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On global solutions of the radial p-Laplace equation
Nonlinear Analysis 70 (2009) 3437–3451
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On global solutions of the radial p-Laplace equationI Andrej A. Kon’kov Department of Differential Equations, Faculty of Mechanics and Mathematics, Moscow Lomonosov State University, Vorobyovy Gory, 119992 Moscow, Russia
article
info
a b s t r a c t
Article history: Received 21 April 2008 Accepted 15 July 2008
We obtain blow-up conditions for solutions of the radial p-Laplace equation. © 2008 Elsevier Ltd. All rights reserved.
Keywords: Nonlinear equations A priori estimates Blow-up phenomenon
1. Introduction We consider the problem 1
d
r n−1 dr
r
n−1
p−2 ! du du = c (r , u), dr dr
u(a) > 0,
du dr
(a) ≥ 0
(1.1)
for the radial p-Laplace operator of dimension n ≥ 1, where p > 1 is a real number and c : [a, ∞) × (0, ∞) → [0, ∞) is a function from the Carathéodory class Kloc ([a, ∞) × R), a > 0 [1]. Let θ > 1 and σ > 1 be some real numbers fixed throughout this paper. We shall assume that inf
ζ ∈(t /θ,θ t )
c (r , ζ ) ≥ f (r )g (t )
(1.2)
for almost all r ≥ a and for all t > 0, where f : [a, ∞) → [0, ∞) belongs to the space L1,loc ([a, ∞)) and g : (0, ∞) → (0, ∞) is a continuous function. Put f (r ) =
ess sup
ξ ∈(r /σ ,r σ )∩[a,∞)
f (ξ ),
r ∈ [a, ∞).
(1.3)
As usual, by Z and N we denote the sets of integers and positive integers, respectively. Also, let Ia = {k ∈ Z : 2k ≥ a}. Definition 1.1. A solution u˜ : [a, b˜ ) → R of (1.1) is called an extension of a solution u : [a, b) → R if b˜ ≥ b and u˜ (r ) = u(r ) for all r ∈ [a, b). Definition 1.2. A solution of problem (1.1) which coincides with every its extension, is said to be non-extendable. It can be shown that every non-extendable solution of (1.1) is either a global solution defined on the whole set [a, ∞) or tends to infinity on a finite interval what is known as the ‘‘blow-up phenomenon’’. Questions treated in the present paper were earlier investigated by a number of authors [1–7]. Below, we obtain conditions guaranteeing that problem (1.1) has no global solutions. Our results can be applied to a wide class of equations including the case when the methods given in [1,2, 5,6] are useless (see Example 2.3). I The research was supported by RFBR, grant 08-01-00819.
E-mail address: [email protected]. 0362-546X/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2008.07.001
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A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
2. Main results Assume that Ω ⊂ [a, ∞) is a measurable set and χΩ is the characteristic function of Ω , i.e.
χΩ (ξ ) =
1, 0,
ξ ∈ Ω, ξ ∈ R \ Ω.
By the capacity of Ω relative to the ordered triple F = (f , p, n) we mean
Z ∞ X capF (Ω ) = inf
ri
dt ri−1
i=1
t n −1
!1/(p−1) (p−1)/p
t
Z
1
ξ n−1 f (ξ )χΩ (ξ )dξ
,
ri−1
where the infimum in the right-hand side is taken over all increasing sequences of real numbers {ri }∞ i=0 such that r0 = a
and
lim ri = ∞.
(2.1)
i→∞
In particular, we write capF (Ω ) = ∞ if
Z ∞ X
ri
dt ri−1
i=1
1
t
Z
t n −1
!1/(p−1) (p−1)/p =∞ ξ n−1 f (ξ )χΩ (ξ )dξ
ri−1
for every increasing sequence of real numbers {ri }∞ i=0 satisfying conditions (2.1). This capacity has the following natural properties. (i) Monotonicity. If Ω1 ⊂ Ω2 , then capF (Ω1 ) ≤ capF (Ω2 ). (ii) Semiadditivity. Suppose that p ≥ 2, then capF (Ω1 ∪ Ω2 ) ≤ capF (Ω1 ) + capF (Ω2 ).
(2.2)
Property (i) is obvious, whereas to prove (ii) it is sufficient to use the estimate
χΩ1 ∪Ω2 (ξ ) ≤ χΩ1 (ξ ) + χΩ2 (ξ ), α
α
ξ ∈ R,
α
and the fact that (h1 + h2 ) ≤ h1 + h2 for all real numbers h1 ≥ 0, h2 ≥ 0, and 0 < α ≤ 1. In the case of 1 < p < 2, inequality (2.2) should be replaced by capF (Ω1 ∪ Ω2 ) ≤ 21/p capF (Ω1 ) + 21/p capF (Ω2 ). Theorem 2.1. Let capF ([a, ∞)) = ∞ and ∞
Z
(g (t )t )−1/p dt < ∞,
(2.3)
1
then problem (1.1) has no global solutions. At first glance it would seem that calculating the above capacity is a difficult task. However, in the majority of cases, it can be successfully estimated. We also prove Theorems 2.2–2.5, which helps one to answer the question whether capF ([a, ∞)) equals infinity. Theorem 2.2. Let p ≥ 2 and ∞
Z
(ξ f (ξ ))1/(p−1) µ−1/p (ξ ) dξ = ∞,
(2.4)
a
where µ(ξ ) = 1 + (ξ p f (ξ ))1/(p−1) , then capF ([a, ∞)) = ∞. Corollary 2.1. Suppose that p ≥ 2. If relations (2.3) and (2.4) are valid, then problem (1.1) has no solutions defined on the whole set [a, ∞). The last statement can be readily obtained from Theorems 2.1 and 2.2.
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
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Theorem 2.3. Suppose that 1 < p < 2 and
ξ
2k
k∈Ia
!1/(p−1)
2k+1
X Z
p−1
f (ξ )ν
−1/p
(ξ )dξ
= ∞,
(2.5)
where ν(ξ ) = 1 + ξ p f (ξ ), then capF ([a, ∞)) = ∞. Corollary 2.2. Let 1 < p < 2 and, moreover, conditions (2.3) and (2.5) be fulfilled, then problem (1.1) has no solutions defined on the whole set [a, ∞). To establish the validity of Corollary 2.2, it is sufficient to combine Theorems 2.1 and 2.3. Theorem 2.4. Suppose that p ≥ 2, ∞
Z
(ξ f (ξ ))1/(p−1) dξ = ∞,
(2.6)
a
and
(r p f (r ))1/(p−1) < ∞, (ξ f (ξ ))1/(p−1) dξ a
lim sup R r r →∞
(2.7)
then capF ([a, ∞)) = ∞. Theorem 2.5. Let 1 < p < 2,
k∈Ia
!1/(p−1)
2k+1
X Z
2k
ξ
p−1
f (ξ )dξ
= ∞,
(2.8)
and lim sup r →∞
(r p f (r ))1/(p−1) P R k∈Ia
2k+1 2k
ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
1/(p−1) < ∞,
(2.9)
then capF ([a, ∞)) = ∞. Theorems 2.4 and 2.5 immediately imply the following two assertions. Corollary 2.3. Suppose that p ≥ 2. If relations (2.3), (2.6) and (2.7) hold, then problem (1.1) has no global solutions. Corollary 2.4. Let 1 < p < 2 and, moreover, conditions (2.3), (2.8) and (2.9) be valid, then problem (1.1) has no solutions defined on the whole set [a, ∞). Theorems 2.1–2.5 will be proved later. Now, we demonstrate the exactness of the above results. Example 2.1. Suppose that g (t ) = t λ and f ( r ) = f0 r s ,
f0 = const > 0.
(2.10)
If λ > p − 1 and s ≥ −p, then in accordance with Corollaries 2.1 and 2.2, problem (1.1) has no solutions defined on the whole set [a, ∞). Further, let f (r ) = f0 r −p logs (1 + r ),
f0 = const > 0,
(2.11)
then the inequalities λ > p − 1 and s ≥ 1 − p imply the absence of global solutions of (1.1). Example 2.2. Suppose that g (t ) = t p−1 logλ (2 + t ) and (2.10) holds. If λ > p and s ≥ −p, then problem (1.1) has no global solutions by Corollaries 2.1 and 2.2. Analogously, if the function f satisfies condition (2.11), then the inequalities λ > p and s ≥ 1 − p guarantee the absence of global solutions of (1.1). It is known that the blow-up conditions given in Examples 2.1 and 2.2 are exact for all n > p > 1 [3,5,6]. Let us consider the case where the function f is not monotone.
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A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
Example 2.3. Assume that g (t ) = t p−1 logλ (2 + t ), λ > p ≥ 2, and
p ∞ X log k
f (r ) = f0
k
k =2
χ[k,k+log−2 k] (r ),
f0 = const > 0.
It does not present any particular problem to verify that both relations (2.3) and (2.4) are fulfilled. In doing so, one can take an arbitrary real number σ > 1 on the right in (1.3), for instance, σ = 2. Thus, applying Corollary 2.1, we conclude that problem (1.1) has no global solutions. Lemma 2.1. Suppose that capF ([a, ∞)) = ∞, then ∞
Z
dt
t n−1
a
t
Z
1
ξ
n −1
f (ξ )dξ
1/(p−1) = ∞.
(2.12)
a
Proof. If (2.12) is not valid, then there exists an increasing sequence of real numbers {ri }∞ i=0 satisfying conditions (2.1) and such that ri
Z
dt
t n −1
ri−1
t
Z
1
ξ n−1 f (ξ )dξ
1/(p−1)
1
<
2i
a
,
i = 2, 3, . . . ,
whence we obviously obtain
Z
ri
dt
ri−1
t n −1
!1/(p−1) (p−1)/p < ξ n−1 f (ξ )dξ
t
Z
1
ri−1
1 2i(p−1)/p
,
i = 2, 3, . . . .
Therefore,
Z ∞ X capF ([a, ∞)) ≤
ri
dt ri−1
i=1
Z
1 t n −1
This contradiction proves the lemma.
!1/(p−1) (p−1)/p < ∞. ξ n−1 f (ξ )dξ
t ri−1
Proof of Theorem 2.1. Assume the converse, then there exists a solution of problem (1.1) defined on the whole set [a, ∞). It is clear from (1.1) that u is a non-decreasing function. We also have lim u(r ) = ∞.
r →∞
In fact, if lim u(r ) < ∞,
(2.13)
r →∞
then inf
ζ ∈[a,∞)
g (u(ζ )) > 0.
Integrating (1.1), we obtain u(r ) = u(a) +
r
Z
dt
a
an − 1 t
t
Z
1
(u0 (a))p−1 + n−1
t n−1
ξ n−1 c (ξ , u)dξ
1/(p−1)
a
for all r ∈ [a, ∞), whence in accordance with (1.2) it can be seen that u(r ) ≥
1/(p−1) Z
inf
ζ ∈[a,∞)
r
g (u(ζ ))
dt
a
1 t n−1
t
Z
ξ n−1 f (ξ )dξ
1/(p−1)
a
for all r ∈ [a, ∞). By Lemma 2.1, this contradicts (2.13). Let us consider a sequence of real numbers {ri }∞ i=0 such that r0 = a and u(ri ) = θ u(ri−1 ), i = 1, 2, . . .. Taking into account (1.1), we have u(ri ) = u(ri−1 ) +
Z
ri
dt ri−1
rin−−11 t n−1
(u (ri−1 )) 0
p−1
+
1 t n −1
Z
!1/(p−1)
t
ξ ri−1
n−1
c (ξ , u)dξ
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3441
for all i ∈ N. This immediately implies the inequality ri
Z
u(ri ) − u(ri−1 ) ≥
dt
t n−1
ri−1
!1/(p−1)
t
Z
1
ξ
n −1
c (ξ , u)dξ
ri−1
for all i ∈ N. Combining the last formula and relation (1.2), we derive u(ri ) − u(ri−1 )
inf
≥
g 1/(p−1) (ζ )
ζ ∈(u(ri−1 ),u(ri ))
ri
Z
dt
t n −1
ri−1
!1/(p−1)
t
Z
1
ξ
n −1
f (ξ )dξ
ri−1
for all i ∈ N. At the same time, u(ri )
Z
u(ri−1 )
(g (t )t )−1/p dt ≥ C
u(ri ) − u(ri−1 )
inf
(p−1)/p
g 1/(p−1) (ζ )
ζ ∈(u(ri−1 ),u(ri ))
with some constant C > 0 depending only on p and θ . Thus, we obtain the estimate
u(ri )
Z
u(ri−1 )
(g (t )t )−1/p dt ≥ C
ri
Z
dt ri−1
t n−1
!1/(p−1) (p−1)/p ξ n−1 f (ξ )dξ
t
Z
1
ri−1
for all i ∈ N, from which it follows that ∞
Z
u(a)
−1/p
(g (t )t )
Z ∞ X dt ≥ C
ri
dt ri−1
i =1
t n −1
!1/(p−1) (p−1)/p
t
Z
1
ξ
n −1
f (ξ )dξ
.
ri−1
This contradicts hypotheses of Theorem 2.1. The proof is completed.
