Sign changing solutions of Lane Emden problems with interior nodal line and semilinear heat equations

0.57 MB·English

by  Francesca De Marchis

SIGN CHANGING SOLUTIONS OF LANE EMDEN PROBLEMS WITH
INTERIOR NODAL LINE AND SEMILINEAR HEAT EQUATIONS
FRANCESCA DE MARCHIS, ISABELLA IANNI, FILOMENA PACELLA
Abstract. We consider the semilinear Lane Emden problem
(cid:26) −∆u=|u|p−1u in Ω
(E )
u=0 on ∂Ω p
where Ω is a smooth bounded simply connected domain in R2, invariant by the action
2
of a finite symmetry group G.
1
We show that if the orbit of each point in Ω, under the action of the group G, has
0
2 cardinality greater than or equal to 4 then, for p sufficiently large, there exists a sign
changing solution of (E ) with two nodal regions whose nodal line does not touch ∂Ω.
t p
c Thisresultisprovedasaconsequenceofananalogousresultfortheassociatedparabolic
O
problem.

3
2

]
P 1. Introduction
A
We consider the semilinear elliptic problem
.
h
t (cid:26)
a −∆u = |u|p−1u in Ω
m (E )
u = 0 on ∂Ω p
[
where Ω ⊆ R2 is a smooth bounded domain and p > 1.
1
v
3 In this paper we address the question of the existence of sign changing solutions of (E )
p
0
with two nodal domains and whose nodal line does not touch ∂Ω. By this we mean that,
3
6 denoting by
.
0 Z = {x ∈ Ω, u (x) = 0} (1.1)
p p
1
2 the nodal set of u , then Z ∩∂Ω = ∅.
p p
1
:
v Obviously if Ω is a ball such a solution exists for any p > 1, just considering the least
i
X energy nodal radial solution of (E ). Moreover, by symmetry considerations, it has been
p
r proved in [3] that the radial one is not the least energy nodal solution in the whole space
a
H1(Ω) and that it has Morse index at least three (actually at least four).
0
Thus it is natural to ask whether a sign changing solution with an interior nodal line
exists for other domains.
2010 Mathematics Subject classification: 35B05, 35B06, 35J91, 35K91.
Keywords: superlinear elliptic boundary value problem, sign-changing solution, parabolic flow, nodal
set.
ResearchpartiallysupportedbyFIRBproject:AnalysisandBeyondandPRIN2009-WRJ3W7grant.
1
2
By imposing some symmetry on Ω we are able to prove the following theorem, assuming,
without loss of generality, that the origin O ∈ Ω.
Theorem 1.1. Assume that Ω is simply connected and invariant under the action of a
finite group G of orthogonal transformations of R2. If |Gx| ≥ 4 for any x ∈ Ω¯ \ {O},
then, for p sufficiently large (E ) admits a sign changing G-symmetric solution u , with
p p
two nodal domains, whose nodal line neither touches ∂Ω, nor passes through the origin.
In the above statement by |Gx| we mean the cardinality of the orbit of the point x under
the action of the group G.
Remark 1.2. Note that by Theorem 3.4 of [2] the above hypothesis on the group is equiv-
alent to ask that G = C or G = D for some h ≥ 4, where C is the cyclic group of order
h h h
h generated by a rotation of 2π and D is the dihedral group of order 2h, generated by a
h h
rotation of 2π and a reflection about a line through the origin.
h
The point of view taken in this article to prove Theorem 1.1 is to study an initial value
problem for the associated semilinear heat equation and prove that, for p large, it is
possible to construct an initial datum for which the solution is global (in time), changes
sign(ateverytime)andthecorrespondingω-limitsetisnonemptyandconsistsofsolutions
of (E ) having the desired properties.
p
More precisely we consider the semilinear parabolic problem

