, b∗) is e iden because R
a p0∈(0,+∞) he e. I
Rb∗
a∗u= +∞ hen we mus ha e Rb∗
a∗ p0−1
p0
= 0 so Rb∗
a∗ p0= +∞.
Finally, because Ra∗
a p0= 0 we ob ain A≤ A <+∞and we ha e
comple ed (a1).
.
426 D. V. P okho o
Le (a) hold. Assume ha a∗< b∗. Fix an a bi a y ∈(a∗, b∗).
Subs i u ing (x) = +∞χ( ,b)(x) + (x)p0−1χ(a, )(x) in o (3), we ob ain
0<Z
a
p01
p
≤CZ
a
u(x)Zx
a
p0q
dx1
q
=C Za∗
a
u(x)Zx
a
p0q
dx+Z
a∗
u(x)Zx
a
p0q
dx!1
q
,
since < +∞a.e. on (a, b∗). I implies
Za∗
a
u(x)Zx
a
p0q
dx < +∞and Z
a∗
u(x)Zx
a
p0q
dx < +∞,
o he wise we ha e a con adic ion 0 <0. Since o x∈(a, a∗) i holds
ha Rx
a p0= 0, we ha e Ra∗
au= 0 and
(4) Z
a∗
p01
p
≤CZ
a∗
u1
qZ
a∗
p0,
because R
a∗ p0∈(0,+∞). I R
a∗u= +∞, hen (4) implies ha R
a∗ p0=
+∞which con adic s wi h he de ini ion o b∗. I R
a∗u=0, hen A( )=0.
I R
a∗u∈(0,+∞), hen i ollows om (4) ha A( )≤C.
Analogously, using he unc ion wi h =b∗we ob ain (4) wi h =
b∗and his inequali y implies ha Rb∗
a∗ p0= +∞in he case Rb∗
a∗u= +∞.
Now i a∗=b∗=b, hen Rb
a = 0 and by pu ing = +∞in o (3)
we ge
+∞ ≤ C Zb
a
u(x)·(+∞)dx!1
q
.
Thus, Rb
au= 0 and all he condi ions o (a1) hold, as well as, in he
case a∗=b∗=a.
Le (a1) hold. Since Ra∗
au=Ra∗
a p0= 0, i yields ha (3) is equi a-
len o he inequali y
(5) Zb
a∗
u(x)Zx
a∗
(y)g(y)1
qdyq
dx ≤C−q Zb
a
gp
q!q
p
o all g∈M+(a, b).
I a∗=b∗=b, hen (5) holds. Le a∗< b∗o a∗=b∗=a. Fix an
a bi a y non nega i e measu able unc ion gon (a, b). I Rb
agp
q= +∞,
Ha dy’s Inequali ies o Nega i e Indices 427
hen we ha e (5). Now le Rb
agp
q<+∞. I b∗< b, hen, by H¨olde ’s
inequali y wi h exponen s pand p0, we ind ha o any x∈(b∗, b)
Zx
a∗
(y)g(y)1
qdy ≥Zx
a∗
p01
p0Zx
a∗
gp
q1
p
= +∞.
Thus,
(6) J1:=Zb
a∗
u(x)Zx
a∗
(y)g(y)1
qdyq
dx=Zb∗
a∗
u(x)Zx
a∗
(y)g(y)1
qdyq
dx.
In pa icula , i a∗=b∗=a, hen (5) holds.
Now le a∗< b∗. Pu
N:= 




