332 CHAPTER 13. NORMS

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 ||·||pin what 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− pp′ = p

(1− 1

p′

)= p 1

p = 1.). ■

It only remains to prove Lemma 13.1.3.Proof of the lemma: Let p′ = q to save on notation and consider the following picture:

b

a

x

t

x = tp−1

t = xq−1

ab ≤∫ a

0

tp−1dt+

∫ b

0

xq−1dx =ap

p+bq

q.

Note equality occurs when ap = bq.Alternate proof of the lemma: For a, b ≥ 0, let b be fixed and

f (a) ≡ 1

pap +

1

qbq − ab, t > 0

If b = 0, it is clear that f (a) ≥ 0 for all a. Then assume b > 0. It is clear since p > 1 thatlima→∞ f (a) = ∞.

f ′ (a) = ap−1 − b

This is negative for small a and then eventually is positive. Consider the minimum value off which must occur at a > 0 thanks to the observation that the function is initially strictlydecreasing. At this point,

0 = f ′ (a) = ap−1 − b = a(p/q) − b