v −∆v = |v|p−1v in Ω×(0,+∞)
 t
v = 0 on ∂Ω×[0,+∞) (P )
p
 v(·,0) = v in Ω
0
and prove the following result.
Theorem 1.3. Let the domain Ω and the group G be as in Theorem 1.1. Then there
exists p > 1 such that for any p ≥ p there exists a G-symmetric function v = v ∈
0 0 0 p,0
H1(Ω) having two nodal domains such that the corresponding solution v (x,t) of (P ) is
0 p p
global, sign changing for every fixed t, the ω-limit set ω(v ) is nonempty and any function
0
u ∈ ω(v ) is a G-symmetric nodal solution of (E ) with two nodal domains and whose
p 0 p
nodal line neither touches ∂Ω, nor passes through the origin.
Obviously the result of Theorem 1.1 follows from Theorem 1.3.
The are several motivations to choose the parabolic approach. A first one is that, the
study of (P ) and, in particular, the analysis of qualitative properties of the solutions of
p
(P ) is interesting in itself. Another one is to show that the sign changing solution of (E )
p p
that we get in Theorem 1.1 arises as a natural evolution, by the heat flow, of a suitable
“initial” function that we construct.
Moreover we believe that the two problems (E ) and (P ), which most of the time are
p p
analyzed in independent ways, have several common features.
One of the important steps in proving Theorem 1.3 is the choice of the initial datum v .
p,0
To do this we select a proper linear combination of the positive radial solution of (E ) in
p
3
the annulus
A := {x ∈ R2 : e−αp < |x| < b} ⊂ Ω (1.2)
p,α
and the negative radial solution of (E ) in the ball
p
B := {x ∈ R2 : |x| < e−αp} ⊂ Ω (1.3)
p,α
for α > 0 and b > 0 suitable chosen. We obtain in this way a function v which is
p,0
G-invariant, has two nodal regions and the nodal line does not intersect ∂Ω.
Then to prove that the trajectory starting from v ends up with a solution of (E ) having
p,0 p
the same properties a good estimate from above of the energy of v , for p large, is crucial.
p,0
This is a delicate and nontrivial point of the proof (see Proposition 2.2). It allows to prove,
together with an estimate from below of the energy in each nodal region, and with the
¯
assumption on the symmetry group, namely |Gx| ≥ 4, ∀x ∈ Ω\{O}, that if the nodal
line of the solution u in ω(v ) touched the boundary or passed through the origin then
p p,0
too many nodal regions would be created and so the energy of u would exceed that of
p
v which is not possible.
p,0
For this last step as well as to show that the solution v (x,t) of (P ) does change sign,
p p
∀t > 0, some topological argument is needed.
We point out that a stronger result would be to show that along the trajectory, i.e.
∀ t ∈ (0,+∞) the solution v (x,t) has always two nodal regions and its nodal line does
p
not touch the boundary or passes through the origin. We believe that this should be true
(for energy reasons!) but we are not able to prove it at this stage.
We also remark that the same energy estimates for v that we obtain here show that,
p.0
under the same assumption as in Theorem 1.1, the least energy nodal solution in the
subspace HG of H1(Ω) of G-invariant functions has two nodal regions and the nodal line