Lemma 2.2. Let p ≥ 2, then a2
Z
dt
Z
1 t n−1
a1
t
ξ
n −1
f (ξ )dξ
1/(p−1)
Z
a2
≥A
1−
a1
1/(p−1)
ξ
a2
a1
(ξ f (ξ ))1/(p−1) dξ
for all real numbers a ≤ a1 < a2 , where the constant A > 0 depends only on p and n. Proof. Consider a sequence of real numbers ti ∈ [a1 , a2 ), i = 0, 1, 2, . . ., defined by induction. We take t0 = a1 . Assume further that ti−1 is already known. We put ti = 2ti−1 if 4ti−1 < a2 and ti = (ti−1 + a2 )/2, otherwise. From the construction of the sequence {ti }∞ i=0 , it is clear that ti−1 < ti < 4ti−1 and ti − ti−1 ≤ 2(ti+1 − ti ) for all i = 1, 2, . . .. Moreover, we have
1/(p−1) ti−1 1/(p−1) (ti − ti−1 )1/(p−1) ≥ B 1 − ti ,
i = 1, 2, . . . ,
a2
(2.14)
with some constant B > 0 depending only on p. Let i be a positive integer, then ti+1
Z
dt
ti
t
Z
1 t n−1
ξ
n −1
f (ξ )dξ
1/(p−1)
Z
≥ C (ti+1 − ti )
a1
ti
!1/(p−1) f (ξ )dξ
,
ti−1
where the constant C > 0 depends only on p and n. At the same time, using the Hölder inequality, one can show that
(ti − ti−1 )
Z
(p−2)/(p−1)
!1/(p−1)
ti
f (ξ )dξ
Z
ti
≥
ti−1
f 1/(p−1) (ξ )dξ .
ti−1
Hence, we obtain ti+1
Z 2
dt
ti
1
Z
t n−1
t
ξ
n −1
f (ξ )dξ
1/(p−1)
1/(p−1)
≥ C (ti − ti−1 )
a1
Z
ti
f 1/(p−1) (ξ )dξ .
ti−1
Combining this with the estimate
(ti − ti−1 )
1/(p−1)
Z
ti
f ti−1
1/(p−1)
(ξ )dξ ≥ B
Z
ti
1−
ti−1
ξ a2
1/(p−1)
(ξ f (ξ ))1/(p−1) dξ ,
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A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
which follows from (2.14), we derive ti+1
Z
dt
t n −1
ti
t
Z
1
ξ n−1 f (ξ )dξ
1/(p−1)
ti
Z
1−
≥A ti−1
a1
1/(p−1)
ξ a2
(ξ f (ξ ))1/(p−1) dξ ,
where A = BC /2. Finally, summing the last inequality over all positive integers i, we complete the proof.
Lemma 2.3. Suppose that a1 < a2 and 0 < α < 1 are some real numbers, then a2
Z
ϕ(ξ )dξ
α
a2
Z ≥B
ϕ(ξ )~ α−1 (ξ )dξ
(2.15)
a1
a1
for every non-negative function ϕ ∈ L1 ([a1 , a2 ]), where
~(ξ ) =
a2
Z ξ
ϕ(ζ )dζ
and the constant B > 0 depends only on α . Remark 2.1. Throughout this paper we assume by definition that 0/0 = 0. In particular, if ~(ξ ) = 0 for some ξ ∈ (a1 , a2 ), then the function ϕ~ α−1 on the right in (2.15) is equal to zero almost everywhere on the interval [ξ , a2 ]. Proof of Lemma 2.3. Without loss of generality, we may assume that ~(ξ ) > 0 for any ξ ∈ (a1 , a2 ); otherwise, we replace a2 by inf{ξ ∈ (a1 , a2 ) : ~(ξ ) = 0}. Consider the sequence of real numbers ξk ∈ [a1 , a2 ], k = 0, 1, . . . , constructed as follows. We put ξ0 = a1 . Further, let ξk−1 be already known. We take ξk ∈ [ξk−1 , a2 ] satisfying the relation ξk
Z
ξk−1
1
ϕ(ξ )dξ =
Z
a2
2 ξk−1
ϕ(ξ )dξ .
It is clear that
Z
a2
ξk
ϕ(ξ )dξ =
Z
ξk ξk−1
ϕ(ξ )dξ = 2−k
Z
a2
ϕ(ξ )dξ ,
k = 1, 2, . . . .
(2.16)
a1
In particular, we have lim ~(ξk ) = 0;
k→∞
therefore, lim ξk = a2 .
k→∞
On the other hand, it can be shown that
Z
ξk ξk−1
ϕ(ξ )~ α−1 (ξ )dξ ≤ ~ α−1 (ξk )
ξk
Z
ξk−1
ϕ(ξ )dξ ,
k = 1, 2, . . . .
By (2.16), this implies the estimate
Z
ξk ξk−1
ϕ(ξ )~ α−1 (ξ )dξ ≤ 2−αk
Z
a2
ϕ(ξ )dξ
α
,
k = 1, 2, . . . .
a1
Summing the last expression over all k ∈ N, we obtain (2.15). The proof is completed.
Proof of Theorem 2.2. Assume that is an increasing sequence of real numbers satisfying conditions (2.1) and σ > 1 is the constant in (1.3). For every i ∈ N we denote ti = ri /σ if σ ri−1 < ri and ti = ri−1 , otherwise. Let i be a positive integer. We obtain
{ri }∞ i=0
Z
ri
1−
2 ri−1
Z
ti
≥
ξ
1/(p−1)
ri
1−
ri−1
ξ ri
!(p−1)/p (ξ f (ξ ))
1/(p−1)
1/(p−1)
dξ
!(p−1)/p 1/(p−1)
(ξ f (ξ ))
dξ
ri
Z +
1−
ti
ξ ri
1/(p−1)
!(p−1)/p 1/(p−1)
(ξ f (ξ ))
dξ
.
(2.17)
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3443
At the same time, ti
Z
ξ
1−
1/(p−1)
ri
ri−1
1/(p−1) Z ti 1 (ξ f (ξ ))1/(p−1) dξ (ξ f (ξ ))1/(p−1) dξ ≥ 1 − σ ri−1
and ri
Z
ξ
1−
1/(p−1)
ri
ti
1/(p−1)
(ξ f (ξ ))
dξ ≥ σ
ri
Z
−1/(p−1)
(ri − ξ )1/(p−1) f 1/(p−1) (ξ )dξ .
ti
Therefore, relation (2.17) implies the inequality ri
Z
1−
A ri−1
1/(p−1)
ri
!(p−1)/p (ξ f (ξ ))
1/(p−1)
!(p−1)/p
ti
Z
ξ
(ξ f (ξ ))
≥
1/(p−1)
dξ
dξ ri
Z +
ri−1
(ri − ξ )1/(p−1) f 1/(p−1) (ξ )dξ
(p−1)/p
,
(2.18)
ti
where the constant A > 0 depends only on p and σ . By Lemma 2.3, we have ri
Z
(ri − ξ )1/(p−1) f 1/(p−1) (ξ )dξ
(p−1)/p
ri
Z ≥B
ti
(ri − ξ )1/(p−1) f 1/(p−1) (ξ )~ −1/p (ξ )dξ
(2.19)
ti
with some constant B > 0 depending only on p, where ri
Z
~(ξ ) =
ξ
(ri − ζ )1/(p−1) f 1/(p−1) (ζ )dζ .
Since
~(ξ ) ≤ (ri − ξ )1/(p−1)
ri
Z
f 1/(p−1) (ζ )dζ ,
ξ
it does not present any particular problem to verify that
(ri − ξ )1/(p−1) ~ −1/p (ξ ) ≥
ri
Z
1 ri − ξ
ξ
f 1/(p−1) (ζ )dζ
−1/p
for all ξ ∈ (ti , ri ). Hence, taking into account the evident estimate ri
Z
1 ri − ξ
ξ
f 1/(p−1) (ζ )dζ ≤ ξ −p/(p−1) µ(ξ )
for all ξ ∈ (ti , ri ), where µ(ξ ) = 1 + (ξ p f (ξ ))1/(p−1) , we derive
(ri − ξ )1/(p−1) ~ −1/p (ξ ) ≥ ξ 1/(p−1) µ−1/p (ξ ),
ξ ∈ (ti , ri ).
(2.20)
Combining (2.19) and (2.20), one can arrive at the conclusion that ri
Z
(ri − ξ )1/(p−1) f 1/(p−1) (ξ )dξ
(p−1)/p
ri
Z ≥B
ti
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ .
ti
Thus, formula (2.18) implies the inequality
Z
ri
1−
C ri−1
Z
ti
≥ ri−1
ξ
1/(p−1)
ri
!(p−1)/p (ξ f (ξ ))
1/(p−1)
dξ
!(p−1)/p (ξ f (ξ ))
1/(p−1)
µ
−1/p
(ξ )dξ
ri
Z +
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ ,
ti
where the constant C > 0 depends only on p and σ . Summing (2.21) over all positive integers i, we obtain
(2.21)
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A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
Z ∞ X
C
1−
Z ∞ X
!(p−1)/p
1/(p−1)
ξ
(ξ f (ξ ))
ri
ri−1
i =1
≥
ri
1/(p−1)
dξ
!(p−1)/p
ti
(ξ f (ξ ))
1/(p−1)
µ
−1/p
(ξ )dξ
∞ Z X
+
ri−1
i=1
ri
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ ,
ti
i =1
whence in accordance with the estimate
Z ∞ X
ti
!(p−1)/p (ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ
∞ Z X
≥
ri−1
i=1
!(p−1)/p
ti
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ
ri−1
i =1
it follows that
Z ∞ X
C
1−
∞ Z X
ti
!(p−1)/p
1/(p−1)
ξ
(ξ f (ξ ))
ri
ri−1
i =1
≥
ri
1/(p−1)
dξ
!(p−1)/p (ξ f (ξ ))
1/(p−1)
µ
−1/p
(ξ )dξ
∞ Z X
+
ri−1
i=1
ri
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ .
(2.22)
ti
i =1
By condition (2.4), at least one of the two summands on the right in (2.22) is equal to infinity; therefore,
Z ∞ X
ri
1−
ri−1
i=1
ξ
1/(p−1)
ri
!(p−1)/p (ξ f (ξ ))
1/(p−1)
dξ
= ∞.
(2.23)
According to Lemma 2.2, relation (2.23) implies the equality
Z ∞ X
ri
dt
t n −1
ri−1
i=1
!1/(p−1) (p−1)/p = ∞. ξ n−1 f (ξ )dξ
t
Z
1
ri−1
Thus, we have capF ([a, ∞)) = ∞. The proof of Theorem 2.2 is completed.
