13.1. THE p NORMS 331

For example, define

||x||1 ≡n∑

i=1

|xi| ,

||x||∞ ≡ max {|xi| , i = 1, · · · , n} ,

or

||x||2 =

(n∑

i=1

|xi|2)1/2

and all three are equivalent norms on∏n

i=1Xi.

13.1 The p Norms

In addition to ||·||1 and ||·||∞ mentioned above, it is common to consider the so called pnorms for x ∈ Cn.

Definition 13.1.1 Let x ∈ Cn. Then define for p ≥ 1,

||x||p ≡

(n∑

i=1

|xi|p)1/p

The following inequality is called Holder’s inequality.

Proposition 13.1.2 For x,y ∈ Cn,

n∑i=1

|xi| |yi| ≤

(n∑

i=1

|xi|p)1/p( n∑

i=1

|yi|p′

)1/p′

The proof will depend on the following lemma.

Lemma 13.1.3 If a, b ≥ 0 and p′ is defined by 1p + 1

p′ = 1, then

ab ≤ ap

p+bp

′

p′.

Proof of the Proposition: If x or y equals the zero vector there is nothing to

prove. Therefore, assume they are both nonzero. Let A = (∑n

i=1 |xi|p)1/p

and B =(∑ni=1 |yi|

p′)1/p′

. Then using Lemma 13.1.3,

n∑i=1

|xi|A

|yi|B

≤n∑

i=1

[1

p

(|xi|A

)p

+1

p′

(|yi|B

)p′]

=1

p

1

Ap

n∑i=1

|xi|p +1

p′1

Bp

n∑i=1

|yi|p′=

1

p+

1

p′= 1

and son∑

i=1

|xi| |yi| ≤ AB =

(n∑

i=1

|xi|p)1/p( n∑

i=1

|yi|p′

)1/p′

. ■

Theorem 13.1.4 The p norms do indeed satisfy the axioms of a norm.