8.3. INNER PRODUCT AND NORMED LINEAR SPACES 187
Then using Lemma 8.3.10,
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
n
∑i=1|xi| |yi| ≤ AB =
(n
∑i=1|xi|p
)1/p( n
∑i=1|yi|p
′)1/p′
.
Theorem 8.3.11 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 8.3.10.Proof of the lemma: Let p′ = q to save on notation and consider the following picture: