74
1 Preliminaries
n
k=1
|x k ||z k |
p−1
≤
n
k=1
|x k |
p
1
p
n
k=1
|z k |
( p−1)q
1
q
=
n
k=1
|x k |
p
1
p
n
k=1
|z k |
p
1
q
Similarly, we obtain
n
k=1
|y k ||z k |
p−1
≤
n
k=1
|y k |
p
1
p
n
k=1
|z k |
p
1
q
Substituting the above two expressions into Eq. (1.8.12), we obtain
n
k=1
|z k |
p
≤
⎡
⎣
n
k=1
|x k |
p
1
p
+
n
k=1
|y k |
p
1
p
⎤
⎦
n
k=1
|z k |
p
1
q
(1.8.13)
The sum terms on the right side of the above expression are positive, both ends
are divided by
n
k=1
|z k |
p
1
q
, and consider 1 −
1
q
=
1
p
, we obtain
n
k=1
|z k |
p
1
p
=
n
k=1
|x k + y k |
p
1
p
≤
n
k=1
|x k |
p
1
p
+
n
k=1
|y k |
p
1
p
(1.8.14)
Let n → ∞, then Eq. (1.8.10) is obtained. Quod erat demonstrandum.
Similarly, the Minkowski inequality can also be written as integral form
t 1
t 0
|z(t)| p dt
1
p =
t 1
t 0
|x(t) + y(t)| p dt
1
p ≤
t 1
t 0
|x(t)| p dt
1
p +
t 1
t 0
|y(t)| p dt
1
p
(1.8.15)
For the inequality (1.8.14) and the inequality (1.8.15), let p = 2, then there is
n
k=1
|z k |
2
1
2
=
n
k=1
|x k + y k |
2
1
2
≤
n
k=1
|x k |
2
1
2
+
n
k=1
|y k |
2
1
2
(1.8.16)
t 1
t 0
|z(t)| 2 dt
1
2 =
t 1
t 0
|x(t) + y(t)| 2 dt
1
2 ≤
t 1
t 0
|x(t)| 2 dt
1
2 +
t 1
t 0
|y(t)| 2 dt
1
2 (1.8.17)
Both the inequality (1.8.16) and the inequality (1.8.17) are called the Cauchy
inequality.