0
neither touches the boundary nor passes through the origin, if p is sufficiently large. We
observe that for this minimality property some assumption on the action of the symmetry
group G is needed. Indeed in 2-dimension the least energy nodal solution of (E ) in the
p
ball has a symmetry hyperplane (see [4, 12]) but the nodal line touches the boundary
([3]).
Another important issue is to have information about the Morse index m(u ) of the
p
solution constructed in Theorem 1.1. As mentioned before for least energy nodal radial
solution in the ball the Morse index is at least three ([3]). By using the same ideas as in
[3], under an additional hypothesis on the symmetry group we also get that m(u ) ≥ 3.
p
Theorem 1.4. Under the same assumptions of Theorem 1.1, if G contains a reflection
with respect to a line T through the origin (namely G = D , for some h ≥ 4) and Ω is
h
convex in the direction orthogonal to T, then m(w ) ≥ 3, where w is any G-symmetric
p p
solution of (E ) whose nodal line does not touch the boundary of Ω.
p
Finallyitwouldbeinterestingtostudytheasymptoticbehaviorasp → ∞ofthesolutions
u constructed in Theorem 1.1. In view of the geometric properties of our solutions we
p
4
believe that the behavior should be similar to that of the radial solutions in the ball
studied in [7]. We plan to analyze this question in a future paper.
Further comments will be delayed to the specific sections.
Theoutlineofthepaperisasfollows.InSection2wedefinethefunctionsusedtoconstruct
theinitialdatumandprovethecrucialenergyestimates.InSection3werecallsomeknown
properties of the semilinear heat flow and prove a preliminary result about sign changing
solutions of (P ). Finally in Section 4 we prove Theorem 1.3 and Theorem 1.4.
p
2. Energy estimates
Let us consider in the Sobolev space H1(Ω) the Nehari manifold
0
N := {u ∈ H1(Ω)\{O} : (cid:107)∇u(cid:107)2 = (cid:107)u(cid:107)p+1}
p 0 2 p+1
and the energy functional
1 1
E (u) := (cid:107)∇u(cid:107)2 − (cid:107)u(cid:107)p+1
p 2 2 p+1 p+1
associated to problem (E ), for p > 1.
p
We start with a simple lemma.
Lemma 2.1. Let u ,u ∈ N , supp u ∩supp u = ∅. Then
1 2 p 1 2
E (t u +t u ) ≤ E (u )+E (u ) for all t ,t ∈ R.
p 1 1 2 2 p 1 p 2 1 2
Proof.
(cid:88)2 (cid:18)t2 |t |p+1 (cid:19)
E (t u +t u ) = E (t u )+E (t u ) = i (cid:107)∇u (cid:107)2 − i (cid:107)u (cid:107)p+1
p 1 1 2 2 p 1 1 p 2 2 2 i 2 p+1 i p+1
i=1
(cid:88)2 (cid:18)t2 |t |p+1(cid:19) (cid:88)2 (cid:18)1 1 (cid:19)
= i − i (cid:107)∇u (cid:107)2 ≤ − (cid:107)∇u (cid:107)2 = E (u )+E (u ).
2 p+1 i 2 2 p+1 i 2 p 1 p 2
i=1 i=1
(cid:3)
Now let us denote by B a ball centered at the origin with radius b > 0 such that B ⊂ Ω
b b
and consider the annulus A and the ball B defined in (1.2) and (1.3).
p,α p,α
The following proposition plays a crucial role in proving our results.
Proposition 2.2. For α > 0 let u be the unique positive radial solution to (E ) in A
p,1,α p p,α
and u be the unique positive radial solution to (E ) in B . Then u ,u ∈ N
p,2,α p p,α p,1,α p,2,α p
and for any (cid:15) > 0 there exists p such that for p ≥ p
(cid:15) (cid:15)