in (k∈Z|k≥log2 Zb∗
a∗
p0!),i Zb∗
a∗
p0
<+∞,
+∞,o he wise,
and cons uc he sequence {ak}k≤Nby he ela ions: Rak
a∗ p0= 2k,
k < N;aN=b∗. Then i ollows om (6) ha
(7) J1=X
k<N Zak+1
ak
u(x)Zx
a∗
(y)g(y)1
qdyq
dx.
No e ha Rak+1
aku < +∞ o all k < N since R
a∗u < +∞ o all ∈
(a∗, b∗) and Rb∗
a∗u= +∞implies ha Rb∗
a∗ p0= +∞.
By applying he H¨olde inequali y wi h exponen s p0and p, we ha e
o all x∈(a∗, b∗)
Zx
a∗
(y)g(y)1
qdy =Zx
a∗
(y)g(y)1
qZy
a∗
p01
pp0−1
pp0
dy
≥ Zx
a∗
(y)p0
Zy
a∗
p0−1
p
dy!1
p0 Zx
a∗
g(y)p
qZy
a∗
p01
p0
dy!1
p
= (p0)1
p0Zx
a∗
p01
p02 Zx
a∗
g(y)p
qZy
a∗
p01
p0
dy!1
p
.

428 D. V. P okho o
Using his ela ion and (7) we es ima e he le pa o (5) as ollows:
J1.X
k<N Zak+1
ak
uZak
a∗
p0q
p02 Zak+1
a∗
g(y)p
qZy
a∗
p01
p0
dy!
q
p
≤X
k<N Zak+1
ak
u(2k)
q
p02
X
j≤k"Zaj+1
aj
gp
q#(2j+1)1
p0

q
p
.
Mo eo e , by applying Minkowski’s inequali y we inally ob ain ha
J1.

X
j<N "Zaj+1
aj
gp
q#(2j+1)1
p0
X
j≤k<N Zak+1
ak
u(2k)
q
p02

p
q


q
p
.A−q

X
j<N "Zaj+1
aj
gp
q#(2j)1
p0
X
j≤k<N
(2k)−q
p0p

p
q


q
p
.A−q"Zb
a
gp
q#q
p
.
The p oo is comple e.
Theo em 2. Le −∞ < p < q < 0,1
:= 1
q−1
p,u, ∈M+(a, b)
and I=I×o I=I×. Then (a)is equi alen o B<+∞, whe e
B:= Zb
a
[(Iu)(x)]
ph(I p0)(x)i
p0
u(x)dx!−1
.
Mo eo e , B ≈ C o he leas possible cons an Cin (3).
P oo : As in Theo em 1 we only conside he case I=I×and no e ha
ini eness o Bis equi alen o he condi ion
(a2)Ra∗
au= 0, R
a∗u < +∞ o all ∈(a∗, b∗), B <+∞and Rb∗
a∗u=+∞
implies Rb∗
a∗ p0= +∞, whe e a∗,b∗be he same as in Theo em 1
and
B:= Zb∗
a∗Zx
a∗
u
pZx
a∗
p0
p0
u(x)dx!−1
.
Ha dy’s Inequali ies o Nega i e Indices 429
Mo eo e , B=B. I (a2) holds, hen B=B, ha is B<+∞.
Con e sely, le B<+∞. Since B ≥ q
−1
A, all condi ions
o (a2), excep B < +∞, ollow om ini eness o Ain he same
way as in he p oo o Theo em 1. I implies he equali y B=B
and (a2) ollows.
The case a∗=b∗can be p o ed analogously wi h he p oo o Theo-
em 1. The e o e, we assume ha a∗< b∗.
Le (a) hold. All condi ions, excep B < +∞, ollow in he same way
as in he p oo o Theo em 1. We only need o show ha Bis ini e.
I Rb∗
a∗u= 0, hen B= 0 <+∞. Le Rb∗
a∗u > 0. Then he e exis
numbe s 1and 2such ha a∗< 1< 2< b∗and R 2
1u > 0. Deno e
˜u:= uχ( 1, 2)and le ˜
Bbe simila o Bwi h ˜uins ead o u. Then
˜
B="Z 2
1Zx
1
u
pZx
a∗
p0
p0
˜u(x)dx#−1
≤Z 1
a∗
p0−1
p0Z 2
1
u−1
q
<+∞,
˜
B > 0 and he inequali y (3) holds wi h ˜uins ead o u:
Zb
a
(x)pdx!1
p
≤C Zb
a
˜u(x)Zx
a
q
dx!1
q
o all ∈M+(a, b).
The las inequali y is equi alen o
(8) Zb∗
a∗
˜u(x)Zx
a∗
(y)g(y)1
qdyq
dx ≤C−q Zb
a
gp
q!q
p
o all g∈M+(a, b).
Pu
g(y)p
q:= (y)p0 Zb∗
y
˜u(z)Zz
a∗
p0q−1
dz!
q
χ(a∗,b∗)(y).
430 D. V. P okho o
Then
Zb
a
gp
q=Zb∗
a∗
(y)p0 Zb∗
y
˜u(z)Zz
a∗
p0q−1
dz!
q
dy =: J2
and since < +∞a.e. on (a∗, b∗) and Rb∗
y˜u(z)Rz
a∗ p0q−1
dz < +∞
o all y∈(a∗, b∗) we ha e
Zb∗
a∗
˜u(x)Zx
a∗
(y)g(y)1
qdyq
dx
≥Zb∗
a∗
˜u(x)"Zb∗
x
˜u(z)Zz
a∗
p0q−1
dz#
pZx
a∗
p0q−1Zx
a∗
(y)p0
dydx
=Zb∗
a∗
(y)p0
Zb∗
y
˜u(x)Zx
a∗
p0q−1"Zb∗
x
˜u(z)Zz
a∗
p0q−1
dz#
p
dx dy =q
J2,
so ha , by (8), q
J2≤C−qJ
q
p
2.
Le {ak}k≤Nbe he same sequence as in he p oo o Theo em 1.
Recall ha Rak+1
ak˜u < +∞ o all k < N and we ha e
J2≤X
k<N Zak+1
ak
p0 Zb∗
ak
˜u(z)Zz
a∗
p0q−1
dz!
q
.X
k<N
2k
X
k≤j<N "Zaj+1
aj
˜u#(2j)q−1

