Ammari H., Benkirane A., Touzani A. (editors) Recent Developments in Nonlinear Analysis
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December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
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For solutions of (P
λ
) we understand critical points of the associated Euler-
Lagrange functional E
λ
∈ C
1
(L
p
(Ω)), given by
E
λ
(v) =
1
p
P (v) − λ
1
q
Q(v) −
1
r
R(v).
As in (cf.
13,19
), we introduce the modified Euler-Lagrange functional defined
on R × L
p
(Ω) by
˜
E
λ
(t, v) = E(tv). If v is an arbitrary element of L
p
(Ω),
∂
t
˜
E
λ
(., v) (resp. ∂
tt
˜
E
λ
(., v))are the first (resp. second) derivative of the real
valued function: t 7→
˜
E
λ
(t, v).
2. Preliminary results
Since the functional
˜
E
λ
is even in t and that we are interested by the
positive solutions, we limit our study for t > 0.
Lemma 2.1. For every v ∈ L
p
(Ω) \ {0}, There is a unique λ(v) > 0 such
that the real valued function t 7→ ∂
˜
E
λ
(t, v) has exactly two positive zeros
(resp. one positive zero ) if 0 < λ < λ(v) (resp. λ = λ(v)). This function
has no zero for λ > λ(v).
Proof : Let v be an arbitrary element of L
p
(Ω) \{0} and let us write
∂
t
˜
E
λ
(t, v) = t
q−1
˜
F
λ
(t, v), where
˜
F
λ
(t, v) = t
p−q
P (v) −λQ(v) −t
r−q
R(v).
Then
∂
tt
˜
E
λ
(t, v) = (q − 1)t
q−2
˜
F
λ
(t, v) + t
q−1
∂
t
˜
F
λ
(t, v),
holds true, with
∂
t
˜
F
λ
(t, v) = t
p−q−1
[(p − q)P (v) − (r − q)t
r−p
R(v)].
It is clair that the real valued function t 7→
˜
F
λ
(t, v) is increasing on ]0, t(v)[,
decreasing on ]t(v), +∞[ and attains its unique maximum for t = t(v),
where
t(v) = (
p − q
r − q
P (v)
R(v)
)
1
r−p
. (2)
Thus, if
˜
F
λ
(t(v), v) > 0 (resp.
˜
F
λ
(t(v), v) = 0), the function t 7→
˜
F
λ
(t, v) has
two positive zeros (resp. one positive zero) and has no zero if
˜
F
λ
(t(v), v) < 0.
On the other hand, a direct computation gives
˜
F
λ
(t(v), v) =
r − p
p −q
(
p − q
r − q
P (v)
R(v)
)
r−q
r−p
R(v) − λQ(v).
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We deduce that
˜
F
λ
(t(v), v) > 0 (resp.
˜
F
λ
(t(v), v) < 0) for λ < λ(v) (resp.
λ > λ(v)) and
˜
F
λ(v)
(t(v), v) = 0, where
λ(v) = ˆc
P
r−q
r−p
(v)
Q(v)R
p−q
r−p
(v)
, (3)
with
ˆc =
r − p
p −q
(
p −q
r − q
)
r−q
r−p
.
Hence, if λ ∈ ]0, λ(v)[, the real valued function t 7→ ∂
t
˜
E
λ
(t, v) has two
positive zeros, denoted by t
1
(v, λ) and t
2
(v, λ), verifying 0 < t
1
(v, λ) <
t(v) < t
2
(v, λ).
Since
˜
F
λ
(t
1
(v, λ), v) =
˜
F
λ
(t
2
(v, λ), v) = 0, ∂
t
˜
F
λ
(t, v) > 0 for t < t(v)
and ∂
t
˜
F
λ
(t, v) < 0 for t > t(v), it follows that
∂
tt
˜
E
λ
(t
1
(v, λ), v) > 0 and ∂
tt
˜
E
λ
(t
2
(v, λ), v) < 0. (4)
This means that the real valued function t 7→
˜
E
λ
(t, v), (t > 0) achieves
its unique local minimum at t = t
1
(v, λ) and its global maximum at t =
t
2
(v, λ).
Lemma 2.2. If we put
ˆ
λ = inf
v∈L
p
(Ω)\{0}
λ(v), then
ˆ
λ > 0.