Lemma 2.4. Suppose that a ≤ a1 < a2 are real numbers and m < 1 + 1/(p − 1) is a positive integer, then a2
Z
t
Z
f (ξ )dξ
dt a1
1/(p−1)
a2
Z
dtf (t )fm (t )
=C
a1
t
Z
a1
f (ξ )dξ
1/(p−1)−m (2.24)
a1
with some constant C > 0 depending only on p and m, where f 1 ( t ) = a2 − t
fk (t ) =
and
a2
Z
f (ξ )fk−1 (ξ )dξ ,
k = 2, 3, . . . .
(2.25)
t
Proof. The proof is by induction on m. It is obvious that a2
Z
t
Z
f (ξ )dξ
dt a1
1/(p−1)
a2
Z =−
a1
dtf10 (t )
a1
=
p−1
f (ξ )dξ
1/(p−1)
a1
Z
1
t
Z
a2
dtf (t )f1 (t )
Z
a1
t
f (ξ )dξ
1/(p−1)−1
.
a1
This proves (2.24) for m = 1. Now, assume that a2
Z
Z
t
f (ξ )dξ
dt a1
1/(p−1)
Z
a2
=B
a1
dtf (t )fm−1 (t )
t
Z
a1
f (ξ )dξ
1/(p−1)−m+1
,
(2.26)
a1
where 1 < m < 1 + 1/(p − 1) is an integer and B > 0 is a real number. Integrating by parts, we have
Z
a2 a1
dtf (t )fm−1 (t )
Z
t
f (ξ )dξ
1/(p−1)−m+1
Z
a2
=−
a1
=
dtfm0 (t )
Z
t
f (ξ )dξ
a1
a1
1
Z
p−1
a2
−m+1 a1
1/(p−1)−m+1
dtf (t )fm (t )
Z
t
f (ξ )dξ a1
1/(p−1)−m
.
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3445
According to (2.26), this readily implies (2.24), where C = (1/(p − 1) − m + 1)B > 0. Lemma 2.4 is completely proved. Lemma 2.5. Let a ≤ a1 < a2 be real numbers and fk , k = 1, 2, . . . , the functions defined by (2.25), then 1/k
fk
(t ) ≥ AF −1/k
a2
Z
f (ξ )dξ ,
t ∈ [a1 , a2 ],
(2.27)
t
for all positive integers k, where F =
sup
ζ ∈(a1 ,a2 )
a2
Z
1 a2 − ζ
ζ
f (ξ )dξ
and A > 0 is a constant depending only on k. Proof. The proof is by induction on k. If k = 1, then (2.27) is trivial. Assume that 1/(k−1)
fk−1
(ξ ) ≥ CF −1/(k−1)
Z
a2
ξ
f (ζ )dζ
(2.28)
with some integer k > 1 for all ξ ∈ [a1 , a2 ], where C > 0 is a real number. Lemma 2.3 implies the estimate 1/k
fk
a2
Z
(t ) =
f (ξ )fk−1 (ξ )dξ
1/k
a2
Z
f (ξ )fk−1 (ξ )~ 1/k−1 (ξ )dξ
≥B
(2.29)
t
t
for all t ∈ [a1 , a2 ], where
~(ξ ) =
Z
a2
ξ
f (ζ )fk−1 (ζ )dζ
and B > 0 is a constant depending only on k. Since fk−1 is a non-increasing function, we obtain
~(ξ ) ≤ fk−1 (ξ )
a2
Z ξ
f (ζ )dζ
for all ξ ∈ [a1 , a2 ], whence in accordance with (2.28) it follows that fk−1 (ξ ) ≥ C 1−1/k F −1/k ~ 1−1/k (ξ ) for all ξ ∈ [a1 , a2 ]. Thus, we have a2
Z
f (ξ )fk−1 (ξ )~ 1/k−1 (ξ )dξ ≥ C 1−1/k F −1/k
a2
Z
t
f (ξ )dξ
t
for all t ∈ [a1 , a2 ]. Combining the last estimate with (2.29), we immediately derive (2.27), where A = BC 1−1/k . The proof is completed. ∞ Lemma 2.6. Let {Vi }∞ i=1 and {Wi }i=1 be denumerable families of measurable subsets of the interval [a, ∞). We put εij = 1 in the ∞ case of mes(Vi ∩ Wj ) 6= 0 and εij = 0, otherwise. If ∪∞ i=1 Wi ⊂ ∪i=1 Vi and, moreover, there exists a positive integer M such that
sup
∞ X
εij ≤ M
(2.30)
εij ≤ M ,
(2.31)
j∈N i=1
and sup
∞ X
i∈N j=1
then M
p/(p−1)
∞ Z X i=1
ψ(ξ )dξ Vi
1/(p−1) ≥
∞ Z X i=1
ψ(ξ )dξ
1/(p−1) (2.32)
Wi
for every measurable function ψ : [a, ∞) → [0, ∞). In particular, if the right-hand side in (2.32) is equal to infinity, then the left-hand side is also equal to infinity.
3446
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
Proof. Assume that j ∈ N. We obviously have ∞ X
εij
Z
ψ(ξ )dξ ≥
Z
ψ(ξ )dξ .
(2.33)
Wj
Vi
i=1
There is k ∈ N such that εkj = 1 and
Z
Z
ψ(ξ )dξ ≥
ψ(ξ )dξ
Vi
Vk
for any i ∈ N satisfying the condition εij = 1. Hence, in accordance with (2.30) we obtain
Z
ψ(ξ )dξ ≥
M
∞ X
Vk
εij
Z
ψ(ξ )dξ .
(2.34)
Vi
i =1
From (2.33) and (2.34), it follows that
Z
ψ(ξ )dξ ≥
M
Z
Vk
ψ(ξ )dξ Wj
or, in other words, M
1/(p−1)
Z
ψ(ξ )dξ
1/(p−1)
!1/(p−1)
Z
ψ(ξ )dξ
≥
.
Wj
Vk
Since ∞ X
εij
Z
ψ(ξ )dξ
1/(p−1)
Z
ψ(ξ )dξ
≥
Vi
i=1
1/(p−1)
,
Vk
this implies the estimate M
1/(p−1)
∞ X
εij
Z
ψ(ξ )dξ
1/(p−1)
ψ(ξ )dξ
≥
.
Wj
Vi
i=1
!1/(p−1)
Z
Finally, summing the last expression over all j ∈ N and taking into account (2.31), we immediately derive (2.32). Lemma 2.6 is completely proved. Proof of Theorem 2.3. Consider an increasing sequence of real numbers {ri }∞ i=0 satisfying conditions (2.1). By Ξ1 we denote the set of positive integers i such that σ ri−1 ≥ ri , where σ > 1 is the constant in (1.3). Also, put Ξ2 = N \ Ξ1 . At first, we assume that i ∈ Ξ1 . Let fk , k = 1, 2, . . ., be the functions defined by (2.25) with a2 = ri . According to Lemma 2.4, we have
Z
ri
Z
!1/(p−1)
t
f (ξ )dξ
dt ri−1
Z
ri
dtf (t )fm (t )
=C
ri−1
Z
ri−1
!1/(p−1)−m
t
f (ξ )dξ
ri−1
for some real number C > 0 depending only on p, where m is the maximal integer such that m < 1 + 1/(p − 1). Since 1/(p − 1) − m ≤ 0, this implies the inequality
Z
ri
Z
!1/(p−1)
t
f (ξ )dξ
dt ri−1
Z
ri
dtf (t )fm (t )
≥C
ri−1
ri−1
ri
Z
!1/(p−1)−m f (ξ )dξ
.
ri−1
At the same time, putting a1 = ri−1 , a2 = ri , and k = m + 1 in Lemma 2.5, we obtain fm+1 (ri−1 ) =
ri
Z
f (t )fm (t )dt ≥
ri−1
where F =
sup
ζ ∈(ri−1 ,ri ) ri
1
−ζ
ri
Z ζ
f (ξ )dξ
A F
Z
ri ri−1
!m+1 f (ξ )dξ
,
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3447
and the constant A > 0 depends only on p. Hence, one can assert that ri
Z
!1/(p−1)
t
Z
f (ξ )dξ
dt
≥
ri−1
ri−1
Z
AC F
ri
!1+1/(p−1) f (ξ )dξ
;
ri−1
therefore,
ri
Z
dt
t n−1
ri−1
!1/(p−1) (p−1)/p Z ξ n−1 f (ξ )dξ ≥ BF 1/p−1
t
Z
1
ri
f (ξ )dξ ,
(2.35)
ri−1
ri−1
where B = σ −(n−1)/p (AC )1−1/p . It does not present any particular problem to verify that F ≤ ξ −p ν(ξ ) for all ξ ∈ (ri−1 , ri ), where ν(ξ ) = 1 + ξ p f (ξ ). Thus, (2.35) implies the estimate
ri
Z
dt
t n−1
ri−1
!1/(p−1) (p−1)/p Z ξ n−1 f (ξ )dξ ≥B
t
Z
1
ri
ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ .
(2.36)
ri−1
ri−1
Now, assume that i ∈ Ξ2 . In this case, we obviously have σ ri−1 < ri . Let mi be the maximal integer satisfying the inequality σ mi ri−1 < ri . From the previous arguments, it follows that ri
Z
dt
t n −1
ri /σ
t
Z
1
ξ
ri /σ
n −1
f (ξ )dξ
1/(p−1) !(p−1)/p
ri
Z ≥B
ri /σ
ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ .
(2.37)
On the other hand, we derive
Z
σ −(k−1) ri
dt σ −k ri
1 t n−1
!1/(p−1)
t
Z
ξ
n −1
f (ξ )dξ
σ −k ri
Z ≥D
σ −(k+1) ri
ri−1
!1/(p−1) ξ
p−1
f (ξ )dξ
(2.38)
for all integers 1 ≤ k ≤ mi − 1, where the constant D > 0 depends only on p, σ , and n. Analogously,
Z
σ −(mi −1) ri
dt
t n −1
σ −mi ri
!1/(p−1)
t
Z
1
ξ
n−1
f (ξ )dξ
Z
σ −mi ri
≥D
ri−1
!1/(p−1) ξ
p−1
f (ξ )dξ
.
(2.39)
ri−1
Relations (2.38) and (2.39) allow us to conclude that
Z
ri
1
dt
t n −1
ri−1
Z
!1/(p−1)
t
ξ
n −1
f (ξ )dξ
≥ DJi ,
(2.40)
ri−1
where
Z
X
Ji =
1≤k≤mi −1
!1/(p−1)
σ −k ri σ −(k+1) ri
ξ
p−1
f (ξ )dξ
Z
σ −mi ri
+
!1/(p−1) ξ
p−1
f (ξ )dξ
.
ri−1
Formula (2.40) implies the inequality
X Z
ri
dt ri−1
i∈Ξ2
1
Z
t n−1
t
!1/(p−1) (p−1)/p X (p−1)/p ξ n−1 f (ξ )dξ ≥ D(p−1)/p Ji ,
ri−1
i∈Ξ2
whence, taking into account the fact that
X
(p−1)/p
Ji
!(p−1)/p ≥
i∈Ξ2
X
,
Ji
i∈Ξ2
we obtain
X Z i∈Ξ2
ri
dt ri−1
1 t n−1
Z
t
ri−1
!1/(p−1) (p−1)/p !(p−1)/p X n −1 (p−1)/p ξ f (ξ )dξ ≥D Ji . i∈Ξ2
3448
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
At the same time, by Lemma 2.6, M p/(p−1)
X
Ji ≥
i∈Ξ2
X Z ω∩[2k ,2k+1 ]
k∈Ia
ξ p−1 f (ξ )dξ
1/(p−1)
,
where ω = ∪i∈Ξ2 [ri−1 , ri /σ ] and M > 0 is some real number depending only on σ . Thus, we have
X Z
dt ri−1
i∈Ξ2
≥
ri
D(p−1)/p M
1
t
Z
t n −1
!1/(p−1) (p−1)/p ξ n−1 f (ξ )dξ
ri−1
X Z ω∩[2k ,2k+1 ]
k∈Ia
ξ
p−1
f (ξ )dξ
1/(p−1) !(p−1)/p
.
(2.41)
Further, relations (2.36) and (2.37) allow one to write
Z ∞ X
ri
dt ri−1
i=1
1
Z
t n −1
t
!1/(p−1) (p−1)/p Z n −1 ≥ B ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ , ξ f (ξ )dξ
(2.42)
τ
ri−1
where τ = [a, ∞) \ ω. If k ∈ Ia , i.e. k ∈ Z and 2k ≥ a, then
(2k+1 − 2k )2−p
Z τ ∩[2k ,2k+1 ]
ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ
p−1
Z
ξ (p−1) f p−1 (ξ )ν −(p−1)
2 /p
2
≥ τ ∩[2k ,2k+1 ]
(ξ )dξ
(2.43)
by the Hölder inequality. Since 2k+1 − 2k ≤ ξ and f p−1 (ξ ) ≥ f (ξ )ξ p(2−p) ν p−2 (ξ ) for almost all ξ ∈ [2k , 2k+1 ], it follows from (2.43) that
Z τ ∩[2k ,2k+1 ]
ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ ≥
Z τ ∩[2k ,2k+1 ]
ξ p−1 f (ξ )ν −1/p (ξ )dξ
1/(p−1)
.
Hence,
Z τ
ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ ≥
X Z τ ∩[2k ,2k+1 ]
k∈Ia
ξ p−1 f (ξ )ν −1/p (ξ )dξ
1/(p−1)
.
According to (2.42), this immediately implies the estimate
Z ∞ X
ri
dt ri−1
i=1
1
Z
t n −1
!1/(p−1) (p−1)/p
t
ξ
n −1
f (ξ )dξ
ri−1
≥B
X Z k∈Ia
τ ∩[2k ,2k+1 ]
ξ
p−1
f (ξ )ν
−1/p
(ξ )dξ
1/(p−1)
.
(2.44)
By condition (2.5), we obtain either
X Z ω∩[2k ,2k+1 ]
k∈Ia
ξ p−1 f (ξ )ν −1/p (ξ )dξ
1/(p−1) =∞
(2.45)
= ∞.
(2.46)
or
X Z k∈Ia
τ ∩[2k ,2k+1 ]
ξ p−1 f (ξ )ν −1/p (ξ )dξ
1/(p−1)
If (2.45) is valid, then
Z ∞ X i=1
ri
dt ri−1
1 t n −1
Z
!1/(p−1) (p−1)/p
t
ξ n−1 f (ξ )dξ
=∞
ri−1
in accordance with formula (2.41); otherwise, the last equality is a consequence of (2.44) and (2.46). Thus, we have capF ([a, ∞)) = ∞. The proof is completed.
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3449
Lemma 2.7. Let η > 1 be a real number. If relation (2.7) is fulfilled, then ηr
Z
r
Z
(ξ f (ξ ))1/(p−1) dξ ≤ A
(ξ f (ξ ))1/(p−1) dξ
a
a
with some constant A > 0 for all large enough r. Proof. By (2.7), there exist real numbers a0 > a and B > 0 such that
(rf (r ))1/(p−1) ≤
B
r
Z
r
(ξ f (ξ ))1/(p−1) dξ
(2.47)
a
for all r > a0 . Integrating (2.47), we obviously have λr
Z
λr
Z
(ξ f (ξ ))1/(p−1) dξ ≤ B
dξ
r
r
≤ B log λ
ξ Z
ξ
Z
(ζ f (ζ ))1/(p−1) dζ
a
λr
(ξ f (ξ ))1/(p−1) dξ
a
for all real numbers λ > 1 and r > a0 . In particular, choosing λ > 1 such that B log λ < 1/2, we obtain λr
Z
(ξ f (ξ ))1/(p−1) dξ ≤
r
λr
Z
1 2
(ξ f (ξ ))1/(p−1) dξ
a
or, in other words, λr
Z
(ξ f (ξ ))
1/(p−1)
dξ ≤ 2
r
Z
a
(ξ f (ξ ))1/(p−1) dξ
a
for all r > a0 . The last estimate allows us to write λk r
Z
(ξ f (ξ ))
1/(p−1)
dξ ≤ 2
k
r
Z
a
(ξ f (ξ ))1/(p−1) dξ
a
for all k ∈ N and r > a0 . Finally, taking k ∈ N satisfying the inequality λk ≥ η, we complete the proof.
Proof of Theorem 2.4. By condition (2.7), there exists a real number B > 0 such that (2.47) is valid for all large enough r. This implies the estimate σr
Z
(r p f (r ))1/(p−1) ≤ σ p/(p−1) B
(ξ f (ξ ))1/(p−1) dξ
a
for all large enough r, where σ > 1 is the constant in (1.3). Hence, in accordance with Lemma 2.7 and relation (2.6) there exist real numbers r0 > a and C > 0 such that
µ(r ) ≤ C
r
Z
(ξ f (ξ ))1/(p−1) dξ
a
for all r > r0 , where µ(r ) = 1 + (r p f (r ))1/(p−1) . Thus, we derive r
Z
(ξ f (ξ ))
1/(p−1)
µ
−1/p
(ξ )dξ ≥
Z
a
r
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ
r0
≥C
−1/p
r
Z
1/(p−1)
(ξ f (ξ ))
dξ
−1/p Z
a
r
(ξ f (ξ ))1/(p−1) dξ .
(2.48)
r0
Since r
Z
r0
(ξ f (ξ ))1/(p−1) dξ ≥
1 2
Z
r
(ξ f (ξ ))1/(p−1) dξ
a
for all large enough r, inequality (2.48) enables one to assert that r
Z a
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ ≥
C −1/p 2
r
Z
(ξ f (ξ ))1/(p−1) dξ
1−1/p
a
for all large enough r. By (2.6), this implies (2.4). To complete the proof it remains to use Theorem 2.2.
3450
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
Lemma 2.8. Let η > 1 be a real number and, moreover, 1 < p < 2. If (2.9) holds, then
ξ
2k
k∈Ia
!1/(p−1)
2k+1
X Z
p−1
f (ξ )χ[a,ηr ] (ξ )dξ
≤A
ξ
2k
k∈Ia
!1/(p−1)
2k+1
X Z
f (ξ )χ[a,r ] (ξ )dξ
p−1
(2.49)
with some constant A > 0 for all large enough r. Proof. According to (2.9), there exist real numbers a0 > a and B > 0 such that B
r p−1 f (r ) ≤
r
X Z
2k+1
2k
k∈Ia
!1/(p−1) p−1 ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
(2.50)
for all r > a0 . Integrating (2.50), we obtain λr
Z
ξ
p−1
f (ξ )dξ ≤ B
λr
Z
X Z
dξ
ξ k∈Ia X Z ≤ B log λ
r
r
k∈Ia
2k+1
ζ
2k
2k+1
ζ
2k
p−1
p−1
!1/(p−1) p−1 f (ζ )χ[a,ξ ] (ζ )dζ
!1/(p−1) p−1 f (ζ )χ[a,λr ] (ζ )dζ
for all real numbers λ > 1 and r > a0 , whence it follows that λr
Z
ξ
p−1
f (ξ )dξ
1/(p−1)
≤ε
r
X Z k∈Ia
!1/(p−1)
2k+1
ξ
2k
p−1
f (ξ )χ[a,λr ] (ξ )dξ
for all r > a0 , where ε = (B log λ)1/(p−1) . Combining this with the evident inequality
X Z
2k
k∈Ia
!1/(p−1)
2k+1
1/(p−1)
≤2
ξ
p−1
f (ξ )χ[a,λr ] (ξ )dξ
X Z
ξ
2k
k∈Ia
!1/(p−1)
2k+1 p−1
λr
Z
+ 21/(p−1)
f (ξ )χ[a,r ] (ξ )dξ
ξ p−1 f (ξ )dξ
1/(p−1)
,
r
we derive
X Z
2k
k∈Ia
!1/(p−1)
2k+1
1/(p−1)
≤2
ξ
p−1
f (ξ )χ[a,λr ] (ξ )dξ
X Z
ξ
2k
k∈Ia
!1/(p−1)
2k+1 p−1
f (ξ )χ[a,r ] (ξ )dξ
+2
1/(p−1)
ε
X Z
2k
k∈Ia
!1/(p−1)
2k+1
ξ
p−1
f (ξ )χ[a,λr ] (ξ )dξ
.
for all r > a0 . If 21/(p−1) ε ≤ 1/2, the last estimate implies that
X Z k∈Ia
!1/(p−1)
2k+1 2k
ξ p−1 f (ξ )χ[a,λr ] (ξ )dξ
≤ 2p/(p−1)
X Z k∈Ia
!1/(p−1)
2k+1
ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
2k
for all r > a0 . Consequently, we have
X Z k∈Ia
!1/(p−1)
2k+1 2k
ξ
p−1
f (ξ )χ[a,λm r ] (ξ )dξ
≤2
mp/(p−1)
k∈Ia
!1/(p−1)
2k+1
X Z
2k
ξ
p−1
f (ξ )χ[a,r ] (ξ )dξ
for all m ∈ N and r > a0 . Thus, to obtain (2.49) it sufficient to take m ∈ N satisfying the inequality λm ≥ η.
Proof of Theorem 2.5. By (2.9), there exists a real number B > 0 such that (2.50) holds for all large enough r. This implies the estimate
r p f (r ) ≤ σ p B
X Z k∈Ia
2k+1 2k
!1/(p−1) p−1 ξ p−1 f (ξ )χ[a,σ r ] (ξ )dξ
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3451
for all large enough r, where σ > 1 is the constant in (1.3). Hence, in accordance with Lemma 2.8 and condition (2.8) there exist real numbers r0 > a and C > 0 such that
ν(r ) ≤ C
X Z
2k+1 2k
k∈Ia
!1/(p−1) p−1 ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
for all r > r0 , where ν(r ) = 1 + r p f (r ). Thus, we have
k∈Ia
!1/(p−1)
2k+1
X Z
2k
ξ
p−1
f (ξ )ν
−1/p
(ξ )χ[a,r ] (ξ )dξ
≥
X Z
ξ
2k
k∈Ia
!1/(p−1)
2k+1 p−1
≥ C −1/(p(p−1))
f (ξ )ν
X Z
!1/(p−1) −1/p ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
ξ
2k
!1/(p−1)
2k+1
X Z k∈Ia
(ξ )χ[r0 ,r ] (ξ )dξ
2k+1 2k
k∈Ia
×
−1/p
p−1
f (ξ )χ[r0 ,r ] (ξ )dξ
.
(2.51)
Since
X Z k∈Ia
!1/(p−1)
2k+1 2k
ξ
p−1
f (ξ )χ[r0 ,r ] (ξ )dξ
≥
Z 1X 2 k∈I a
!1/(p−1)
2k+1
ξ
2k
p−1
f (ξ )χ[a,r ] (ξ )dξ
for all large enough r, inequality (2.51) enables one to assert that
X Z k∈Ia
2k+1 2k
!1/(p−1) ξ p−1 f (ξ )ν −1/p (ξ )χ[a,r ] (ξ )dξ
≥
C −1/(p(p−1))
2
X Z k∈Ia
2k+1 2k
!1/(p−1) (p−1)/p ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
for all large enough r. By condition (2.8), this implies (2.5). To complete the proof it remains to use Theorem 2.3.
References [1] [2] [3] [4] [5]
I.T. Kiguradze, T.A. Chanturiya, Asymptotic Properties of Solutions of Nonautonomous Ordinary Differential Equations, Kluwer, Dordrecht, 1993. I.T. Kiguradze, G.G. Kvinikadze, On strongly increasing solutions of nonlinear ordinary differential equations, Ann. Mat. Pure Appl. 130 (1982) 67–87. A.A. Kon’kov, On properties of solutions of elliptic inequalities with a nonlinearity in the principal part, Dokl. Russ. Acad. Sci. 383 (1) (2002) 15–19. A.A. Kon’kov, Comparison theorems for elliptic inequalities with a non-linearity in the principal part, J. Math. Anal. Appl. 325 (2007) 1013–1041. E. Mitidieri, S.I. Pohozaev, Nonexistence of positive solutions for quasilinear elliptic problems on Rn , Proc. V.A. Steklov Inst. Math. 227 (18) (1999) 192–222. [6] E. Mitidieri, S.I. Pohozaev, A priori estimates and blow-up of solutions to nonlinear partial differential equations and inequalities, Proc. V.A. Steklov Inst. Math. 234 (2001) 3–283. [7] L. Veron, Comportement asymptotique des solutions d’equations elliptiques semi-lineaires dans Rn , Ann. Math. Pure Appl. 127 (1981) 25–50.