q
.
Pu ing β
q
j:= (2j)
p0Raj+1
aj˜u
q, we ind ha
J2.X
k<N
2k
X
k≤j<N
βj(2j)−q


q
.X
k<N
β
q
k
by he disc e e Ha dy inequali y (see e.g. [2]) and since o any ixed n∈Z

X
k≤n
2k

q

X
j≥nh(2j)−q
ip
q

q
p
.1
Ha dy’s Inequali ies o Nega i e Indices 431
holds. Con e sely,
J2≥X
k<N "Zak
ak−1
p0# Zb∗
ak
˜u(z)Zz
a∗
p0q−1
dz!
q
&X
k<N
2kZak+1
ak
˜u
q
(2k)(q−1)
q=X
k<N
β
q
k.
Analogously,
˜
B− .X
k<N Zak+1
ak
˜uZak+1
a∗
˜u
p
(2k)
p0
≤X
k<N
(2k)
p0
X
j≤kZaj+1
aj
˜u

q
=X
k<N
(2k)
p0
X
j≤k
βj(2j)−q
p0

q
.X
k<N
β
q
k
and
˜
B− &X
k<N Zak+1
ak
˜u(x)Zx
a∗
˜u
p
dx(2k)
p0
≥X
k<N
(2k)
p0Zak+1
ak
˜u(x)Zx
ak
˜u
p
dx =q
X
k<N
β
q
k.
Thus,
˜
B− .J2≤C−qJ
q
p
2.C−q˜
B− q
p,
ha is ˜
B.C, since ˜
B∈(0,+∞). By now le ing 1→a∗and 2→b∗,
we conclude ha B < +∞and he p oo o he implica ion (a)⇒(a2)
is comple e.
Con e sely, assume ha (a2) holds. Fix any non-nega i e measu -
able unc ion gon (a, b). By a guing simila as in he p oo ha (a1)
438 D. V. P okho o
e e y x∈(a, b∗) by he de ini ion o b∗. The e o e, we ha e
Zb
a
(x)pdx!1
p
≤ Zb∗
a
(x)pdx!1
p
≤ Zb∗
a
g(x)p (x)−pdx!1
p
= Zb
a
g(x)p (x)−pdx!1
p
≤C Zb∗
a
u(x) [(Ig)(x)]qdx!1
q
=C Zb∗
a
u(x) [(I( ))(x)]qdx!1
q
=C Zb
a
u(x) [(I( ))(x)]qdx!1
q
.
The p oo o he case I=I×is comple e and he case I=I×can be
p o ed analogously.
P oposi ion 3. Le p, q ∈(−∞,0) and I=I×o I=I×. Then he
inequali y (13) is equi alen o
(15) Zb
a
g(y)p (y)−pdy!1
p
≤C Zb
ahu(x)1
q(Ig)(x)iq
dx!1
q
o all g∈M+(a, b).
P oo : This p oposi ion can be p o ed in he same way as he p oo o
P oposi ion 2. We only no e ha bo h (15) and (13) implies u < +∞
a.e. on (a, b) (see he p oo o P oposi ion 1) and, consequen ly, o almos
all x∈(a, b) he equali y (Ih)(x) = +∞implies u(x)1
q(Ih)(x) = +∞.