Proof : By Sobolev injection theorem, we have X ,→ L
q
(Ω) and X ,→
L
r
(Ω). Thus there exists two positive constants c
1
and c
2
such that
||Λv||
q
≤ c
1
||v||
p
et ||Λv||
r
≤ c
2
||v||
p
.
Then (3) implies for every v ∈ L
p
(Ω) \ {0}
λ(v) ≥
ˆc
c
q
1
c
r(p−q)
r−p
2
> 0.
Consider λ ∈]0,
ˆ
λ[ and let (v
n
) be minimizing sequence of v 7→
˜
E
λ
(t
1
(v, λ), v) in L
p
(Ω) \ {0} (resp. of v 7→
˜
E
λ
(t
2
(v, λ), v)).
Put V
n
= t
1
(v
n
, λ)v
n
and W
n
= t
2
(v
n
, λ)v
n
.
Lemma 2.3. The sequences (V
n
) and (W
n
) verify :
(i) lim sup
n→+∞
||V
n
||
p
< +∞ (resp. lim sup
n→+∞
||W
n
||
p
< +∞ )
(ii) lim inf
n→+∞
||V
n
||
p
> 0 (resp. lim inf
n→+∞
||W
n
||
p
> 0)
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222
Proof : (i) We know that ∂
t
˜
E
λ
[t
1
(v
n
, λ), v
n
) = 0.
Hence
||V
n
||
p
p
= λ||ΛV
n
||
q
q
+ ||ΛV
n
||
r
r
. (5)
Suppose that there is a subsequence of (V
n
), still denoted by (V
n
) such that
lim
n→+∞
||V
n
||
p
= +∞. Us r > q, there exist a constant c > 0 such that
||ΛV
n
||
q
≤ c||ΛV
n
||
r
. Then the relation (5) implies that lim
n→+∞
||ΛV
n
||
r
=
+∞. The fact that 0 < q < r enables us to deduce: ||ΛV
n
||
q
q
= o
n
(||ΛV
n
||
r
r
).
Then
||V
n
||
p
p
= ||ΛV
n
||
r
r
(1 + o
n
(1)),
and
E
λ
(V
n
) = ||ΛV
n
||
r
r
(
1
p
−
1
r
+ o
n
(1)).
which implies that E
λ
(V
n
) tends to +∞ as n goes to +∞ and this is
impossible.
The same arguments with a minimizing sequence (v
n
) of v 7→
˜
E
λ
(t
2
(v, λ), v)
show that lim sup
n→+∞
||W
n
||
p
< +∞.
(ii) Relation (5) and the fact that ∂
tt
˜
E
λ
[t
1
(v
n
, λ), v
n
) > 0, implies
(p − 1)||V
n
||
p
p
− λ(q − 1)||ΛV
n
||
q
q
− (r − 1)||ΛV
n
||
r
r
> 0. (6)
If we combine (5) and (6), we obtain for every n ∈ N
λ(p − q)||ΛV
n
||
q
q
+ (p − r)||ΛV
n
||
r
r
> 0.
So
E
λ
(V
n
) = λ
q − p
pq
Q(V
n
) +
r − p
pr
R(V
n
)
≤
−1
pq
(λ(p − q)Q(V
n
) + (p −r)R(v
n
))
< 0.
Suppose that there is a subsequence of (V
n
), still denoted by (V
n
)
such that lim
n→+∞
||V
n
||
p
= 0. By Sobolev injection theorem we de-
duce that lim
n→+∞
||ΛV
n
||
q
= 0 and lim
n→+∞
||ΛV
n
||
r
= 0. It follows that
lim
n→+∞
E
λ
(V
n
) = 0, i.e inf
v∈L
p
(Ω)\{0}
˜
E
λ
(t
1
(v, λ), v) = 0, which is impossible
since
˜
E
λ
(t
1
(v
n
, λ), v
n
) < 0 for every n.
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Let (v
n
) be a minimizing sequence of v 7→
˜
E
λ
(t
2
(v), v) in L
p
(Ω) \{0}.
Sinse ∂
t
˜
E
λ
(t
2
(v
n
), v
n
) = 0 and ∂
tt
˜
E
λ
(t
2
(v
n
), v
n
) < 0, it follows that
||W
n
||
p
p
−λ||ΛW
n
||
q
q
− ||ΛW
n
||
r
r
= 0,
(p − 1) ||W
n
||
p
p
− λ(q − 1)||ΛW
n
||
q
q
− (r − 1)||ΛW
n
||
r
r
< 0.
