5.2. UNIFORM CONVEXITY AND SMOOTHNESS OF LP
5.2
121
Uniform Convexity and Uniform Smoothness
of Lp , 1 < p < 1
Let (⌦, ⌃, µ) be a measure space. The first goal of this section is to prove
the following
Theorem 5.2.1. Let 1 < p < 1 and denote the modulus of uniform convexity of Lp (µ) by p . Then for any 1 < p < 1 there is a cp > 0 so that.
(
cp "2 if 1 < p < 2
p (")
cp "p if 2 < p < 1.
Lemma 5.2.2. Assume ⇠, ⌘ 2 R
a) If 2  p < 1, then
b) If 0 < p  2
|⇠ + ⌘|p + |⇠
⌘|p
2(|⇠|p + |⌘|p ).
|⇠ + ⌘|p + |⇠
⌘|p  2(|⇠|p + |⌘|p ).
If p 6= 2 equality in (a) and (b) only holds if either ⇠ or ⌘ is zero.
Proof. If p = 2 we have equality by the binomial formula.
If 2 < p < 1 and ↵, 2 R, we apply Hölder’s inequality to the function
{1, 2} ! {↵2 ,
2
}, 1 7! ↵2 , 2 7!
2
,
the counting measure on {1, 2}, and the exponents p/2 and p/(p
↵2 +
(5.4)
2
 (|↵|p + | |p )2/p 2(p
|↵|p + | |p
(↵2 +
2)/p
2 p/2 (2 p)/2
)
2
2).
, and, thus,
.
If 0 < p < 2 we can replace p by 4/p and obtain
|↵|4/p + | |4/p
(↵2 +
2 2/p (p 2)/p
)
2
,
and if we replace |↵| and | | by |↵|p/2 and | |p/2 respectively, we obtain
(5.5)
|↵|2 + | |2
p
p
(|↵|p + | |p )2/p 2(p
(2 p)/2
|↵| + | |  2
2
2)/p
2 p/2
(|↵| + | | )
, or
.
122
CHAPTER 5. LP -SPACES
Since
0
⇠2
1
⌘2 + ⇠2
we derive that
|⇠|p
(|⌘|2 + |⇠|2 )p/2
(5.6)
(

⇠2
⌘ 2 +⇠ 2
⇠2
⌘ 2 +⇠ 2
if 2 < p
if 2 > p.
Forming similar inequalities by exchanging the roles of ⌘ and ⇠ and adding
them we get
(
 (|⌘|p + |⇠|p )p/2 if 2 < p
(5.7)
|⌘|p + |⇠|p
(|⌘|p + |⇠|p )p/2 if 2 > p.
Note that equality in (5.7) can only hold if ⌘ = 0 or ⇠ = 0.
Letting now ↵ = |⇠ + ⌘| and = |⇠ ⌘| we deduce from (5.4) and (5.7)
if p > 2
|⇠ + ⌘|p + |⇠
⌘|p
|⇠ + ⌘|2 + |⇠
= 2 ⇠2 + ⌘
2 p/2
⌘|2
p/2 (2 p)/2
2
2 |⌘|p + |⇠|p ,
which finishes the proof of part (a), while part (b) follows from applying
(5.5) and (5.7).
Corollary 5.2.3. Let 0 < p < 1 and f, g 2 Lp (µ). Then
(
2(kf kpp + kgkpp ) if p 2
p
p
kf + gkp + kf gkp
 2(kf kpp + kgkpp ) if p  2.
If p 6= 2 equality only holds if f · g = 0 µ-almost everywhere.
Lemma 5.2.4. Let 1 < p < 2. Then there is a positive constant C = C(p)
so that
⇣ s t 2
s + t 2 ⌘1/2 ⇣ |s|p + |t|p ⌘1/p
(5.8)
+