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Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
On global solutions of the radial p-Laplace equationI Andrej A. Kon’kov Department of Differential Equations, Faculty of Mechanics and Mathematics, Moscow Lomonosov State University, Vorobyovy Gory, 119992 Moscow, Russia
article
info
a b s t r a c t
Article history: Received 21 April 2008 Accepted 15 July 2008
We obtain blow-up conditions for solutions of the radial p-Laplace equation. © 2008 Elsevier Ltd. All rights reserved.
Keywords: Nonlinear equations A priori estimates Blow-up phenomenon
1. Introduction We consider the problem 1
d
r n−1 dr
r
n−1
p−2 ! du du = c (r , u), dr dr
u(a) > 0,
du dr
(a) ≥ 0
(1.1)
for the radial p-Laplace operator of dimension n ≥ 1, where p > 1 is a real number and c : [a, ∞) × (0, ∞) → [0, ∞) is a function from the Carathéodory class Kloc ([a, ∞) × R), a > 0 [1]. Let θ > 1 and σ > 1 be some real numbers fixed throughout this paper. We shall assume that inf
ζ ∈(t /θ,θ t )
c (r , ζ ) ≥ f (r )g (t )
(1.2)
for almost all r ≥ a and for all t > 0, where f : [a, ∞) → [0, ∞) belongs to the space L1,loc ([a, ∞)) and g : (0, ∞) → (0, ∞) is a continuous function. Put f (r ) =
ess sup
ξ ∈(r /σ ,r σ )∩[a,∞)
f (ξ ),
r ∈ [a, ∞).
(1.3)
As usual, by Z and N we denote the sets of integers and positive integers, respectively. Also, let Ia = {k ∈ Z : 2k ≥ a}. Definition 1.1. A solution u˜ : [a, b˜ ) → R of (1.1) is called an extension of a solution u : [a, b) → R if b˜ ≥ b and u˜ (r ) = u(r ) for all r ∈ [a, b). Definition 1.2. A solution of problem (1.1) which coincides with every its extension, is said to be non-extendable. It can be shown that every non-extendable solution of (1.1) is either a global solution defined on the whole set [a, ∞) or tends to infinity on a finite interval what is known as the ‘‘blow-up phenomenon’’. Questions treated in the present paper were earlier investigated by a number of authors [1–7]. Below, we obtain conditions guaranteeing that problem (1.1) has no global solutions. Our results can be applied to a wide class of equations including the case when the methods given in [1,2, 5,6] are useless (see Example 2.3). I The research was supported by RFBR, grant 08-01-00819.
E-mail address: [email protected]. 0362-546X/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2008.07.001
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A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
2. Main results Assume that Ω ⊂ [a, ∞) is a measurable set and χΩ is the characteristic function of Ω , i.e.
χΩ (ξ ) =
1, 0,
ξ ∈ Ω, ξ ∈ R \ Ω.
By the capacity of Ω relative to the ordered triple F = (f , p, n) we mean
Z ∞ X capF (Ω ) = inf
ri
dt ri−1
i=1
t n −1
!1/(p−1) (p−1)/p
t
Z
1
ξ n−1 f (ξ )χΩ (ξ )dξ
,
ri−1
where the infimum in the right-hand side is taken over all increasing sequences of real numbers {ri }∞ i=0 such that r0 = a
and
lim ri = ∞.
(2.1)
i→∞
In particular, we write capF (Ω ) = ∞ if
Z ∞ X
ri
dt ri−1
i=1
1
t
Z
t n −1
!1/(p−1) (p−1)/p =∞ ξ n−1 f (ξ )χΩ (ξ )dξ
ri−1
for every increasing sequence of real numbers {ri }∞ i=0 satisfying conditions (2.1). This capacity has the following natural properties. (i) Monotonicity. If Ω1 ⊂ Ω2 , then capF (Ω1 ) ≤ capF (Ω2 ). (ii) Semiadditivity. Suppose that p ≥ 2, then capF (Ω1 ∪ Ω2 ) ≤ capF (Ω1 ) + capF (Ω2 ).
(2.2)
Property (i) is obvious, whereas to prove (ii) it is sufficient to use the estimate
χΩ1 ∪Ω2 (ξ ) ≤ χΩ1 (ξ ) + χΩ2 (ξ ), α
α
ξ ∈ R,
α
and the fact that (h1 + h2 ) ≤ h1 + h2 for all real numbers h1 ≥ 0, h2 ≥ 0, and 0 < α ≤ 1. In the case of 1 < p < 2, inequality (2.2) should be replaced by capF (Ω1 ∪ Ω2 ) ≤ 21/p capF (Ω1 ) + 21/p capF (Ω2 ). Theorem 2.1. Let capF ([a, ∞)) = ∞ and ∞
Z
(g (t )t )−1/p dt < ∞,
(2.3)
1
then problem (1.1) has no global solutions. At first glance it would seem that calculating the above capacity is a difficult task. However, in the majority of cases, it can be successfully estimated. We also prove Theorems 2.2–2.5, which helps one to answer the question whether capF ([a, ∞)) equals infinity. Theorem 2.2. Let p ≥ 2 and ∞
Z
(ξ f (ξ ))1/(p−1) µ−1/p (ξ ) dξ = ∞,
(2.4)
a
where µ(ξ ) = 1 + (ξ p f (ξ ))1/(p−1) , then capF ([a, ∞)) = ∞. Corollary 2.1. Suppose that p ≥ 2. If relations (2.3) and (2.4) are valid, then problem (1.1) has no solutions defined on the whole set [a, ∞). The last statement can be readily obtained from Theorems 2.1 and 2.2.
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3439
Theorem 2.3. Suppose that 1 < p < 2 and
ξ
2k
k∈Ia
!1/(p−1)
2k+1
X Z
p−1
f (ξ )ν
−1/p
(ξ )dξ
= ∞,
(2.5)
where ν(ξ ) = 1 + ξ p f (ξ ), then capF ([a, ∞)) = ∞. Corollary 2.2. Let 1 < p < 2 and, moreover, conditions (2.3) and (2.5) be fulfilled, then problem (1.1) has no solutions defined on the whole set [a, ∞). To establish the validity of Corollary 2.2, it is sufficient to combine Theorems 2.1 and 2.3. Theorem 2.4. Suppose that p ≥ 2, ∞
Z
(ξ f (ξ ))1/(p−1) dξ = ∞,
(2.6)
a
and
(r p f (r ))1/(p−1) < ∞, (ξ f (ξ ))1/(p−1) dξ a
lim sup R r r →∞
(2.7)
then capF ([a, ∞)) = ∞. Theorem 2.5. Let 1 < p < 2,
k∈Ia
!1/(p−1)
2k+1
X Z
2k
ξ
p−1
f (ξ )dξ
= ∞,
(2.8)
and lim sup r →∞
(r p f (r ))1/(p−1) P R k∈Ia
2k+1 2k
ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
1/(p−1) < ∞,
(2.9)
then capF ([a, ∞)) = ∞. Theorems 2.4 and 2.5 immediately imply the following two assertions. Corollary 2.3. Suppose that p ≥ 2. If relations (2.3), (2.6) and (2.7) hold, then problem (1.1) has no global solutions. Corollary 2.4. Let 1 < p < 2 and, moreover, conditions (2.3), (2.8) and (2.9) be valid, then problem (1.1) has no solutions defined on the whole set [a, ∞). Theorems 2.1–2.5 will be proved later. Now, we demonstrate the exactness of the above results. Example 2.1. Suppose that g (t ) = t λ and f ( r ) = f0 r s ,
f0 = const > 0.
(2.10)
If λ > p − 1 and s ≥ −p, then in accordance with Corollaries 2.1 and 2.2, problem (1.1) has no solutions defined on the whole set [a, ∞). Further, let f (r ) = f0 r −p logs (1 + r ),
f0 = const > 0,
(2.11)
then the inequalities λ > p − 1 and s ≥ 1 − p imply the absence of global solutions of (1.1). Example 2.2. Suppose that g (t ) = t p−1 logλ (2 + t ) and (2.10) holds. If λ > p and s ≥ −p, then problem (1.1) has no global solutions by Corollaries 2.1 and 2.2. Analogously, if the function f satisfies condition (2.11), then the inequalities λ > p and s ≥ 1 − p guarantee the absence of global solutions of (1.1). It is known that the blow-up conditions given in Examples 2.1 and 2.2 are exact for all n > p > 1 [3,5,6]. Let us consider the case where the function f is not monotone.
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A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
Example 2.3. Assume that g (t ) = t p−1 logλ (2 + t ), λ > p ≥ 2, and
p ∞ X log k
f (r ) = f0
k
k =2
χ[k,k+log−2 k] (r ),
f0 = const > 0.
It does not present any particular problem to verify that both relations (2.3) and (2.4) are fulfilled. In doing so, one can take an arbitrary real number σ > 1 on the right in (1.3), for instance, σ = 2. Thus, applying Corollary 2.1, we conclude that problem (1.1) has no global solutions. Lemma 2.1. Suppose that capF ([a, ∞)) = ∞, then ∞
Z
dt
t n−1
a
t
Z
1
ξ
n −1
f (ξ )dξ
1/(p−1) = ∞.
(2.12)
a
Proof. If (2.12) is not valid, then there exists an increasing sequence of real numbers {ri }∞ i=0 satisfying conditions (2.1) and such that ri
Z
dt
t n −1
ri−1
t
Z
1
ξ n−1 f (ξ )dξ
1/(p−1)
1
<
2i
a
,
i = 2, 3, . . . ,
whence we obviously obtain
Z
ri
dt
ri−1
t n −1
!1/(p−1) (p−1)/p < ξ n−1 f (ξ )dξ
t
Z
1
ri−1
1 2i(p−1)/p
,
i = 2, 3, . . . .
Therefore,
Z ∞ X capF ([a, ∞)) ≤
ri
dt ri−1
i=1
Z
1 t n −1
This contradiction proves the lemma.
!1/(p−1) (p−1)/p < ∞. ξ n−1 f (ξ )dξ
t ri−1
Proof of Theorem 2.1. Assume the converse, then there exists a solution of problem (1.1) defined on the whole set [a, ∞). It is clear from (1.1) that u is a non-decreasing function. We also have lim u(r ) = ∞.
r →∞
In fact, if lim u(r ) < ∞,
(2.13)
r →∞
then inf
ζ ∈[a,∞)
g (u(ζ )) > 0.
