Linear Algebra, Theory And Applications, 2012a

344 NORMS FOR FINITE DIMENSIONAL VECTOR SPACES
in what follows. Note also that p p ′
= p − 1. Then using the Holder inequality,
||x + y|| p =
≤
=
≤
n∑
|x i + y i | p
i=1
n∑
|x i + y i | p−1 |x i | +
i=1
n∑
|x i + y i | p p ′ |x i | +
i=1
i=1
n∑
|x i + y i | p−1 |y i |
i=1
n∑
|x i + y i | p p ′ |y i |
i=1
(
∑ n
) 1/p
′
⎡( n
) 1/p (
∑
n
) ⎤ 1/p
∑
|x i + y i | p ⎣ |x i | p + |y i | p ⎦
i=1
= ||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 − 1 ′
p
= p 1 ′ p =1. .
It only remains to prove Lemma 14.1.3.
Proofofthelemma:Let p ′ = q to save on notation and consider the following picture:
b
x
x = t p−1
t = x q−1
i=1
a
t
ab ≤
∫ a
0
t p−1 dt +
Note equality occurs when a p = b q .
Alternate proof of the lemma: Let
∫ b
0
x q−1 dx = ap
p + bq
q .
f (t) ≡ 1 p (at)p + 1 ( ) q b
,t>0
q t
You see right away it is decreasing for a while, having an asymptote at t = 0 and then
reaches a minimum and increases from then on. Take its derivative.
( ) q−1 ( )
f ′ (t) =(at) p−1 b −b
a +
t t 2
Set it equal to 0. This happens when
t p+q = bq
a p . (14.5)