.
C
2
2
Proof. We can assume without loss of generality that s = 1 > |t| and need
therefore to show that for some C > 0 and all t 2 [ 1, 1] we have
(5.9)
⇣1
C
t ⌘2
 (t) =
⇣ 1 + |t|p ⌘2/p
2
⇣ 1 + t ⌘2
2
.
5.2. UNIFORM CONVEXITY AND SMOOTHNESS OF LP
123
Since is strictly positive on [ 1, 0] we only need to find C so that (5.8)
holds for all t 2 [0, 1]. Since ⇠ 7! ⇠ 1/p is strictly concave it follows for all
0<t<1
⇣ 1 tp ⌘2/p ⇣ 1
t ⌘2
+
>
+
,
2
2
2 2
we only need to show that
(5.10)
lim
t!1
(t)
> 0.
(1 t)2
We compute
i
d2
d h (2/p)+1
1
p (2/p) 1 p 1
(t)
=
2
(1
+
t
)
t
(1
+
t)
dt2
dt
2
= 2 (2/p)+1 (2 p)(1 + tp )(2/p) 2 t2p 2
1
+ 2 (2/p)+1 (p 1)(1 + tp )(2/p) 1 tp 2
2
and thus
d
(t)
dt
d2
(t)
dt2
t=1
= (2
t=1
= 0, and
p)(1/2) + (p
1)
(1/2) = (p
1)/2 > 0
Applying now twice the L’Hospital rule, we deduce our wanted inequality
(5.10)
Via integrating, Lemma 5.2.4 yields
Corollary 5.2.5. If 1 < p  2 and f, g 2 Lp (µ) and if C = C(p) is as in
Lemma 5.2.4, it follows
(5.11)
⇣ f
g
C
2
+
f + g 2 ⌘1/2
2
p
⇣ |f |p + |g|p ⌘1/p

2
=2
1/p
⇣
p
kf kp + kgkp
⌘1/p
.
Proposition 5.2.6. If 1  p < q < 1 and fj 2 Lp , j = 1, 2 . . . then
n
X
j=1
kfi kqp
!1/q

n
⇣X
j=1
|fj |q
⌘1/q
.
p
124
CHAPTER 5. LP -SPACES
Proof. We can assume without loss of generality that
n
X
j=1
kfi kqp = 1.
We estimate
n
⇣X
j=1
|fj |
q
⌘1/q
n
X
=
j=1
p
kfj kqp
⇣ |f | ⌘q
j
kfj kp
!1/q
=
p
!p/q !1/p
⇣ |f | ⌘q
j
kfj kp
j=1
!
n
⇣ |f | ⌘p 1/p
X
j
kfj kqp
kfj kp
n
X
kfj kqp
j=1
p
p
(We use the concavity of the function ⇠ 7! ⇠ p/q .)
n
X
j=1
kfj kqp
⇣ |f | ⌘p
j
kfj kp
1/p
= 1,
1
which proves our claim.
Proof of Theorem 5.2.1. For 2  p < 1 we will deduce our claim from
Corollary 5.2.3. For f, g 2 Lp (µ), with kf k = kgk = 1, we deduce from the
first inequality in Corollary 5.2.3
2p =
1⇥
k(f + g)
2
g)kp + k(f + g) + (f
(f
g)kp ]
kf + gkp + kf
and thus, using the approximation (2p +⇠)1/p = 2+ p1 21
that
kf + gk  2p
kf
gkp )1/p = 2
1 1
2
p
p
kf
p ⇠ +o(⇠),
gkp + o(kf
2
p
+
f +g
2
2 ⌘1/2
p

⇣ f g 2 f +g 2 ⌘1/2
+
C
2
we deduce
gkp ),
which implies our claim.
Now assume that 1 < p < 2. Let f, g 2 SLp (µ) with " = kf
Let C = C(p) be the constant in Corollary 5.2.5.
We deduce from Proposition 5.2.6 and Corollary 5.2.5 that
⇣ f g
C
gkp
p
gkp > 0.
5.2. UNIFORM CONVEXITY AND SMOOTHNESS OF LP

⇣ |f |p + |g|p ⌘1/p
=2
Solving for k(f + g)/2kp leads to
r
f +g
f g
 1
2
C
p
2
1/p
2
p
⇣
p
kf kp + kgkp
=1
125
⌘1/p
= 1.
"2
+ o("2 )
2C
which implies our claim.
Exercises
1. We say that a Banach space X is finitely representable in a Banach
space Y , if for every finite dimensional subspace F of X and every " > 0
there is a finite dimensional subspace E of Y , so that dBM (E, F ) <
1 + ".
Show that if Y is uniform convex and X is finitely representable in Y ,
then X is also uniformly convex.
2.⇤ We say that a Banach space X is crudely finitely representable in
a Banach space Y , if there is a constant K so that for every finite
dimensional subspace F of X there is a finite dimensional subspace E
of Y , so that dBM (E, F ) < 1 + ".
Show that if Y is uniform convex and X is finitely representable in Y ,
then there is an equivalent norm on X which turns X into a uniformly
convex space.