Combining the two last inequalities and by Sobolev injection theorem there
exist a constant c
0
such that for every n we have
(p −q)||W
n
||
p
p
< (r − q)||ΛW
n
||
r
r
≤ c
0
||W
n
||
r
p
.
Hence
(p − q) ≤ c
0
||W
n
||
r−p
p
.
Now, suppose that there is a subsequence of (W
n
), still denoted by (W
n
)
such that lim
n→+∞
||W
n
||
p
= 0. This implies that p−q ≤ 0. which is impossible
since p > q.
Lemma 2.4. The functionals v 7→
˜
E
λ
(t
1
(v, λ), v) and v 7→
˜
E
λ
(t
2
(v, λ), v)
are bonded below in L
p
(Ω).
Proof : Let (v
n
) be a minimizing sequence of the functional v 7→
˜
E
λ
(t
1
(v, λ), v).
We know that ∂
t
˜
E
λ
(t
1
(v
n
, λ), v
n
) = 0, then
[t
1
(v
n
, λ)]
p
||v
n
||
p
p
= λ[t
1
(v
n
, λ)]
q
||Λv
n
||
q
q
+ [t
1
(v
n
, λ)]
r
||Λv
n
||
r
r
.
Hence
˜
E
λ
(t
1
(v
n
, λ), v
n
) = λ(
1
p
−
1
q
)[t
1
(v
n
, λ)]
q
||Λv
n
||
q
q
+(
1
p
−
1
r
)[t
1
(v
n
, λ)]
r
||Λv
n
||
r
r
.
As p < r, we conclude that
˜
E
λ
(t
1
(v
n
, λ), v
n
) ≥ λ(
1
p
−
1
q
)[t
1
(v
n
, λ)]
q
||Λv
n
||
q
q
. (7)
Sobolev injection of X in L
q
(Ω) and the fact that lim sup
n→+∞
||V
n
||
p
< +∞,
implies that there exists c and k positive such that for every n in N, we
have ||V
n
||
p
< k. and ||ΛV
n
||
q
≤ c||V
n
||
p
< kc. As q < p, the inequality (7)
implies
˜
E
λ
(t
1
(v
n
, λ), v
n
) > (
1
p
−
1
q
)λk
q
c
q
.
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We show by the same method that the functional v 7→
˜
E
λ
(t
2
(v, λ), v) is
bonded bellow.
Put
α
1
(λ) = inf
v∈L
p
(Ω)\{0}
˜
E
λ
(t
1
(v, λ), v). (8)
α
2
(λ) = inf
v∈L
p
(Ω)\{0}
˜
E
λ
(t
2
(v, λ), v). (9)
We have the following lemma:
Lemma 2.5. If λ ∈]0,
ˆ
λ[, then
α
1
(λ) = inf
v∈S,v≥0
˜
E
λ
(t
1
(v, λ), v) and α
2
(λ) = inf
v∈S,v≥0
˜
E
λ
(t
2
(v, λ), v),
where S is the unit sphere of L
p
(Ω).
Proof : Let t > 0. If ∂
t
˜
E
λ
(t, v) > 0, then t ∈]t
1
(v, λ), t
2
(v, λ)[.
Since |Λv| ≤ Λ|v|, we deduce that
∂
t
˜
E
λ
(t
i
(|v|, λ), v) ≥ ∂
t
˜
E
λ
(t
i
(|v|, λ), |v|) = 0, i = 1, 2.
It follows that ]t
1
(|v|, λ), t
2
(|v|, λ)[ ⊆ ]t
1
(v, λ), t
2
(v, λ)[.
Hence, t
1
(|v|, λ) ≥ t
1
(v, λ).
Using the fact that t 7→
˜
E
λ
(t, |v|) is decreasing on ]0, t
1
(|v|, λ)], we get
˜
E
λ
(t
1
((v, λ), |v|) ≥
˜
E
λ
(t
1
(|v|, λ), |v|)
and since |Λv| ≤ Λ|v|, we get
˜
E
λ
(t
1
(v, λ), v) ≥
˜
E
λ
(t
1
(v, λ), |v|).
