Adding this to 4.4 yields 2.11, since φ
′′ p
r ≥ 0: see Lemma 3.1 v. 6
◦
The condition 2.12. If z
p
| y| |x| 0, then
V
p y y
|x|, | y| − V
p y
|x|, | y| | y|
〈 y
′
, k 〉
2
= pp − 2| y|
p −2
〈 y
′
, k 〉
2
≤ 0 and
V
p y
|x|, | y| | y|
|k|
2
= p| y|
p −2
|k|
2
. Therefore, combining this with 4.2 and 4.3 we see that the left-hand side of 2.12 is not larger
than − p| y|
p −2
|h|
2
− |k|
2
− p[p − 1z
−p p
|x|
p −2
− | y|
p −2
]|h|
2
≤ −p| y|
p −2
|h|
2
− |k|
2
, 4.6
as needed. Here in the last passage we have used the estimates z
−1 p
|x| | y| and z
2 p
≤ p − 1 see Corollary 3.3. On the other hand, if 0 z
p
| y| |x|, then recall the function F
p
given by 3.7. A little calculation yields
V
p y y
|x|, | y| − V
p y
|x|, | y| | y|
〈 y
′
, k 〉
2
= α
p
| y|
p −2
F
p
r〈 y
′
, k 〉
2
, which is nonpositive by means of Lemma 3.2 here r =
|x|| y|, as before. Furthermore, by 3.3, V
y
|x|, | y| | y|
= α
p
| y|
p −2
h pφ
p
r − rφ
′ p
r i
|k|
2
= −α
p
| y|
p −2
φ
′′ p
r|k|
2
, which, combined with 4.4 and 4.5 implies that the left-hand side of 2.12 does not exceed
α
p
| y|
p −2
φ
′′ p
r|h|
2
− |k|
2
. 7
◦
The bound 2.3. In view of the above reasoning, we have to take c
1
x, y = p| y|
p −2
and c
2
x, y = −α
p
| y|
p −2
φ
′′ p
r. It is evident that the estimate 2.3 is valid for this choice of c
i
, i = 1, 2.
4.2 The case 2 p 3
Let V
p
: [0, ∞ × [0, ∞ → R be given by
V
p
x, y = α
p
x
p
φ
p
yx if y
≥ z
p
x ,
y
p
− z
p p
x
p
if y ≤ z
p
x ,
where α
p
= z
p
φ
p −1
z
p −1
. 4.7
Let S
1
= {x, y ∈ H × H : 0 | y| z
p
|x|} and
S
2
= {x, y ∈ H × H : | y| z
p
|x| 0}. 541
As in the previous case, we will verify that V
p
enjoys the requirements listed in Section 2. 1
◦
Regularity . Clearly, we have that V
p
is continuous, of class C
1
on 0, ∞ × 0, ∞ and of class C
2
on 0, ∞ × 0, ∞ \ {x, y : y = z
p
x }. Hence U
p
given by 2.7 has the necessary smoothness. In addition, it is easy to see that the first order derivative of U
p
is bounded on bounded sets. 2
◦
The growth condition 2.8. This is guaranteed by the asymptotics 3.5.
3
◦
The inequality 2.9 This is obvious: V
p
x, y ≤ x
p
1 − z
p p
≤ 0 if x ≥ y ≥ 0. 4
◦
The majorization 2.10. This can be established exactly in the same manner as in the previous
case. In fact one can show that the majorization is strict provided y z
p
x . The details are left to
the reader. 5
◦
The condition 2.11. Note that if 0
| y| z
p
|x|, then
V
p x x
|x|, | y| − V
p x
|x|, | y| |x|
〈x
′
, h 〉
2
= −pp − 2z
p p
|x|
p −2
〈x
′
, h 〉
2
≤ 0 4.8
and V
p x
|x|, | y| |x|
|h|
2
= −pz
p p
|x|
p −2
|h|
2
≤ 0, 4.9
so 2.11 follows. Suppose then, that | y| z
p
|x| 0 and recall F
p
introduced in Lemma 3.2. By means of this lemma, after some straightforward computations, one gets
V
p x x
|x|, | y| − V
p x
|x|, | y| |x|
〈x
′
, h 〉
2
= α
p
|x|
p −2
F
p
r 〈x
′
, h 〉
2
≤ 0, 4.10
where we have set r = | y||x|. Moreover, by 3.3,
V
p x
|x|, | y| |x|
|h|
2
= α
p
h pφ
p
r − rφ
′ p
r i
|x|
p −2
|h|
2
= −α
p
φ
′′ p
r |x|
p −2
|h|
2
4.11 is nonpositive; this completes the proof of 2.11.
6
◦
The condition 2.12. If 0
| y| z
p
|x|, then
V
p y y
|x|, | y| − V
p y
|x|, | y| | y|
〈 y
′
, k 〉
2
= pp − 2| y|
p −2
〈 y
′
, k 〉
2
≤ pp − 2| y|
p −2
|k|
2
and V
p y
|x|, | y| | y|
|k|
2
= p| y|
p −2
|k|
2
. Therefore, combining this with 4.8 and 4.9 we see that the left-hand side of 2.12 can be
bounded from above by pp
− 1| y|
p −2
|k|
2
− pz
p p
|x|
p −2
|h|
2
≤ pp − 1z
p −2
p
|x|
p −2
|k|
2
− pz
p p
|x|
p −2
|h|
2
≤ −pz
p p
|x|
p −2
|h|
2
− |k|
2
. 4.12
542
Here in the latter passage we have used Corollary 3.3. If | y| z
p
|x| 0, then, again using the notation r =
| y||x|,
V
p y y
|x|, | y| − V
p y
|x|, | y| | y|
〈 y
′
, k 〉
2
= α
p
|x|
p −2
h φ
′′ p
r − r
−1
φ
′ p
r i
〈 y
′
, k 〉
2
≤ α
p
|x|
p −2
h φ
′′ p
r − r
−1
φ
′ p
r i
|k|
2
and V
p y
|x|, | y| | y|
|k|
2
= α
p
|x|
p −2
r
−1
φ
′ p
r|k|
2
. Combining this with 4.10 and 4.11, we get that the left-hand side of 2.12 does not exceed
−α
p
|x|
p −2
φ
′′ p
r|h|
2
− |k|
2
. 7
◦
The bound 2.3. By the above considerations, we are forced to take c
1
x, y = pz
p p
|x|
p −2
and c
2
x, y = α
p
|x|
p −2
φ
′′ p
r and it is clear that the condition is satisfied. This establishes 1.4 for 2 p 3.
4.3 The case p