Integrating (1.1), we obtain u(r ) = u(a) +
r
Z
dt
a
an − 1 t
t
Z
1
(u0 (a))p−1 + n−1
t n−1
ξ n−1 c (ξ , u)dξ
1/(p−1)
a
for all r ∈ [a, ∞), whence in accordance with (1.2) it can be seen that u(r ) ≥
1/(p−1) Z
inf
ζ ∈[a,∞)
r
g (u(ζ ))
dt
a
1 t n−1
t
Z
ξ n−1 f (ξ )dξ
1/(p−1)
a
for all r ∈ [a, ∞). By Lemma 2.1, this contradicts (2.13). Let us consider a sequence of real numbers {ri }∞ i=0 such that r0 = a and u(ri ) = θ u(ri−1 ), i = 1, 2, . . .. Taking into account (1.1), we have u(ri ) = u(ri−1 ) +
Z
ri
dt ri−1
rin−−11 t n−1
(u (ri−1 )) 0
p−1
+
1 t n −1
Z
!1/(p−1)
t
ξ ri−1
n−1
c (ξ , u)dξ
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3441
for all i ∈ N. This immediately implies the inequality ri
Z
u(ri ) − u(ri−1 ) ≥
dt
t n−1
ri−1
!1/(p−1)
t
Z
1
ξ
n −1
c (ξ , u)dξ
ri−1
for all i ∈ N. Combining the last formula and relation (1.2), we derive u(ri ) − u(ri−1 )
inf
≥
g 1/(p−1) (ζ )
ζ ∈(u(ri−1 ),u(ri ))
ri
Z
dt
t n −1
ri−1
!1/(p−1)
t
Z
1
ξ
n −1
f (ξ )dξ
ri−1
for all i ∈ N. At the same time, u(ri )
Z
u(ri−1 )
(g (t )t )−1/p dt ≥ C
u(ri ) − u(ri−1 )
inf
(p−1)/p
g 1/(p−1) (ζ )
ζ ∈(u(ri−1 ),u(ri ))
with some constant C > 0 depending only on p and θ . Thus, we obtain the estimate
u(ri )
Z
u(ri−1 )
(g (t )t )−1/p dt ≥ C
ri
Z
dt ri−1
t n−1
!1/(p−1) (p−1)/p ξ n−1 f (ξ )dξ
t
Z
1
ri−1
for all i ∈ N, from which it follows that ∞
Z
u(a)
−1/p
(g (t )t )
Z ∞ X dt ≥ C
ri
dt ri−1
i =1
t n −1
!1/(p−1) (p−1)/p
t
Z
1
ξ
n −1
f (ξ )dξ
.
ri−1
This contradicts hypotheses of Theorem 2.1. The proof is completed.
Lemma 2.2. Let p ≥ 2, then a2
Z
dt
Z
1 t n−1
a1
t
ξ
n −1
f (ξ )dξ
1/(p−1)
Z
a2
≥A
1−
a1
1/(p−1)
ξ
a2
a1
(ξ f (ξ ))1/(p−1) dξ
for all real numbers a ≤ a1 < a2 , where the constant A > 0 depends only on p and n. Proof. Consider a sequence of real numbers ti ∈ [a1 , a2 ), i = 0, 1, 2, . . ., defined by induction. We take t0 = a1 . Assume further that ti−1 is already known. We put ti = 2ti−1 if 4ti−1 < a2 and ti = (ti−1 + a2 )/2, otherwise. From the construction of the sequence {ti }∞ i=0 , it is clear that ti−1 < ti < 4ti−1 and ti − ti−1 ≤ 2(ti+1 − ti ) for all i = 1, 2, . . .. Moreover, we have
1/(p−1) ti−1 1/(p−1) (ti − ti−1 )1/(p−1) ≥ B 1 − ti ,
i = 1, 2, . . . ,
a2
(2.14)
with some constant B > 0 depending only on p. Let i be a positive integer, then ti+1
Z
dt
ti
t
Z
1 t n−1
ξ
n −1
f (ξ )dξ
1/(p−1)
Z
≥ C (ti+1 − ti )
a1
ti
!1/(p−1) f (ξ )dξ
,
ti−1
where the constant C > 0 depends only on p and n. At the same time, using the Hölder inequality, one can show that
(ti − ti−1 )
Z
(p−2)/(p−1)
!1/(p−1)
ti
f (ξ )dξ
Z
ti
≥
ti−1
f 1/(p−1) (ξ )dξ .
ti−1
Hence, we obtain ti+1
Z 2
dt
ti
1
Z
t n−1
t
ξ
n −1
f (ξ )dξ
1/(p−1)
1/(p−1)
≥ C (ti − ti−1 )
a1
Z
ti
f 1/(p−1) (ξ )dξ .
ti−1
Combining this with the estimate
(ti − ti−1 )
1/(p−1)
Z
ti
f ti−1
1/(p−1)
(ξ )dξ ≥ B
Z
ti
1−
ti−1
ξ a2
1/(p−1)
(ξ f (ξ ))1/(p−1) dξ ,
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A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
which follows from (2.14), we derive ti+1
Z
dt
t n −1
ti
t
Z
1
ξ n−1 f (ξ )dξ
1/(p−1)
ti
Z
1−
≥A ti−1
a1
1/(p−1)
ξ a2
(ξ f (ξ ))1/(p−1) dξ ,
where A = BC /2. Finally, summing the last inequality over all positive integers i, we complete the proof.
Lemma 2.3. Suppose that a1 < a2 and 0 < α < 1 are some real numbers, then a2
Z
ϕ(ξ )dξ
α
a2
Z ≥B
ϕ(ξ )~ α−1 (ξ )dξ
(2.15)
a1
a1
for every non-negative function ϕ ∈ L1 ([a1 , a2 ]), where
~(ξ ) =
a2
Z ξ
ϕ(ζ )dζ
and the constant B > 0 depends only on α . Remark 2.1. Throughout this paper we assume by definition that 0/0 = 0. In particular, if ~(ξ ) = 0 for some ξ ∈ (a1 , a2 ), then the function ϕ~ α−1 on the right in (2.15) is equal to zero almost everywhere on the interval [ξ , a2 ]. Proof of Lemma 2.3. Without loss of generality, we may assume that ~(ξ ) > 0 for any ξ ∈ (a1 , a2 ); otherwise, we replace a2 by inf{ξ ∈ (a1 , a2 ) : ~(ξ ) = 0}. Consider the sequence of real numbers ξk ∈ [a1 , a2 ], k = 0, 1, . . . , constructed as follows. We put ξ0 = a1 . Further, let ξk−1 be already known. We take ξk ∈ [ξk−1 , a2 ] satisfying the relation ξk
Z
ξk−1
1
ϕ(ξ )dξ =
Z
a2
2 ξk−1
ϕ(ξ )dξ .
It is clear that
Z
a2
ξk
ϕ(ξ )dξ =
Z
ξk ξk−1
ϕ(ξ )dξ = 2−k
Z
a2
ϕ(ξ )dξ ,
k = 1, 2, . . . .
(2.16)
a1
In particular, we have lim ~(ξk ) = 0;
k→∞
therefore, lim ξk = a2 .
k→∞
On the other hand, it can be shown that
Z
ξk ξk−1
ϕ(ξ )~ α−1 (ξ )dξ ≤ ~ α−1 (ξk )
ξk
Z
ξk−1
ϕ(ξ )dξ ,
k = 1, 2, . . . .
By (2.16), this implies the estimate
Z
ξk ξk−1
ϕ(ξ )~ α−1 (ξ )dξ ≤ 2−αk
Z
a2
ϕ(ξ )dξ
α
,
k = 1, 2, . . . .
a1
Summing the last expression over all k ∈ N, we obtain (2.15). The proof is completed.
Proof of Theorem 2.2. Assume that is an increasing sequence of real numbers satisfying conditions (2.1) and σ > 1 is the constant in (1.3). For every i ∈ N we denote ti = ri /σ if σ ri−1 < ri and ti = ri−1 , otherwise. Let i be a positive integer. We obtain
{ri }∞ i=0
Z
ri
1−
2 ri−1
Z
ti
≥
ξ
1/(p−1)
ri
1−
ri−1
ξ ri
!(p−1)/p (ξ f (ξ ))
1/(p−1)
1/(p−1)
dξ
!(p−1)/p 1/(p−1)
(ξ f (ξ ))
dξ
ri
Z +
1−
ti
ξ ri
1/(p−1)
!(p−1)/p 1/(p−1)
(ξ f (ξ ))
dξ
.
(2.17)
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3443
At the same time, ti
Z
ξ
1−
1/(p−1)
ri
ri−1
1/(p−1) Z ti 1 (ξ f (ξ ))1/(p−1) dξ (ξ f (ξ ))1/(p−1) dξ ≥ 1 − σ ri−1
and ri
Z
ξ
1−
1/(p−1)
ri
ti
1/(p−1)
(ξ f (ξ ))
dξ ≥ σ
ri
Z
−1/(p−1)
(ri − ξ )1/(p−1) f 1/(p−1) (ξ )dξ .
ti
Therefore, relation (2.17) implies the inequality ri
Z
1−
A ri−1
1/(p−1)
ri
!(p−1)/p (ξ f (ξ ))
1/(p−1)
!(p−1)/p
ti
Z
ξ
(ξ f (ξ ))
≥
1/(p−1)
dξ
dξ ri
Z +
ri−1
(ri − ξ )1/(p−1) f 1/(p−1) (ξ )dξ
(p−1)/p
,
(2.18)
ti
where the constant A > 0 depends only on p and σ . By Lemma 2.3, we have ri
Z
(ri − ξ )1/(p−1) f 1/(p−1) (ξ )dξ
(p−1)/p
ri
Z ≥B
ti
(ri − ξ )1/(p−1) f 1/(p−1) (ξ )~ −1/p (ξ )dξ
(2.19)
ti
with some constant B > 0 depending only on p, where ri
Z
~(ξ ) =
ξ
(ri − ζ )1/(p−1) f 1/(p−1) (ζ )dζ .
Since
~(ξ ) ≤ (ri − ξ )1/(p−1)
ri
Z
f 1/(p−1) (ζ )dζ ,
ξ
it does not present any particular problem to verify that
(ri − ξ )1/(p−1) ~ −1/p (ξ ) ≥
ri
Z
1 ri − ξ
ξ
f 1/(p−1) (ζ )dζ
−1/p
for all ξ ∈ (ti , ri ). Hence, taking into account the evident estimate ri
Z
1 ri − ξ
ξ
f 1/(p−1) (ζ )dζ ≤ ξ −p/(p−1) µ(ξ )
for all ξ ∈ (ti , ri ), where µ(ξ ) = 1 + (ξ p f (ξ ))1/(p−1) , we derive
(ri − ξ )1/(p−1) ~ −1/p (ξ ) ≥ ξ 1/(p−1) µ−1/p (ξ ),
ξ ∈ (ti , ri ).
(2.20)
Combining (2.19) and (2.20), one can arrive at the conclusion that ri
Z
(ri − ξ )1/(p−1) f 1/(p−1) (ξ )dξ
(p−1)/p
ri
Z ≥B
ti
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ .
ti
Thus, formula (2.18) implies the inequality
Z
ri
1−
C ri−1
Z
ti
≥ ri−1
ξ
1/(p−1)
ri
!(p−1)/p (ξ f (ξ ))
1/(p−1)
dξ
!(p−1)/p (ξ f (ξ ))
1/(p−1)
µ
−1/p
(ξ )dξ
ri
Z +
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ ,
ti
where the constant C > 0 depends only on p and σ . Summing (2.21) over all positive integers i, we obtain
(2.21)
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A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
Z ∞ X
C
1−
Z ∞ X
!(p−1)/p
1/(p−1)
ξ
(ξ f (ξ ))
ri
ri−1
i =1
≥
ri
1/(p−1)
dξ
!(p−1)/p
ti
(ξ f (ξ ))
1/(p−1)
µ
−1/p
(ξ )dξ
∞ Z X
+
ri−1
i=1
ri
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ ,
ti
i =1
whence in accordance with the estimate
Z ∞ X
ti
!(p−1)/p (ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ
∞ Z X
≥
ri−1
i=1
!(p−1)/p
ti
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ
ri−1
i =1
it follows that
Z ∞ X
C
1−
∞ Z X
ti
!(p−1)/p
1/(p−1)
ξ
(ξ f (ξ ))
ri
ri−1
i =1
≥
ri
1/(p−1)
dξ
!(p−1)/p (ξ f (ξ ))
1/(p−1)
µ
−1/p
(ξ )dξ
∞ Z X
+
ri−1
i=1
ri
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ .
(2.22)
ti
i =1
By condition (2.4), at least one of the two summands on the right in (2.22) is equal to infinity; therefore,
Z ∞ X
ri
1−
ri−1
i=1
ξ
1/(p−1)
ri
!(p−1)/p (ξ f (ξ ))
1/(p−1)
dξ
= ∞.
