MEASURE AND INTEGRATION: LECTURE 24
Inequalities
Generalized Minkowski inequality. Let Rn = R� × Rm and z =
(x, y) ∈ Rn . If Rn → C is measurable, then
�
|f (x, y)|p dx : Rm → R = �fy �Lp (R� ) : Rm → R
R�
n
is R ­measurable for 1 ≤ p < ∞.
Assume that
�
�fy �Lp (R� ) dy < ∞.
Rm
�
Then for a.e. x ∈ R , fx (y) : Rm → C is in L1 (Rm ). Let
�
F (x) =
fx (y) dy.
Rm
�
�
Then F (x) : R → C is R ­measurable and we have
�
�F �Lp (R� ) ≤
�fy �Lp (R� ) dy.
Rm
Note this is
�p
�1/p �
��
�1/p
�� � �
�
�
p
�
�
≤
|f (x, y)| dx
dy.
�
m f (x, y) dy
� dx
�
m
�
R
R
R
R
We could replace by X, Y σ­finite measure space and Y = {p1 , . . . , pn },
dy the counting measure and get old Minkowski’s.
Proof. We have
�
|F (x)| ≤
|fx (y)| dy,
Rm
so without loss of generality, assume f
theorem applies; so now let p > 1.
Define g : R� × Rm → R≥0 by
⎧
−1/p�
⎪
⎨f (x, y) �fy �Lp (R� )
g(x, y) = 0
⎪
⎩
∞
Then, for each y,
Date: December 9, 2003.
1
≥ 0. If p = 1, then Fubini’s
if 0 < �fy �p < ∞;
if �fy �p = 0;
if �fy �p = ∞.
2
MEASURE AND INTEGRATION: LECTURE 24
(1) f (x, y) ≤ g(x, y) �fy �p1/p
(2)
�
for a.e. x, and
�gy �Lp (R� ) = �fy �1/p
p .
Then
�
F (x) =
f (x, y) dy
Rm
�
≤
Rm
�
g(x, y) �fy �p1/p dy
�1/p�
��
≤ �gx �Lp (Rm )
�fy �p dy
�
= �gx �Lp (Rm ) · C 1/p ,
�
where C = Rm �fy �p dy.
We now use Fubini’s theorem:
�F (x)�pLp (R� )
≤C
p/p�
� �
�gx �pLp (Rm )
�
dx
�
�
��
p−1
p
=c
g(x, y) dy dx
R�
Rm
�
� ��
p−1
p
=c
g(x, y) dx dy
Rm
R�
�
p−1
=c
�gy �pLp (R� ) dy
m
�R
= cp−1
�fy �Lp (R� ) dy
Rm
�
p−1
p
�fy �Lp (R� ) dy.
=c c=c =
R�
Rm
Thus,
�
�F (x)�Lp (R� ) ≤
Rm
�fy �Lp (R� ) dy
and so
��
�
�
�
�
�
f
(x,
y)
dy
�
m
�
p
R
L (R� )
�
≤
Rm
�fy �Lp (R� ) dy.
�
MEASURE AND INTEGRATION: LECTURE 24
3
Application. Let f ∈ L1 (Rn and g ∈ Lp (Rn ). Then f ∗ g ∈ Lp (Rn )
since
�p
�� ��
�1/p
�
�
�
�
�
n f (y)g(x − y) dy
�
dx
Rn
R
�1/p
� ��
p
≤
|f (y)g(x − y)| dx
dy
Rn
Rn
��
�
|f (y)|
=
�1/p
p
|g(x − y)| dx
Rn
dy
Rn
�
=
Rn
|
f (y)| �g�p dy
= �f �1 �g�p .
Distribution functions. Suppose f : X → [0, ∞] and let µ{f > t} =
µ(x | f (x) > t}.
Theorem 0.1.
�
∞
�
f dµ =
µ{f > t} dt
X
and
�
0
∞
�
p
µ{f > t}tp−1 dt
f dµ = p
X
0
More generally, if ϕ is differentiable, then
�
� ∞
ϕ ◦ f dµ =
µ{f > t}ϕ� (t) dt.
X
0
Proof. We have
�
��
�
|f | dx =
|f (x)|
�
dt
Rn
Rn
dx
0
�
�
=
χ[0,f (x)] (t) dt dx
n
�R �
R
=
χ[0,f (x)] (t) dx dt
Rn
R
Then
�
∞
�
p
µ{|f |p > t} dt
|f | dx =
Rn
�
0 ∞
=
µ{|t| > t1/p } dt
0
�
=p
0
∞
µ{|f | > s}sp−1 ds,
4
MEASURE AND INTEGRATION: LECTURE 24
letting s = t1/p , so sp = t and dt = psp−1 ds.
�
Marcinkiewicz interpolation. Recall the maximal function
�
1
|f (y)| dy.
M f (x) = sup
0<r<∞ λ(B(x, r)) B(x,r)
Note if f ∈ L∞ , then �M f �∞ ≤ �f �∞ . Thus, M maps L∞ into itself:
M : L ∞ → L∞ .
On the other hand, by Hardy­Littlewood, if f ∈ L1 , then
(0.1)
µ{M f > t} ≤
3n
�f �1 ,
t
and M maps L1 to weak L1 .
Using a method called Marcinkiewicz interpolation, we prove the
following.
Theorem 0.2. Let 1 < p < ∞ and f ∈ Lp . Then M f ∈ Lp , and
(0.2)
�M f �p ≤ C(n, p) �f �p ,
where C(n, p) is bounded as p → ∞ and C(n, p) → ∞ as p → 1.
Proof. Observe that M f = M |f |, so assume f ≥ 0. Choose a constant
0 < c < 1 (we will choose the best c later). For t ∈ (0, ∞), write
f = gt + ht , where
�
f (x) f (x) > ct;
gt (x) =
0
f (x) ≤ ct.
So, 0 ≤ ht (x) ≤ ct for every x, and thus ht ∈ L∞ . We have
M f ≤ M gt + M ht ≤ M gt + ct
from (0.2). Thus, M f − ct ≤ M gt , o if M f (x) > t, then (1 − c)t ≤
M gt (x).
Let Et = {f > ct}. Then
λ{M f > t} ≤ λ{M gt > (1 − c)t}
3n
�gt �1
≤
from (0.1)
(1 − c)t
�
3n
=
f dx.
(1 − c)t Et
MEASURE AND INTEGRATION: LECTURE 24
5
Thus,
�
p
�
∞
λ{M f > t}tp−1 dt
0
��
�
� ∞
3n p
p−2
t
f dx dt
≤
1−c 0
Et
�
� �
� f (x)/c
3n p
p−2
=
f (x)
t
dt dx
1 − c Rn
0
�
�p−1
�
1
f (x)
3n p
dx
=
f (x)
1 − c Rn
p−1
c
�
3n p
c1−p
=
·
f (x)p dx
1 − c p − 1 Rn
= C(n, p) �f �pp .
(M f ) dx = p
Rn
Thus,
1/p
3n pc1−p
�M f �p ≤
�f �p .
(1 − c)(p − 1)
�
��
�
→1 as p→∞
�
Choose c = 1/p = (p − 1)/p; this gives the best constant.