Hence we conclude that
˜
E
λ
(t
1
(|v|, λ), |v|) ≤
˜
E
λ
(t
1
(v, λ), v).
Since |Λv| ≤ Λ|v| and the function t 7→
˜
E
λ
(t, v) is creasing on
[t
1
(v, λ), t
2
(v, λ)], we obtain
˜
E
λ
(t
2
(|v|, λ), |v|) ≤
˜
E
λ
(t
2
(|v|, λ), v)
≤
˜
E
λ
(t
2
(v, λ), v).
Finally, we have showed that for every v ∈ L
p
(Ω) \ {0}
˜
E
λ
(t
i
(|v|, λ), |v|) ≤
˜
E
λ
(t
i
(v, λ), v), where i = 1, 2. (10)
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Moreover, for every γ > 0 , we get
˜
E
λ
(γt,
v
γ
) =
˜
E
λ
(t, v),
∂
t
˜
E
λ
(γt,
v
γ
) =
1
γ
∂
t
˜
E
λ
(t, v),
∂
tt
˜
E
λ
(γt,
v
γ
) =
1
γ
2
∂
tt
˜
E
λ
(t, v).
It follows that
t
1
(v, λ) =
1
γ
t
1
(
v
γ
, λ), (11)
t
2
(v, λ) =
1
γ
t
2
(
v
γ
, λ). (12)
By the virtue of (10), (11) and (12), we conclude that
α
1
(λ) = inf
v∈S,v≥0
˜
E
λ
(t
1
(v, λ), v), (13)
α
2
(λ) = inf
v∈S,v≥0
˜
E
λ
(t
2
(v, λ), v), (14)
where S is the unit sphere of L
p
(Ω).
Lemma 2.6. Let (v
n
) ⊂ S be a minimizing sequence of (13) (resp. of (14)).
Then, (V
n
) := (t
1
(v
n
, λ)v
n
) (resp. (W
n
) := (t
2
(v
n
, λ)v
n
)) are Palais-Smale
sequences for the functional E
λ
.
Proof : We will show this lemma only for the sequence (V
n
), the proof for
(W
n
) can be done in the same way.
Let λ ∈]0,
ˆ
λ[. Then lim
n→+∞
E
λ
(V
n
) = α
1
(λ).
Now we show that lim
n→+∞
E
0
λ
(V
n
) = 0.
Notice that for every v ∈ L
p
(Ω) \ {0}, we have ∂
t
˜
E
λ
(t
1
(v, λ), v) = 0 and
∂
tt
˜
E
λ
(t
1
(v, λ), v) 6= 0. The implicit function theorem implies that the func-
tional v 7→ t
1
(v, λ) is C
1
since
˜
E
λ
is. Let us introduce the C
1
functional
f
1,λ
defined on S by
f
1,λ
(v) =
˜
E
λ
(t
1
(v, λ), v) = E
λ
(t
1
(v, λ)v).
Hence
α
1
(λ) = inf
v∈S
f
1,λ
(v) = inf
v∈S,v≥0
f
1,λ
(v) and lim
n→+∞
f
1,λ
(v
n
) = α
1
(λ).
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Using the Ekeland variational principle on the complete manifold (S, || ||
p
)
to the functional f
1,λ
, we conclude that
|f
0
1,λ
(v
n
)(ϕ)| ≤
1
n
||ϕ||
p
, for every ϕ ∈ T
v
n
S,
where T
v
n
S is the tangent space to S at the point v
n
.
Moreoever, since ∂
t
˜
E
λ
(t
1
(v
n
, λ), v
n
) ≡ 0, then for every ϕ ∈ T
v
n
S, one has
f
0
1,λ
(v
n
)(ϕ) = ∂
t
˜
E
λ
(t
1
(v
n
, λ), v
n
)∂
v
t
1
(v
n
, λ)(ϕ)
+∂
v
˜
E
λ
(t
1
(v
n
, λ), v
n
)(ϕ)
= ∂
v
˜
E
λ
(t
1
(v
n
, λ), v
n
)(ϕ),
where ∂
v
t
1
(v
n
, λ) denotes the derivative of t
1
(., λ) with respect to its first
variable at the point (v
n
, λ).
Furthermore, let
P : L
p
(Ω)\{0} → R × S
v 7→ (P
1
(v), P
2
(v)) = (k v k
p
,
v
k v k
p
).