(2.23)
According to Lemma 2.2, relation (2.23) implies the equality
Z ∞ X
ri
dt
t n −1
ri−1
i=1
!1/(p−1) (p−1)/p = ∞. ξ n−1 f (ξ )dξ
t
Z
1
ri−1
Thus, we have capF ([a, ∞)) = ∞. The proof of Theorem 2.2 is completed.
Lemma 2.4. Suppose that a ≤ a1 < a2 are real numbers and m < 1 + 1/(p − 1) is a positive integer, then a2
Z
t
Z
f (ξ )dξ
dt a1
1/(p−1)
a2
Z
dtf (t )fm (t )
=C
a1
t
Z
a1
f (ξ )dξ
1/(p−1)−m (2.24)
a1
with some constant C > 0 depending only on p and m, where f 1 ( t ) = a2 − t
fk (t ) =
and
a2
Z
f (ξ )fk−1 (ξ )dξ ,
k = 2, 3, . . . .
(2.25)
t
Proof. The proof is by induction on m. It is obvious that a2
Z
t
Z
f (ξ )dξ
dt a1
1/(p−1)
a2
Z =−
a1
dtf10 (t )
a1
=
p−1
f (ξ )dξ
1/(p−1)
a1
Z
1
t
Z
a2
dtf (t )f1 (t )
Z
a1
t
f (ξ )dξ
1/(p−1)−1
.
a1
This proves (2.24) for m = 1. Now, assume that a2
Z
Z
t
f (ξ )dξ
dt a1
1/(p−1)
Z
a2
=B
a1
dtf (t )fm−1 (t )
t
Z
a1
f (ξ )dξ
1/(p−1)−m+1
,
(2.26)
a1
where 1 < m < 1 + 1/(p − 1) is an integer and B > 0 is a real number. Integrating by parts, we have
Z
a2 a1
dtf (t )fm−1 (t )
Z
t
f (ξ )dξ
1/(p−1)−m+1
Z
a2
=−
a1
=
dtfm0 (t )
Z
t
f (ξ )dξ
a1
a1
1
Z
p−1
a2
−m+1 a1
1/(p−1)−m+1
dtf (t )fm (t )
Z
t
f (ξ )dξ a1
1/(p−1)−m
.
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3445
According to (2.26), this readily implies (2.24), where C = (1/(p − 1) − m + 1)B > 0. Lemma 2.4 is completely proved. Lemma 2.5. Let a ≤ a1 < a2 be real numbers and fk , k = 1, 2, . . . , the functions defined by (2.25), then 1/k
fk
(t ) ≥ AF −1/k
a2
Z
f (ξ )dξ ,
t ∈ [a1 , a2 ],
(2.27)
t
for all positive integers k, where F =
sup
ζ ∈(a1 ,a2 )
a2
Z
1 a2 − ζ
ζ
f (ξ )dξ
and A > 0 is a constant depending only on k. Proof. The proof is by induction on k. If k = 1, then (2.27) is trivial. Assume that 1/(k−1)
fk−1
(ξ ) ≥ CF −1/(k−1)
Z
a2
ξ
f (ζ )dζ
(2.28)
with some integer k > 1 for all ξ ∈ [a1 , a2 ], where C > 0 is a real number. Lemma 2.3 implies the estimate 1/k
fk
a2
Z
(t ) =
f (ξ )fk−1 (ξ )dξ
1/k
a2
Z
f (ξ )fk−1 (ξ )~ 1/k−1 (ξ )dξ
≥B
(2.29)
t
t
for all t ∈ [a1 , a2 ], where
~(ξ ) =
Z
a2
ξ
f (ζ )fk−1 (ζ )dζ
and B > 0 is a constant depending only on k. Since fk−1 is a non-increasing function, we obtain
~(ξ ) ≤ fk−1 (ξ )
a2
Z ξ
f (ζ )dζ
for all ξ ∈ [a1 , a2 ], whence in accordance with (2.28) it follows that fk−1 (ξ ) ≥ C 1−1/k F −1/k ~ 1−1/k (ξ ) for all ξ ∈ [a1 , a2 ]. Thus, we have a2
Z
f (ξ )fk−1 (ξ )~ 1/k−1 (ξ )dξ ≥ C 1−1/k F −1/k
a2
Z
t
f (ξ )dξ
t
for all t ∈ [a1 , a2 ]. Combining the last estimate with (2.29), we immediately derive (2.27), where A = BC 1−1/k . The proof is completed. ∞ Lemma 2.6. Let {Vi }∞ i=1 and {Wi }i=1 be denumerable families of measurable subsets of the interval [a, ∞). We put εij = 1 in the ∞ case of mes(Vi ∩ Wj ) 6= 0 and εij = 0, otherwise. If ∪∞ i=1 Wi ⊂ ∪i=1 Vi and, moreover, there exists a positive integer M such that
sup
∞ X
εij ≤ M
(2.30)
εij ≤ M ,
(2.31)
j∈N i=1
and sup
∞ X
i∈N j=1
then M
p/(p−1)
∞ Z X i=1
ψ(ξ )dξ Vi
1/(p−1) ≥
∞ Z X i=1
ψ(ξ )dξ
1/(p−1) (2.32)
Wi
for every measurable function ψ : [a, ∞) → [0, ∞). In particular, if the right-hand side in (2.32) is equal to infinity, then the left-hand side is also equal to infinity.
3446
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
Proof. Assume that j ∈ N. We obviously have ∞ X
εij
Z
ψ(ξ )dξ ≥
Z
ψ(ξ )dξ .
(2.33)
Wj
Vi
i=1
There is k ∈ N such that εkj = 1 and
Z
Z
ψ(ξ )dξ ≥
ψ(ξ )dξ
Vi
Vk
for any i ∈ N satisfying the condition εij = 1. Hence, in accordance with (2.30) we obtain
Z
ψ(ξ )dξ ≥
M
∞ X
Vk
εij
Z
ψ(ξ )dξ .
(2.34)
Vi
i =1
From (2.33) and (2.34), it follows that
Z
ψ(ξ )dξ ≥
M
Z
Vk
ψ(ξ )dξ Wj
or, in other words, M
1/(p−1)
Z
ψ(ξ )dξ
1/(p−1)
!1/(p−1)
Z
ψ(ξ )dξ
≥
.
Wj
Vk
Since ∞ X
εij
Z
ψ(ξ )dξ
1/(p−1)
Z
ψ(ξ )dξ
≥
Vi
i=1
1/(p−1)
,
Vk
this implies the estimate M
1/(p−1)
∞ X
εij
Z
ψ(ξ )dξ
1/(p−1)
ψ(ξ )dξ
≥
.
Wj
Vi
i=1
!1/(p−1)
Z
Finally, summing the last expression over all j ∈ N and taking into account (2.31), we immediately derive (2.32). Lemma 2.6 is completely proved. Proof of Theorem 2.3. Consider an increasing sequence of real numbers {ri }∞ i=0 satisfying conditions (2.1). By Ξ1 we denote the set of positive integers i such that σ ri−1 ≥ ri , where σ > 1 is the constant in (1.3). Also, put Ξ2 = N \ Ξ1 . At first, we assume that i ∈ Ξ1 . Let fk , k = 1, 2, . . ., be the functions defined by (2.25) with a2 = ri . According to Lemma 2.4, we have
Z
ri
Z
!1/(p−1)
t
f (ξ )dξ
dt ri−1
Z
ri
dtf (t )fm (t )
=C
ri−1
Z
ri−1
!1/(p−1)−m
t
f (ξ )dξ
ri−1
for some real number C > 0 depending only on p, where m is the maximal integer such that m < 1 + 1/(p − 1). Since 1/(p − 1) − m ≤ 0, this implies the inequality
Z
ri
Z
!1/(p−1)
t
f (ξ )dξ
dt ri−1
Z
ri
dtf (t )fm (t )
≥C
ri−1
ri−1
ri
Z
!1/(p−1)−m f (ξ )dξ
.
ri−1
At the same time, putting a1 = ri−1 , a2 = ri , and k = m + 1 in Lemma 2.5, we obtain fm+1 (ri−1 ) =
ri
Z
f (t )fm (t )dt ≥
ri−1
where F =
sup
ζ ∈(ri−1 ,ri ) ri
1
−ζ
ri
Z ζ
f (ξ )dξ
A F
Z
ri ri−1
!m+1 f (ξ )dξ
,
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3447
and the constant A > 0 depends only on p. Hence, one can assert that ri
Z
!1/(p−1)
t
Z
f (ξ )dξ
dt
≥
ri−1
ri−1
Z
AC F
ri
!1+1/(p−1) f (ξ )dξ
;
ri−1
therefore,
ri
Z
dt
t n−1
ri−1
!1/(p−1) (p−1)/p Z ξ n−1 f (ξ )dξ ≥ BF 1/p−1
t
Z
1
ri
f (ξ )dξ ,
(2.35)
ri−1
ri−1
where B = σ −(n−1)/p (AC )1−1/p . It does not present any particular problem to verify that F ≤ ξ −p ν(ξ ) for all ξ ∈ (ri−1 , ri ), where ν(ξ ) = 1 + ξ p f (ξ ). Thus, (2.35) implies the estimate
ri
Z
dt
t n−1
ri−1
!1/(p−1) (p−1)/p Z ξ n−1 f (ξ )dξ ≥B
t
Z
1
ri
ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ .
(2.36)
ri−1
ri−1
Now, assume that i ∈ Ξ2 . In this case, we obviously have σ ri−1 < ri . Let mi be the maximal integer satisfying the inequality σ mi ri−1 < ri . From the previous arguments, it follows that ri
Z
dt
t n −1
ri /σ
t
Z
1
ξ
ri /σ
n −1
f (ξ )dξ
1/(p−1) !(p−1)/p
ri
Z ≥B
ri /σ
ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ .
(2.37)
On the other hand, we derive
Z
σ −(k−1) ri
dt σ −k ri
1 t n−1
!1/(p−1)
t
Z
ξ
n −1
f (ξ )dξ
σ −k ri
Z ≥D
σ −(k+1) ri
ri−1
!1/(p−1) ξ
p−1
f (ξ )dξ
(2.38)
for all integers 1 ≤ k ≤ mi − 1, where the constant D > 0 depends only on p, σ , and n. Analogously,
Z
σ −(mi −1) ri
dt
t n −1
σ −mi ri
!1/(p−1)
t
Z
1
ξ
n−1
f (ξ )dξ
Z
σ −mi ri
≥D
ri−1
!1/(p−1) ξ
p−1
f (ξ )dξ
.
(2.39)
ri−1
Relations (2.38) and (2.39) allow us to conclude that
Z
ri
1
dt
t n −1
ri−1
Z
!1/(p−1)
t
ξ
n −1
f (ξ )dξ
≥ DJi ,
(2.40)
ri−1
where
Z
X
Ji =
1≤k≤mi −1
!1/(p−1)
σ −k ri σ −(k+1) ri
ξ
p−1
f (ξ )dξ
Z
σ −mi ri
+
!1/(p−1) ξ
p−1
f (ξ )dξ
.
ri−1
Formula (2.40) implies the inequality
X Z
ri
dt ri−1
i∈Ξ2
1
Z
t n−1
t
!1/(p−1) (p−1)/p X (p−1)/p ξ n−1 f (ξ )dξ ≥ D(p−1)/p Ji ,
ri−1
i∈Ξ2
whence, taking into account the fact that
X
(p−1)/p
Ji
!(p−1)/p ≥
i∈Ξ2
X
,
Ji
i∈Ξ2
we obtain
X Z i∈Ξ2
ri
dt ri−1
1 t n−1
Z
t
ri−1
!1/(p−1) (p−1)/p !(p−1)/p X n −1 (p−1)/p ξ f (ξ )dξ ≥D Ji . i∈Ξ2
3448
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
At the same time, by Lemma 2.6, M p/(p−1)
X
Ji ≥
i∈Ξ2
X Z ω∩[2k ,2k+1 ]
k∈Ia
ξ p−1 f (ξ )dξ
1/(p−1)
,
where ω = ∪i∈Ξ2 [ri−1 , ri /σ ] and M > 0 is some real number depending only on σ . Thus, we have
X Z
dt ri−1
i∈Ξ2
≥
ri
D(p−1)/p M
1
t
Z
t n −1
!1/(p−1) (p−1)/p ξ n−1 f (ξ )dξ
ri−1
X Z ω∩[2k ,2k+1 ]
k∈Ia
ξ
p−1
f (ξ )dξ
1/(p−1) !(p−1)/p
.
