5.5. UNCONDITIONALITY OF HAAR BASIS
5.5
137
The Haar basis is Unconditional in Lp [0, 1], 1 <
p<1
Theorem 5.5.1. (Unconditionality of the Haar basis in Lp )
(p)
Let 1 < p < 1. Then (ht ) is an unconditional basis of Lp [0, 1]. More
precisely, for any two families (at )t2T and (bt )t2T in c00 (T ) with |at |  |bt |,
for all t 2 T , it follows that
X
(5.16)
(p)
 (p⇤
a t ht
t2T
where
⇤
p = max p,
p
p
1
=
1)
X
(p)
bt ht
,
t2T
(
p
p/(p
1)
if p 2
if p  2
We will prove the theorem for 2 < p < 1. For p = 2 it is clear since
is orthonormal and for 1 < p < 2 it follows from Propostion 3.4.5 by
duality (note that p⇤ = q ⇤ if p1 + 1q = 1).
We first need the following Lemma
(2)
(ht )
Lemma 5.5.2. Let 2 < p < 1 and define
(5.17) v : C ⇥ C ! [0, 1),
(5.18) u : C ⇥ C ! [0, 1),
(x, y) 7! |y|p
(p
1)p |x|p , and
(x, y) 7! ↵p |x| + |y|
⇣
1 ⌘p 1
with ↵p = p 1
.
p
p 1
|y|
(p
1)|x|
Then it follows for x, y, a, b 2 C, with |a|  |b|
(5.19)
(5.20)
v(x, y)  u(x, y)
u( x, y) = u(x, y)
(5.21)
u(0, 0) = 0
(5.22)
u(x + a, y + b) + u(x
a, y
b)  2u(x, y)
Proof. Let x, y, a, b 2 C, |a|  |b| be given. (5.20) and (5.21) are trivially satisfied. Since u and v are both p-homogeneous (i.e. u(↵ · x, ↵y) = |↵|p u(x, y)
for ↵ 2 C) we can assume that |x| + |y| = 1 in order to show (5.19). Thus
the inequality (put s = |x|) reduces to show
(5.23) F (s) = ↵p (1 ps) (1 s)p +(p 1)p sp
0 for 0  s  1 and 2  p.
138
CHAPTER 5. LP -SPACES
In order to verify (5.23), first show that F (0) > 0. Indeed, by concavity of
ln x it follows that
ln p = ln (p
1) + 1 < ln(p
1) +
1
p
1
,
and, thus,
ln(p
1) + 1 = ln(p
1) +
1
p
1
p
p
+
2
p
> ln p +
1
p
2
p 2
> ln p +
.
1
p
Integrating both sides of
ln(x
x
1) + 1 > ln x +
2
x
.
from x = 2 to p > 2, implies that
(p
1) ln(p
1) > (p
2) ln p
and, thus,
1)p
(p
which yields
⇣
↵p = p 1
1 ⌘p
p
1
1
> pp
=
2
,
1)p
(p
pp
1
>1
2
and thus the claim that F (0) > 0.
Secondly we claim that F (1) > 0. Indeed,
F (1) =
(p
1)p
pp
2
+ (p
p
1) = (p
1)
p 2
"
(p
1)
2
1
pp
#
> 0,
Thirdly, we compute the first and second derivative of F and get
F 0 (s) =
00
F (s) =
↵p p + p(1
p(p
1)(1
s)p
s)
1
p 2
+ (p
1)p psp
+ (p
p+1
1)
1
ps
, and
p 2
and deduce that F ( p1 ) = F 0 ( p1 ) = 0, F 00 ( p1 ) > 0 and that F 00 (s) vanishes for
exactly one value of s (because it is the di↵erence of an increasing functions
and a decreasing function). Thus, F (s) cannot have more points at which it
vanishes and it follows that F (s) 0 for all s 2 [0, 1] and we deduce (5.19).
Finally we need to show (5.22). We can (by density argument) assume
that x and a as well as y and b are linear independent as two-dimensional
5.5. UNCONDITIONALITY OF HAAR BASIS
139
vectors over R. This implies that |x + ta| and |y + tb| can never vanish, and,
thus, that the function
G : R ! R,
t 7! t = u(x + ta, y + tb),
is infinitely often di↵erentiable.
We compute the second derivative of G at 0, getting
"
G00 (0) = ↵p
p(p
1) |a|2
|b|2 |x| + |y|
p(p
2) |b|2
<(h
1)(p
⇣
p(p
y
, bi2 )|y|
|y|
p 2
1
|x| + |y|
p 1
⌘2
x
y
2) <(h , ai) + <(h , bi) |x| |x| + |y|
|x|
|y|
p 3
#
.