Ha dy’s Inequali ies o Nega i e Indices 439
P oposi ion 4. Le p, q ∈(−∞,0) and I=I×o I=I×. Then he
inequali y
(16) Zb
a
[g(y) (y)−1]pdy!1
p
≤C Zb
a
u(x) [(Ig)(x)]qdx!1
q
o all g∈M+(a, b)
holds, i mes{x∈(a, b)| (x) = +∞} >0o (3) holds.
P oo : Su iciency. Le (3) hold. Fix an a bi a y measu able unc-
ion g≥0. Pu (x) := g(x) (x)−1in o (3). Then we ha e (16) since
(x) (x)≤g(x). Le mes{x∈(a, b)| (x) = +∞} >0. Then
Zb
a
[g(y) (y)−1]pdy ≥Z{x∈(a,b)| (x)=+∞}
[g(y) (y)−1]pdy = +∞
o any measu able unc ion g≥0. Hence, he le pa o (16) is ze o.
Necessi y. Le mes{x∈(a, b)| (x) = +∞} = 0 and (16) holds.
Fix an a bi a y measu able unc ion ≥0 and le he unc ion 0be
in eg able on (a, b) and 0(x)∈(0,+∞), x∈(a, b). Deno e E:= {x∈
(a, b)| (x) = 0}and Ec:= (a, b) E. Pu gn= χEc+1
n 0χE,n∈N
in o (16). Then we ob ain
Zb
a
(x)pdx!1
p
≤ZEc
(x)pdx1
p
= Zb
a
[gn(x) (x)−1]pdx!1
p
≤C Zb
a
u(x) [(Ign)(x)]qdx!1
q
o all n∈N. Since u(x)[(Ign)(x)]q↑u(x)[(I( ))(x)]qas n→ ∞ o
e e y x∈(a, b) he Mono one Con e gence Theo em implies (3).
440 D. V. P okho o
P oposi ion 5. Le p, q ∈(−∞,0) and I=I×o I=I×. Then he
inequali y
(17) Zb
a
[g(y) (y)−1]pdy!1
p
≤C Zb
ahu(x)1
q(Ig)(x)iq
dx!1
q
o all g∈M+(a, b)
holds, i mes{x∈(a, b)| (x) = +∞} >0o (13) holds.
P oo : P oposi ion 5 can be p o ed in a comple ely simila way as P opo-
si ion 4, so we lea e ou he de ails.
Rema k. I is clea ha i p, q ∈(0,1), hen (14), (15), (16) and (17) a e
equi alen , and (3) is equi alen o (13).
P oposi ion 6. Le p, q ∈(0,1) and I=I×. Then (16) holds, i
Rb∗
u= +∞ o all ∈(a, a∗),Rb
b∗ (x)−pdx = 0 and
(18) Zb∗
a∗
(x)pdx!1
p
≤C Zb∗
a∗
u(x)Zx
a∗
q
dx!1
q
o all ∈M+(a∗,b∗)
holds, whe e a∗:= in { ∈(a, b]|mes{x∈( , b)| (x) = 0}= 0}and
b∗:= in { ∈[a∗, b]|Rb
u= 0}.
P oo : Necessi y. Fix any ∈(a, a∗). The e exis s γ∈( , a∗) such ha
