1.8 Some Inequalities in Common Use
73
≤
⎡
⎣
n
k=1
(x k − z k ) 2
⎤
⎦
1
2
⎡
⎣
n
k=1
(x k − y k ) 2
⎤
⎦
1
2
+
⎡
⎣
n
k=1
(z k − y k ) 2
⎤
⎦
1
2
⎡
⎣
n
k=1
(x k − y k ) 2
⎤
⎦
1
2
=
⎧
⎪ ⎨
⎪ ⎩
⎡
⎣
n
k=1
(x k − z k ) 2
⎤
⎦
1
2
+
⎡
⎣
n
k=1
(z k − y k ) 2
⎤
⎦
1
2
⎫
⎪ ⎬
⎪ ⎭
⎡
⎣
n
k=1
(x k − y k ) 2
⎤
⎦
1
2
Thereby there is
n
k=1
(x k − y k )
2
1
2
≤
n
k=1
(x k − z k )
2
1
2
+
n
k=1
(z k − y k )
2
1
2
(1.8.8)
or
d(x, y) ≤ d(x, z) + d(z, y)
(1.8.9)
This shows that d(x, y) satisfies the triangle inequality.
Let C be a complex number field, p ≥ 1, the series {x k } and series {y k } satisfy x k ,
y k ∈ C, with
∞
k=1
|x k |
p
< ∞,
∞
k=1
|y k |
p
< ∞, then there is the following inequality
∞
k=1
|x k + y k |
p
1
p
≤
∞
k=1
|x k |
p
1
p
+
∞
k=1
|y k |
p
1
p
(1.8.10)
Equation (1.8.10) is called the Minkowski inequality. The inequality was found
by Minkowski in 1896.
Proof When p = 1, it can be known from the properties of modulus of a complex
number that for arbitrary k ∈ N, there is |x k + y k | ≤ |x k |+|y k |, therefore Eq. (1.8.10)
holds.
Let p > 1, z k = x k + y k , k ∈ N, then there is
|z k |
p
= |x k + y k ||z k |
p−1
≤ (|x k | + |y k |)|z k |
p−1
(1.8.11)
For arbitrary n ∈ N, sum the two sides, we obtain
n
k=1
|z k ||z k |
p−1
≤
n
k=1
|x k ||z k |
p−1
+
n
k=1
|y k ||z k |
p−1
(1.8.12)
The Hölder inequality is applied to the first term of the above expression, and take
notice that ( p − 1)q = p, we obtain