10.5. TIETZE EXTENSION THEOREM 269

Proposition 10.5.8 For x,y ∈ Cn,∑ni=1 |xi| |yi| ≤ (∑n

i=1 |xi|p)1/p(

∑ni=1 |yi|p

′)1/p′

.

The proof will depend on the following lemma.

Lemma 10.5.9 If a,b≥ 0 and p′ is defined by 1p +

1p′ = 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= (∑ni=1 |xi|p)1/p and B=

(∑

ni=1 |yi|p

′)1/p′

.Then using Lemma 10.5.9,

n

∑i=1

|xi|A|yi|B≤

n

∑i=1

[1p

(|xi|A

)p

+1p′

(|yi|B

)p′]

=1p

1Ap

n

∑i=1|xi|p +

1p′

1Bp

n

∑i=1|yi|p

′=

1p+

1p′

= 1

and so ∑ni=1 |xi| |yi| ≤ AB = (∑n

i=1 |xi|p)1/p(

∑ni=1 |yi|p

′)1/p′

■

Theorem 10.5.10 The p norms do indeed satisfy the axioms of a norm.

Proof: It is obvious that ||·||p does indeed satisfy most of the norm axioms. The onlyone that is not clear is the triangle inequality. To save notation write ||·|| in place of ||·||p inwhat follows. Note also that p

p′ = p−1. Then using the Holder inequality,

||x+y||p =n

∑i=1|xi + yi|p ≤

n

∑i=1|xi + yi|p−1 |xi|+

n

∑i=1|xi + yi|p−1 |yi|

=n

∑i=1|xi + yi|

pp′ |xi|+

n

∑i=1|xi + yi|

pp′ |yi|

≤

(n

∑i=1|xi + yi|p

)1/p′( n

∑i=1|xi|p

)1/p

+

(n

∑i=1|yi|p

)1/p

= ||x+y||p/p′(||x||p + ||y||p

)so dividing by ||x+y||p/p′ , it follows ||x+y||p ||x+y||−p/p′ = ||x+y|| ≤ ||x||p +||y||p

(p− p

p′ = p(

1− 1p′

)= p 1

p = 1). ■

It only remains to prove Lemma 10.5.9.Proof of the lemma: Let p′ = q to save on notation and consider the following picture:

b

a

x

t

x = t p−1

t = xq−1