e2α−1
pE (u ) ≤ 4πe +(cid:15),
p p,1,α
α
pE (u ) ≤ 4πe4α+1 +(cid:15).
p p,2,α
5
Remark 2.3. The estimates given in Proposition 2.2 are accurate and give a sharp upper
bound on the minimal energy of G-invariant sign-changing functions in N , as needed to
p
prove our results. Their proof is long and technically complicated and therefore we postpone
it to the end of the section. However even a rough estimate of the minimal energy seems
not easy to get.
Corollary 2.4 (Energy upper bound).
There exists α¯ > 0 such that for any (cid:15) > 0 there exists p such that
(cid:15)
pE (t u +t u ) ≤ 4.97 ·4πe+ (cid:15) for all t ,t ∈ R, for p ≥ p .
p 1 p,1,α¯ 2 p,2,α¯ 1 2 (cid:15)
Proof. Let α > 0, since u ∈ N , i = 1,2 and supp u ∩ supp u = ∅, by Lemma
p,i,α p p,1,α p,2,α
2.1
pE (t u +t u ) ≤ pE (u )+pE (u ) for all t ,t ∈ R.
p 1 p,1,α 2 p,2,α p p,1,α p p,2,α 1 2
Wedenotebyα¯ thepointofminimumin(0,+∞)ofthescalarfunctionf(α) := e2α−1+e4α.
α
Let (cid:15) > 0, using Proposition 2.2 one has for p ≥ p
(cid:15)
(cid:0) (cid:1)
pE (u )+pE (u ) = 4πeminf(α)+(cid:15) ≤ 4πef 1 +(cid:15) ≤ 4.97 ·4πe+(cid:15).
p p,1,α¯ p p,2,α¯ 5
α>0
(cid:3)
Proposition 2.5 (Energy lower bound). Let (u ) be a family of nodal solutions of prob-
p p
lem (E ). Then for any (cid:15) > 0 there exists p such that for p ≥ p ,
p (cid:15) (cid:15)
pE (u χ ) ≥ 4πe−(cid:15)
p p Dp
where D ⊂ Ω is any nodal domain of u and χ is the characteristic function of D . In
p p Dp p
particular if u has k nodal domains then for any (cid:15) > 0
p
pE (u ) ≥ k4πe−(cid:15),
p p
for p sufficiently large.
Proof. Since u χ ∈ N one has that
p Dp p
pE (u χ ) ≥ inf(pE ).
p p Dp p
Np
The conclusion follows passing to the limit and observing that
(cid:18) (cid:19)
lim infpE = 4πe (2.1)
p
p→∞ Np
(see [1, Lemma 2.1] and also [14]). (cid:3)
Proof of Proposition 2.2. It is enough to prove that
(cid:90) 8πe2α
p |∇u |2dx ≤ +o (1), as p → +∞, (2.2)
p,1,α p
α
A
p,α
6
(cid:90)
p |∇u |2dx = 8πe·e4α +o (1), as p → +∞. (2.3)
p,2,α p
B
p,α
Indeed, since u ∈ N , i = 1,2, one has that
p,i,α p
(cid:18) (cid:19)
1 1 1
pE (u ) = p − (cid:107)∇u (cid:107)2 ≤ p(cid:107)∇u (cid:107)2, for any p > 1.
p p,i,α 2 p+1 p,i,α 2 2 p,i,α 2
STEP 1: we prove (2.2).
1
It is easy to see that u = αp−1z , where z is a minimizer for
p,1,α p p p
(cid:82)
|∇u|2
A
I := inf p,α
p u ∈ H01,r(Ap,α) (cid:16)(cid:82) up+1(cid:17)p+21
A
u (cid:54)= 0 p,α
(cid:82) (cid:82)
and α = |∇z |2/ zp+1. Hence one has
p A p A p
p,α p,α
 p+1
(cid:82) p−1
(cid:90) (cid:90) |∇z |2
2 A p
p |∇u |2dx = pαp−1 |∇z |2dx = p p,α 
p,1,α p p (cid:16) (cid:17) 2 
Ap,α Ap,α (cid:82) zp+1 p+1
A p
p,α
 p+1
(cid:82) p−1
|∇ω |2
A p
≤ p p,α  ,
(cid:16) (cid:17) 2 
(cid:82) ωp+1 p+1
A p
p,α
where ω , introduced in [6], is defined as follows:
p
(cid:40)
2 αp+log|x| e−αp ≤ |x| ≤ b1e−1αp
2 2
ω (x) = ω (|x|) =
p p αp+logb log b b21e−12αp ≤ |x| ≤ b
|x|
Since
(cid:90) 8π (cid:90) b 1 8π
|∇ω |2dx = dr = ,
p (αp+logb)2 r αp+logb
Ap,α e−αp
we get
(cid:90) p+1
(8π)p−1 1
p |∇u |2dx ≤ p =: M . (2.4)
Ap,α p,1,α (αp+logb)pp−+11 ((cid:82)A ωpp+1)p−21 p,α,b
p,α
7
To estimate M one needs to compute or at least to control
p,α,b
 