(2.41)
Further, relations (2.36) and (2.37) allow one to write
Z ∞ X
ri
dt ri−1
i=1
1
Z
t n −1
t
!1/(p−1) (p−1)/p Z n −1 ≥ B ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ , ξ f (ξ )dξ
(2.42)
τ
ri−1
where τ = [a, ∞) \ ω. If k ∈ Ia , i.e. k ∈ Z and 2k ≥ a, then
(2k+1 − 2k )2−p
Z τ ∩[2k ,2k+1 ]
ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ
p−1
Z
ξ (p−1) f p−1 (ξ )ν −(p−1)
2 /p
2
≥ τ ∩[2k ,2k+1 ]
(ξ )dξ
(2.43)
by the Hölder inequality. Since 2k+1 − 2k ≤ ξ and f p−1 (ξ ) ≥ f (ξ )ξ p(2−p) ν p−2 (ξ ) for almost all ξ ∈ [2k , 2k+1 ], it follows from (2.43) that
Z τ ∩[2k ,2k+1 ]
ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ ≥
Z τ ∩[2k ,2k+1 ]
ξ p−1 f (ξ )ν −1/p (ξ )dξ
1/(p−1)
.
Hence,
Z τ
ξ p−1 f (ξ )ν 1/p−1 (ξ )dξ ≥
X Z τ ∩[2k ,2k+1 ]
k∈Ia
ξ p−1 f (ξ )ν −1/p (ξ )dξ
1/(p−1)
.
According to (2.42), this immediately implies the estimate
Z ∞ X
ri
dt ri−1
i=1
1
Z
t n −1
!1/(p−1) (p−1)/p
t
ξ
n −1
f (ξ )dξ
ri−1
≥B
X Z k∈Ia
τ ∩[2k ,2k+1 ]
ξ
p−1
f (ξ )ν
−1/p
(ξ )dξ
1/(p−1)
.
(2.44)
By condition (2.5), we obtain either
X Z ω∩[2k ,2k+1 ]
k∈Ia
ξ p−1 f (ξ )ν −1/p (ξ )dξ
1/(p−1) =∞
(2.45)
= ∞.
(2.46)
or
X Z k∈Ia
τ ∩[2k ,2k+1 ]
ξ p−1 f (ξ )ν −1/p (ξ )dξ
1/(p−1)
If (2.45) is valid, then
Z ∞ X i=1
ri
dt ri−1
1 t n −1
Z
!1/(p−1) (p−1)/p
t
ξ n−1 f (ξ )dξ
=∞
ri−1
in accordance with formula (2.41); otherwise, the last equality is a consequence of (2.44) and (2.46). Thus, we have capF ([a, ∞)) = ∞. The proof is completed.
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3449
Lemma 2.7. Let η > 1 be a real number. If relation (2.7) is fulfilled, then ηr
Z
r
Z
(ξ f (ξ ))1/(p−1) dξ ≤ A
(ξ f (ξ ))1/(p−1) dξ
a
a
with some constant A > 0 for all large enough r. Proof. By (2.7), there exist real numbers a0 > a and B > 0 such that
(rf (r ))1/(p−1) ≤
B
r
Z
r
(ξ f (ξ ))1/(p−1) dξ
(2.47)
a
for all r > a0 . Integrating (2.47), we obviously have λr
Z
λr
Z
(ξ f (ξ ))1/(p−1) dξ ≤ B
dξ
r
r
≤ B log λ
ξ Z
ξ
Z
(ζ f (ζ ))1/(p−1) dζ
a
λr
(ξ f (ξ ))1/(p−1) dξ
a
for all real numbers λ > 1 and r > a0 . In particular, choosing λ > 1 such that B log λ < 1/2, we obtain λr
Z
(ξ f (ξ ))1/(p−1) dξ ≤
r
λr
Z
1 2
(ξ f (ξ ))1/(p−1) dξ
a
or, in other words, λr
Z
(ξ f (ξ ))
1/(p−1)
dξ ≤ 2
r
Z
a
(ξ f (ξ ))1/(p−1) dξ
a
for all r > a0 . The last estimate allows us to write λk r
Z
(ξ f (ξ ))
1/(p−1)
dξ ≤ 2
k
r
Z
a
(ξ f (ξ ))1/(p−1) dξ
a
for all k ∈ N and r > a0 . Finally, taking k ∈ N satisfying the inequality λk ≥ η, we complete the proof.
Proof of Theorem 2.4. By condition (2.7), there exists a real number B > 0 such that (2.47) is valid for all large enough r. This implies the estimate σr
Z
(r p f (r ))1/(p−1) ≤ σ p/(p−1) B
(ξ f (ξ ))1/(p−1) dξ
a
for all large enough r, where σ > 1 is the constant in (1.3). Hence, in accordance with Lemma 2.7 and relation (2.6) there exist real numbers r0 > a and C > 0 such that
µ(r ) ≤ C
r
Z
(ξ f (ξ ))1/(p−1) dξ
a
for all r > r0 , where µ(r ) = 1 + (r p f (r ))1/(p−1) . Thus, we derive r
Z
(ξ f (ξ ))
1/(p−1)
µ
−1/p
(ξ )dξ ≥
Z
a
r
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ
r0
≥C
−1/p
r
Z
1/(p−1)
(ξ f (ξ ))
dξ
−1/p Z
a
r
(ξ f (ξ ))1/(p−1) dξ .
(2.48)
r0
Since r
Z
r0
(ξ f (ξ ))1/(p−1) dξ ≥
1 2
Z
r
(ξ f (ξ ))1/(p−1) dξ
a
for all large enough r, inequality (2.48) enables one to assert that r
Z a
(ξ f (ξ ))1/(p−1) µ−1/p (ξ )dξ ≥
C −1/p 2
r
Z
(ξ f (ξ ))1/(p−1) dξ
1−1/p
a
for all large enough r. By (2.6), this implies (2.4). To complete the proof it remains to use Theorem 2.2.
3450
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
Lemma 2.8. Let η > 1 be a real number and, moreover, 1 < p < 2. If (2.9) holds, then
ξ
2k
k∈Ia
!1/(p−1)
2k+1
X Z
p−1
f (ξ )χ[a,ηr ] (ξ )dξ
≤A
ξ
2k
k∈Ia
!1/(p−1)
2k+1
X Z
f (ξ )χ[a,r ] (ξ )dξ
p−1
(2.49)
with some constant A > 0 for all large enough r. Proof. According to (2.9), there exist real numbers a0 > a and B > 0 such that B
r p−1 f (r ) ≤
r
X Z
2k+1
2k
k∈Ia
!1/(p−1) p−1 ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
(2.50)
for all r > a0 . Integrating (2.50), we obtain λr
Z
ξ
p−1
f (ξ )dξ ≤ B
λr
Z
X Z
dξ
ξ k∈Ia X Z ≤ B log λ
r
r
k∈Ia
2k+1
ζ
2k
2k+1
ζ
2k
p−1
p−1
!1/(p−1) p−1 f (ζ )χ[a,ξ ] (ζ )dζ
!1/(p−1) p−1 f (ζ )χ[a,λr ] (ζ )dζ
for all real numbers λ > 1 and r > a0 , whence it follows that λr
Z
ξ
p−1
f (ξ )dξ
1/(p−1)
≤ε
r
X Z k∈Ia
!1/(p−1)
2k+1
ξ
2k
p−1
f (ξ )χ[a,λr ] (ξ )dξ
for all r > a0 , where ε = (B log λ)1/(p−1) . Combining this with the evident inequality
X Z
2k
k∈Ia
!1/(p−1)
2k+1
1/(p−1)
≤2
ξ
p−1
f (ξ )χ[a,λr ] (ξ )dξ
X Z
ξ
2k
k∈Ia
!1/(p−1)
2k+1 p−1
λr
Z
+ 21/(p−1)
f (ξ )χ[a,r ] (ξ )dξ
ξ p−1 f (ξ )dξ
1/(p−1)
,
r
we derive
X Z
2k
k∈Ia
!1/(p−1)
2k+1
1/(p−1)
≤2
ξ
p−1
f (ξ )χ[a,λr ] (ξ )dξ
X Z
ξ
2k
k∈Ia
!1/(p−1)
2k+1 p−1
f (ξ )χ[a,r ] (ξ )dξ
+2
1/(p−1)
ε
X Z
2k
k∈Ia
!1/(p−1)
2k+1
ξ
p−1
f (ξ )χ[a,λr ] (ξ )dξ
.
for all r > a0 . If 21/(p−1) ε ≤ 1/2, the last estimate implies that
X Z k∈Ia
!1/(p−1)
2k+1 2k
ξ p−1 f (ξ )χ[a,λr ] (ξ )dξ
≤ 2p/(p−1)
X Z k∈Ia
!1/(p−1)
2k+1
ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
2k
for all r > a0 . Consequently, we have
X Z k∈Ia
!1/(p−1)
2k+1 2k
ξ
p−1
f (ξ )χ[a,λm r ] (ξ )dξ
≤2
mp/(p−1)
k∈Ia
!1/(p−1)
2k+1
X Z
2k
ξ
p−1
f (ξ )χ[a,r ] (ξ )dξ
for all m ∈ N and r > a0 . Thus, to obtain (2.49) it sufficient to take m ∈ N satisfying the inequality λm ≥ η.
Proof of Theorem 2.5. By (2.9), there exists a real number B > 0 such that (2.50) holds for all large enough r. This implies the estimate
r p f (r ) ≤ σ p B
X Z k∈Ia
2k+1 2k
!1/(p−1) p−1 ξ p−1 f (ξ )χ[a,σ r ] (ξ )dξ
A.A. Kon’kov / Nonlinear Analysis 70 (2009) 3437–3451
3451
for all large enough r, where σ > 1 is the constant in (1.3). Hence, in accordance with Lemma 2.8 and condition (2.8) there exist real numbers r0 > a and C > 0 such that
ν(r ) ≤ C
X Z
2k+1 2k
k∈Ia
!1/(p−1) p−1 ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
for all r > r0 , where ν(r ) = 1 + r p f (r ). Thus, we have
k∈Ia
!1/(p−1)
2k+1
X Z
2k
ξ
p−1
f (ξ )ν
−1/p
(ξ )χ[a,r ] (ξ )dξ
≥
X Z
ξ
2k
k∈Ia
!1/(p−1)
2k+1 p−1
≥ C −1/(p(p−1))
f (ξ )ν
X Z
!1/(p−1) −1/p ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
ξ
2k
!1/(p−1)
2k+1
X Z k∈Ia
(ξ )χ[r0 ,r ] (ξ )dξ
2k+1 2k
k∈Ia
×
−1/p
p−1
f (ξ )χ[r0 ,r ] (ξ )dξ
.
(2.51)
Since
X Z k∈Ia
!1/(p−1)
2k+1 2k
ξ
p−1
f (ξ )χ[r0 ,r ] (ξ )dξ
≥
Z 1X 2 k∈I a
!1/(p−1)
2k+1
ξ
2k
p−1
f (ξ )χ[a,r ] (ξ )dξ
for all large enough r, inequality (2.51) enables one to assert that
X Z k∈Ia
2k+1 2k
!1/(p−1) ξ p−1 f (ξ )ν −1/p (ξ )χ[a,r ] (ξ )dξ
≥
C −1/(p(p−1))
2
X Z k∈Ia
2k+1 2k
!1/(p−1) (p−1)/p ξ p−1 f (ξ )χ[a,r ] (ξ )dξ
for all large enough r. By condition (2.8), this implies (2.5). To complete the proof it remains to use Theorem 2.3.
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