Inspecting each term we deduce (recall that |a|
|b|) from the Cauchy
00
inequality that G (0) < 0. Since for t 6= 0 it follows that G00 (t) = G̃00 (0)
where
G̃(s) : R ! R, s 7! u(x
tb +sb),
|+
{zta} +sa, |y +
{z }
x̃
we deduce that
yields
G00 (t)
ỹ
 0 for all t 2 R. Thus, G is a concave function which
1
[u(x + a, y + b) + u(x
2
which proves (5.22).
a, y
1
b)] = [G(1) + G( 1)]  G(0) = u(x, y),
2
Proof of Theorem 5.5.1. Assume that h̃n is normalized in L1 so that hn =
(p)
h̃n /kh̃n kp is a linear reordering of (hP
with the
t )t 2 T which is compatible
Pn
n
order on T . For n 2 N let fn =
i=1 ai h̃i and gn =
i=1 bi h̃i , where
(ai )ni=1 , (bi )ni=1 in R, with |aj | |bj |, for j = 1, 2 . . . n, we need to show that
kgn kp  (1 p⇤ )kfn k. The fact that we are considering the normalization in
L1 [0, 1] instead of the normalization in Lp [0, 1] (i.e. h̃n instead of hn ) will
not e↵ect the outcome. We deduce from (5.19) that
Z 1
Z 1
p
p
p
kgn k
(p 1) kfn k =
v(fn (t), gn (t)) dt 
u(fn (t), gn (t)) dt.
0
A+
0
Let A = supp(h̃n ),
= A \ {h̃n > 0} and A = A \ {h̃n < 0}. Since fn
and gn 1 are constant on A we deduce
Z 1
u(fn (t), gn (t)) dt
0
1
140
CHAPTER 5. LP -SPACES
=
Z
u(fn
+
+
=

Z
1 (t), gn 1 (t)) dt
[0,1]\A
Z
ZA
+
u(fn
1 (t)
+ an , gn
1 (t)
+ bn ) dt
u(fn
1 (t)
an , gn
1 (t)
bn ) dt
A
u(fn
1 (t), gn 1 (t)) dt
u(fn
1 (t)
u(fn
1 (t), gn
[0,1]\A
1
+
2
Z
Z
+ an , gn
A
[0,1]\A
[By (5.22)]
Z 1
=
u(fn
1 (t)
+ bn ) + u(fn 1 (t) an , gn
Z
u(fn 1 (t), gn 1 (t)) dt
1 (t)) dt +
1 (t)
bn ) dt
A
1 (t), gn 1 (t)) dt
0
Iterating this argument yields
Z 1
Z
u(fn (t), gn (t)) dt 
0
1
u(f1 (t), g1 (t)) dt
0
= u(a1 , b1 )
1
= u(a1 , b1 ) + u( a1 , b1 ) [By (5.20)]
2
 u(0, 0) = 0 [By (5.21) and (5.22)],
which implies our claim that kgn k  (p
1)kfn k.
From the unconditionality of the Haar basis and Khintchine’s Theorem
we now can deduce the following equivalent representation of the norm on
Lp .
Theorem 5.5.3 (The square-function norm). Let 1 < p < 1 and let (fn )
be an unconditional basic sequence in Lp [0, 1]. For example (fn ) could be
a linear ordering of the Haar basis. Then there is a constant C
1, only
depending on the unconditionality constant of (fi ) and the constants
A
P1 p and
Bp in Khintchine’s Inequality (Lemma 5.3.3) so that for any g = i=1 ai fi 2
span(fi : i 2 N) it follows that
1
1 X ⇣ 2 2 ⌘1/2
|ai | |fi |
C
i=1
p
 kgkp  C
1 ⇣
X
i=1
|ai |2 |fi |2
⌘1/2
p
,
5.5. UNCONDITIONALITY OF HAAR BASIS
141
which means that k · kp is on span(fi : i 2 N) equivalent to the norm
1 ⇣
X
|||f ||| =
i=1
|ai |2 |fi |2
⌘1/2
p
1
X
=
i=1
|ai |2 |fi |2
1/2
p/2
.
Proof. For two positive numbers A and B and c > 0 we write: A ⇠c B if
1
c A  B  cA. Let Kp be the Khintchine constant for Lp , i.e the smallest
number so that for the Rademacher sequence (rn )
1
X
a i ri
⇠Kp
p
i=1
1
⇣X
i=1
|ai |2
⌘1/2
for (ai ) ⇢ K,
and let bu be the unconditionality constant of (fi ), i.e.
1
X
i ai fi
⇠ bu
p
i=1
1
X
ai fi
i=1
p
for (ai ) ⇢ K and ( i ) ⇢ {±1}.
We consider Lp [0, 1] in a natural way as subspace of Lp [0, 1]2 , with
f˜(s, t) := f (s) for f 2 Lp [0, 1]. Then let rn (t) = rn (s, t) be the n-th
Rademacher function action on the second coordinate, i.e
rn (s, t) = sign(sin(2n ⇡t)), (s, t) 2 [0, 1]2 .
It follows from the bu -unconditionality for any (aj )m
j=1 ⇢ K, that
m
X
p
aj fj (·)
p
j=1
⇠
bpu
=
Z
m
X
aj fj (·)rj (t)
j=1
1
m
X
0
j=1
aj fj (·)
p
p
⇠bpu
=
=
Z
Z
Z
0
0
0
1Z 1
0
1
0
1Z 1
m
X
j=1
p
aj fj (s)rj (t)
j=1
and integrating over all t 2 [0, 1] implies
m
X
p
m
X
aj fj (s)rj (t)
j=1
m
X
aj fj (s)rj (t)
j=1
aj fj (s)rj (·)
p
p
!p
!p
ds
⌘p
ds for all t 2 [0, 1],
ds dt
dt ds(By Theorem of Fubini)
142
CHAPTER 5. LP -SPACES
⇠Kpp
Z
0
m
1⇣X
j=1
|aj fj (s)|2
⌘p/2
which proves our claim using C = Kp bu .
ds =
m
⇣X
j=1
|aj fj |2
⌘1/2
p
,