(cid:90) 2p+2π (cid:90) b12e−12αp (cid:90) b b 
ωp+1 =  logp+1(eαpr)rdr+ logp+1( )rdr
Ap,α p (αp+logb)p+1  e−αp b12e−12αp r 
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125)
=:I =:I
1,p,α,b 2,p,α,b
 
2p+2π (cid:90) b12e21αp (cid:90) 1 1
= e−2αp rlogp+1rdr+b2 rlogp+1 dr.
(αp+logb)p+1 1 b−12e−12αp r
This is easier if p ∈ N because one can compute explicitly the two integrals. Indeed for
n ∈ N we have
 
(cid:90) 2n+2π (cid:90) b12e12αn (cid:90) 1
ωn+1 = e−2αn rlogn+1rdr+(−1)n+1b2 rlogn+1rdr
Ap,α n (αn+logb)n+1 1 b−21e−12αn
(2.5)
and so iterating (n+1) times the following integration by parts
(cid:90) (cid:26) s2 logks− k (cid:82) slogk−1ds k = 2,...,n+1
slogksds = 2 2
s2 logs− s2 k = 1
2 4
and observing that logkb−1e−1αn = (−1)klogkb1e1αn = (−1)k(αn+logb)k we get
2 2 2 2
2k
(cid:90) b12e12αn e−αnb(n+1)! (cid:88)n+1 (αn+logb)n+1−k e−2αn(n+1)!
e−2αn rlogn+1rdr = (−1)k +(−1)n
2n+2 (n+1−k)! 2n+2
1
k=0
(cid:90) 1 e−αnb(n+1)! (cid:88)n+1 (αn+logb)n+1−k b2(n+1)!
(−1)n+1b2 rlogn+1rdr = − +(−1)n
b−21e−12αn 2n+2 k=0 (n+1−k)! 2n+2
So substituting into (2.5) and denoting by [t] the integer part of a number t ∈ R,
(cid:90) e−αnb(n+1)!π (cid:88)n+1 (αn+logb)n+1−k (n+1)!π
ωn+1 = [(−1)k −1] + [(−1)ne−2αn +b2]
n (αn+logb)n+1 (n+1−k)! (αn+logb)n+1
A
p,α k=0
[n]
e−αnb(n+1)!π (cid:88)2 (αn+logb)n−2m (n+1)!π
= −2 + [(−1)ne−2αn +b2]
(αn+logb)n+1 (n−2m)! (αn+logb)n+1
m=0
 −2e−αnb(n+1)!π (cid:80)n2 (αn+logb)n−2m + (n+1)!π (e−2αn +b2) n even
 (αn+logb)n+1 m=0 (n−2m)! (αn+logb)n+1
=
 −2e−αnb(n+1)!π (cid:80)n−21 (αn+logb)n−2m + (n+1)!π (b2 −e−2αn) n odd
(αn+logb)n+1 m=0 (n−2m)! (αn+logb)n+1
8
 −2e−αnb(n+1)!π (cid:80)n2 (αn+logb)2h + (n+1)!π (e−2αn +b2) n even
 (αn+logb)n+1 h=0 (2h)! (αn+logb)n+1
=
 −2e−αnb(n+1)!π (cid:80)n−21 (αn+logb)2h+1 + (n+1)!π (b2 −e−2αn) n odd
(αn+logb)n+1 h=0 (2h+1)! (αn+logb)n+1
 (cid:104) (cid:105)
 −2e−αnb(n+1)!π be−αn−b−1e−αn −(cid:80)+∞ (αn+logb)2h + (n+1)!π (e−2αn +b2) n even
 (αn+logb)n+1 2 h=n+2 (2h)! (αn+logb)n+1
2
=
(cid:104) (cid:105)
 −2e−αnb(n+1)!π be−αn−b−1e−αn −(cid:80)+∞ (αn+logb)2h+1 + (n+1)!π (b2 −e−2αn) n odd
(αn+logb)n+1 2 h=n+1 (2h+1)! (αn+logb)n+1
2
 2e−αnb(n+1)!π (cid:80)+∞ (αn+logb)2h n even
 (αn+logb)n+1 h=n+2 (2h)!
2
=
 2e−αnb(n+1)!π (cid:80)+∞ (αn+logb)2h+1 n odd
(αn+logb)n+1 h=n+1 (2h+1)!
2
e−αnb(n+1)!π (αn+logb)n+2 e−αnb(αn+logb)
≥ 2 = 2π .
(αn+logb)n+1 (n+2)! (n+2)
From the latter estimate and (2.4) we immediately deduce that
8πe2α
lim M ≥ . (2.6)
n,α,b
n→+∞ α
In general, for p ∈ R, p > 1, we can rewrite M as follows
p,α,b
p+1 p+1
(8π)p−1e2α (αp+logb)p−1
M = p (2.7)
p,α,b α2 (cid:16) (cid:17) 2 (cid:16) (cid:17) 2
eα(p−1) p−1 (αp+logb)p+1(cid:82) ωp+1 p−1
αp−1 Ap,α p
(8π)pp−+11e2α (cid:18)eα(cid:19)p−21 (cid:18) logb(cid:19)pp−+11 1
= 1+ .
α2−pp−+11 α αp  p−21
 1 eαp (cid:90) 
 (αp+logb)p+1 ωp+1
pp αp p 
 A 
p,α
(cid:124) (cid:123)(cid:122) (cid:125)
:=Q
p,α,b
By virtue of (2.6), we know that lim Q ≥ 1, thus if we show that, for p ∈ R,
n→+∞ n,α,b
p→+∞
Q → 1 we are done.
p,α,b
To do so we write Q in a more explicit form
p,α,b
 