Applying H¨older’s inequality, we get for every (v, ϕ) ∈ L
p
(Ω)\{0}×L
p
(Ω) :
kP
0
2
(v)(ϕ)k
p
≤ 2
kϕk
p
kvk
p
.
From lemma 2.3 and by the fact that kV
n
k
p
= t(v
n
, λ), there exists positive
constant C such that
t
1
(v
n
, λ) ≥ C, ∀n ∈ N.
Hence for every ϕ ∈ L
p
(Ω), we obtain
|E
0
(V
n
)(ϕ)| = |∂
t
˜
E
λ
(P
1
(V
n
), P
2
(V
n
))P
0
1
(V
n
)(ϕ)
+∂
v
˜
E
λ
(P
1
(V
n
), P
2
(V
n
))P
0
2
(V
n
)(ϕ)|
= |∂
v
˜
E
λ
(t(v
n
), v
n
)P
0
2
(V
n
)(ϕ)|
= |f
0
1,λ
(v
n
)P
0
2
(V
n
)(ϕ)|
≤
1
n
k P
0
2
(V
n
)(ϕ) k
p
≤
2
n
k ϕ k
p
C
We easily conclude that
lim
n→+∞
E
0
(V
n
) = 0 in L
p
0
(Ω).
Remark 2.1. Until now, the minimizing sequences we consider are in S
and are nonnegative.
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3. Existence results
Theorem 3.1. Let 1 < q < p < r < p
∗
2
and λ ∈ ]0,
ˆ
λ[. Then the problem
(P
λ
) has at least two positive solutions.
Proof : We will use the notations of the previous lemmas.
Since the sequences (V
n
) and (W
n
) are Palais-Smale for the functional E
λ
,
it suffices to show that E
λ
(0 < λ <
ˆ
λ) satisfy Palais-Smale condition.
By lemma 2.3, we deduce that (V
n
) is bonded in L
p
(Ω). Passing if nec-
essary to a subsequence, we get
V
n
* V
1
in L
p
(Ω),
ΛV
n
* ΛV
1
in X,
ΛV
n
→ ΛV
1
in L
r
(Ω), (and in L
q
(Ω)).
(15)
On the other hand we have,
hN
p
(V
n
), V
n
− V
1
i = h E
0
λ
(V
n
), V
n
− V
1
i + λ
Z
Ω
N
q
(ΛV
n
)(ΛV
n
− ΛV
1
)dx
+
Z
Ω
N
r
(ΛV
n
)(ΛV
n
− ΛV )dx.
Moreover, E
0
λ
(V
n
) → 0, N
q
(ΛV
n
) → N
q
(ΛV
1
) and N
r
(ΛV
n
) → N
r
(ΛV
1
).
Then hN
p
(V
n
), V
n
− V
1
i → 0.
The fact that N
p
is (S+) type implies that V
n
→ V
1
in L
p
(Ω).
We know that for any minimizing sequence (v
n
) of (13), there is a sub-
sequence still denoted by (v
n
) such that V
n
= t
1
(v
n
, λ)v
n
and t
1
(v
n
, λ) =
||V
n
||
p
. Hence
t
1
(v
n
, λ) → ||V
1
||
p
= t
1
,
which implies that
v
n
→ V
1
/t
1
= v
1
, and t
1
= t
1
(v
1
, λ),
where v
1
∈ S.
In the same way, for any minimizing sequence (v
n
) ⊂ S of (14), passing
if necessary to a subsequence, there is t
2
∈ ]0, +∞[ such that
t
2
(v
n
, λ)v
n
→ t
2
in R,
v
n
→ v
2
= V
2
/t
2
,
where V
2
is the limit of the sequence (W
n
) := (t
2
(v
n
, λ)v
n
) in L
p
(Ω) and
t
2
= ||V
2
||
p
= t
2
(v
2
, λ).
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228
At this stage, it is easy to see that V
1
6= V
2
. Indeed, since
∂
tt
˜
E
λ
(t
1
(v
1
, λ), v
1
) > 0 and ∂
tt
˜
E
λ
(t
2
(v
2
, λ), v
2
) < 0, it follows that
∂
tt
E
λ
(t
1
, V
1
/t
1
) > 0 and ∂
tt
E
λ
(t
2
, V
2
/t
2
) < 0. This achieves the proof.