mes{x∈( , γ)| (x) = 0}>0. Pu g:= χ( ,γ)in o (16). Then we
ob ain
+∞=Zγ
(x)−pdx1
p
≤C Zb
a
u(x)Zx
a
gq
dx!1
q
=C Zb∗
u(x)Zx
gq
dx!1
q
≤C(γ− ) Zb∗
u!1
q
,
ha is, Rb∗
u= +∞. Inse g:= χ(b∗,b)in o (16) and we ind ha
Rb
b∗ (x)−pdx = 0.
Ha dy’s Inequali ies o Nega i e Indices 441
Now ix any measu able ≥0. I Rb∗
a∗u(x)Rx
a∗ qdx = +∞,
hen (18) holds. Le Rb∗
a∗u(x)Rx
a∗ qdx < +∞. Deno e
E:= x∈(a∗,b∗)|u(x)Zx
a∗
q
<+∞.
By he de ini ion o b∗ o any ξ∈(a∗,b∗) he e exis s x∈(ξ, b∗)∩E
such ha u(x)6= 0. Then Rx
a∗ < +∞. Since ξwas aken a bi a y we
ha e < +∞a.e. on (a∗,b∗) and
mes{x∈(a∗,b∗)| (x)6= 0, (x) = +∞} = 0.
This ela ion and > 0 a.e. on (a∗,b∗) imply ha (y) (y) (y)−1= (y)
o almos e e y y∈(a∗,b∗). I emains only o pu g:= χ(a∗,b∗)
in o (16) o ob ain (18).
Su iciency. Fix any measu able g≥0. Le Ra∗
ag= 0. Inse (x) =
g(x) (x)−1,x∈(a∗,b∗), in o (18). Since ≤gand Rb
b∗ (x)−pdx = 0
we ob ain
Zb
a
[g(y) (y)−1]pdy!1
p
= Zb∗
a∗
[g(y) (y)−1]pdy!1
p
≤C Zb∗
a∗
u(x)Zx
a∗
gq
dx!1
q
=C Zb
a
u(x)Zx
a
gq
dx!1
q
.
I Ra∗
ag > 0, hen he e exis s a numbe ∈(a, a∗) such ha R
ag > 0
and
Zb
a
u(x)Zx
a
gq
dx!1
q
≥ Zb∗
u!1
qZ
a
g= +∞
by he i s condi ion.
Finally, in a simila way we can p o e he ollowing s a emen :
442 D. V. P okho o
P oposi ion 7. Le p, q ∈(0,1) and I=I×. Then (16) holds, i
R
¯
au= +∞ o all ∈(¯
b, b),R¯
a
a (x)−pdx = 0 and
(19) Z¯
b
¯
a
(x)pdx!1
p
≤C Z¯
b
¯
a
u(x)"Z¯
b
x
#q
dx!1
q
o all ∈M+(¯
a,¯
b)
holds, whe e ¯
b:= sup{ ∈[a, b)|mes{x∈(a, )| (x) = 0}= 0}and
¯
a:= sup{ ∈[a, ¯
b]|R
au= 0}.
Acknowledgemen . The au ho exp esses his deep g a i ude o P o-
esso La s-E ik Pe sson o d awing he au ho ’s a en ion o his p ob-
lem and o ui ul discussions. He also hanks P o esso Vladimi
D. S epano o aluable commen s and ema ks.
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