eαp2p+2 (cid:90) b12e−21αp (cid:90) b b 
Q = π  logp+1(eαpr)rdr+ logp+1( )rdr, (2.8)
p,α,b ppαp  e−αp b12e−21αp r 
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125)
=:I =:I
1,p,α,b 2,p,α,b
9
and we claim that for p sufficiently large
d
(I +I ) ≥ 0. (2.9)
1,p,α,b 2,p,α,b
dp
Once (2.9) is established the thesis follows easily. Indeed for p (cid:29) 1
eα[p]2[p]+2
Q ≥ π [I +I ]
p,α,b ([p]+1)([p]+1)α([p]+1) 1,p,α,b 2,p,α,b
1 (cid:18) [p] (cid:19)[p] 1
≥ Q min{1, }
[p],α,b
α [p]+1 [p]+1
and so
(cid:18) (cid:19) 2 (cid:18) (cid:19)2[p] (cid:18) (cid:19) 2
2 2 1 p−1 [p] p−1 1 p−1 p→+∞
(Qp,α,b)p−1 ≥ (Q[p],α,b)p−1 min{1, } −→ 1.
α [p]+1 [p]+1
Finally it remains to prove (2.9).
Differentiating we obtain:
 
d d (cid:90) b21e12αp
I1,p,α,b s==eαpr e−2αp logp+1(s)sds
dp dp
1

(cid:90) b21e12αp 1 1 1
= e−2αp−2α logp+1(s)sds+ αbeαp( αp+ logb)p+1
2 2 2
1

(cid:90) b12e12αp
+ logp+1(s)log(log(s))sds
1
αbe−αp (cid:90) b12e12αp
= (αp+logb)p+1 +(p+1)αe−2αp logp(s)sds
2p+2
1
αbe−αp (cid:90) b12e12αp
− (αp+logb)p+1 +e−2αp logp+1(s)log(log(s))sds
2p+1
1
αbe−αp (cid:90) b12e12αp
= − (αp+logb)p+1 +e−2αp logp+1(s)log(log(s))sds
2p+2
1
(cid:90) b12e12αp
+(p+1)αe−2αp logp(s)sds; (2.10)
1
d αbe−αp (cid:90) b b b
I = (αp+logb)p+1 + logp+1( )log(log( ))rdr
dp 2,p,α,b 2p+2 b21e−12αp r r
s=b αbe−αp (cid:90) b21e21αp 1
=r (αp+logb)p+1 +b2 logp+1(s)log(log(s)) ds.(2.11)
2p+2 s3
1
10
Then
d (cid:90) e b2
(I +I ) = logp+1(s)log(log(s))(e−2αps+ )ds
dp 1,p,α,b 2,p,α,b s3
1
(cid:124) (cid:123)(cid:122) (cid:125)
=:A
(cid:90) b12e12αp b2
+ logp+1(s)log(log(s))(e−2αps+ )ds
s3
e
(cid:124) (cid:123)(cid:122) (cid:125)
=:B
(cid:90) b12e12αp
+(p+1)αe−2αp logp(s)sds (2.12)
1
(cid:124) (cid:123)(cid:122) (cid:125)
=:C
Let us estimate separately, for p large, the three terms appearing above in formula (2.12).
e−2αps+b2≤e−2αp+b2,s∈[1,e] (cid:90) e
A s3 ≥ (e−2αp +b2) logp+1(s)log(log(s))ds
1
(cid:90) 1
t=l=og(s) (e−2αp +b2) tp+1log(t)etdt
0
tp+1et≤e,t∈[0,1] (cid:90) 1 p(cid:29)1
≥ (e−2αp +b2) log(t)dt ≥ −2b2e; (2.13)
0
(cid:90) b12e12αp b2
B ≥ logp+1(s)log(log(s)) ds
s3
e
p(cid:29)1 (cid:90) 2e2 1
≥ b2 log(2) logp+1(s) ds
s3
e2
r=1 (cid:90) e−2 1
=s b2 log(2) logp+1( )rdr
r
1e−2
2
(cid:90) e−2 1
≥ b2 log(2) 2p+1 e−2dr ≥ b2 log(2)e−42p−1; (2.14)
2
1e−2
2
C ≥ 0. (2.15)
At last collecting estimates (2.13), (2.14) and (2.15) it is immediate to see that for any
α > 0, for any b > 0 and for any p sufficiently large
d
(I +I ) ≥ 0.
1,p,α,b 2,p,α,b
dp
The proof of (2.2) is thereby complete.
STEP 2: proof of (2.3).