In the sequel the solutions V
1
and V
2
of (P
0
λ
), for λ ∈]0,
ˆ
λ[, will be de-
noted by V
1,λ
and V
2,λ
. Also, t
1,λ
, t
2,λ
, v
1,λ
and v
2,λ
will stand for t
1
(v
1
, λ),
t
2
(v
2
, λ), v
1
and v
2
respectively.
Theorem 3.2. Let 1 < q < p < r < p
∗
2
. Then
(i) E
λ
(V
1,λ
) < 0 for λ ∈ ]0,
ˆ
λ[,
(ii)
E
λ
(V
2,λ
) > 0 for λ ∈ ]0, λ
0
[,
E
λ
(V
2,λ
) < 0 for λ ∈ ]λ
0
,
ˆ
λ[,
where
λ
0
=
q
r
(
r
p
)
r−q
r−p
ˆ
λ.
Proof : (i) Let us recall that ∂
t
˜
E
λ
(t
1,λ
, v
1,λ
) = 0 and ∂
tt
˜
E
λ
(t
1,λ
, v
1,λ
) >
0. Then
P (V
1,λ
) − λQ(V
1,λ
) − R(V
1,λ
) = 0,
(p − 1)P (V
1,λ
) − λ(q − 1)Q(V
1,λ
) − (r −1)R(V
1,λ
) > 0.
Using the fact that 1 < q < p < r, we get
λ(p − q)Q(V
1,λ
) + (p − r)R(V
1,λ
) > 0.
Hence
E
λ
(V
1,λ
) = λ
q − p
pq
Q(V
1,λ
) +
r − p
pr
R(V
1,λ
)
≤
−1
pq
(λ(p − q)Q(V
1,λ
) + (p − r)R(v
1,λ
))
< 0.
(ii) Let v be an arbitrary element of L
p
(Ω) \ {0} and let us write
˜
E
λ
(t, v) = t
q
˜
G
λ
(t, v), where
˜
G
λ
(t, v) =
t
p−q
p
P (v) −
λ
q
Q(v) −
t
r−q
r
R(v).
It follows that
∂
t
˜
E
λ
(t, v) = qt
q−1
˜
G
λ
(t, v) + t
q
∂
˜
G
λ
(t, v),
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
229
with
∂
t
˜
G
λ
(t, v) = t
p−q−1
(
p − q
p
P (v) −
r − q
r
t
r−p
R(v)).
It is clear that the real valued function t →
˜
G
λ
(t, v) is increasing on
]0, t
0
(v)[, decreasing on ]t
0
(v), +∞[ and attains its unique maximum for
t = t
0
(v), where
t
0
(v) = (
r
p
)
1
r−p
t(v), (16)
and t(v) is defined by the relation (2).
On the other hand, a direct computation gives
˜
G
λ
(t
0
(v), v) =
1
r
(
r
p
)
r−q
r−p
r − p
p −q
(
p −q
r − q
P (v)
R(v)
)
r−q
r−p
R(v) − λ
Q(v)
q
.
Similarly,
˜
G
λ
(t
0
(v), v) > 0 (resp.
˜
G
λ
(t
0
(v), v) < 0) if λ < λ
0
(v) (resp.
λ > λ
0
(v)) and
˜
G
λ
0
(v)
(t
0
(v), v) = 0, where
λ
0
(v) =
q
r
(
r
p
)
r−q
r−p
λ(v), (17)
with λ(v) given by (3). Thus, we get
˜
E
λ
(t
0
(v), v) > 0 if λ < λ
0
(v),
˜
E
λ
(t
0
(v), v) = 0 if λ = λ
0
(v),
˜
E
λ
(t
0
(v), v) < 0 if λ > λ
0
(v).
(18)
First, since the function
]0, 1[ → R
t →
ln t
1 −t
is increasing, then for every real numbers x and y such that 0 < x < y, one
has
ln(
1
x
) >
1 −x
1 −y
ln(
1
y
) = ln(
1
y
)
1−x
1−y
,
and consequently
0 < x(1/y)
1−x
1−y
< 1.
In the particular case x =
q
r
and y =
p
r
, we get
0 <
q
r
(
r
p
)
r−q
r−p
< 1,