400 CHAPTER 15. THE Lp SPACES

area between the t axis and the curve is at least as large as ab. Using beginning calculus,this is equivalent to the following inequality.

ab≤∫ a

0t p−1dt +

∫ b

0xq−1dx =

ap

p+

bq

q.

The above picture represents the situation which occurs when p > 2 because the graph ofthe function is concave up. If 2 ≥ p > 1 the graph would be concave down or a straightline. You should verify that the same argument holds in these cases just as well. In fact,the only thing which matters in the above inequality is that the function x = t p−1 be strictlyincreasing.

Note equality occurs when ap = bq.Here is an alternate proof.

Lemma 15.1.4 For a,b≥ 0,

ab≤ ap

p+

bq

qand equality occurs when if and only if ap = bq.

Proof: If b = 0, the inequality is obvious. Fix b > 0 and consider

f (a)≡ ap

p+

bq

q−ab.

Then f ′ (a) = ap−1 − b. This is negative when a < b1/(p−1) and is positive when a >b1/(p−1). Therefore, f has a minimum when a = b1/(p−1). In other words, when ap =bp/(p−1) = bq since 1/p+1/q = 1. Thus the minimum value of f is

bq

p+

bq

q−b1/(p−1)b = bq−bq = 0.

It follows f ≥ 0 and this yields the desired inequality.Proof of Holder’s inequality: If either

∫| f |pdµ or

∫|g|pdµ equals ∞, the inequality

15.1.1 is obviously valid because ∞ ≥ anything. If either∫| f |pdµ or

∫|g|pdµ equals 0,

then f = 0 a.e. or that g = 0 a.e. and so in this case the left side of the inequality equals0 and so the inequality is therefore true. Therefore assume both

∫| f |pdµ and

∫|g|pdµ are

less than ∞ and not equal to 0. Let(∫| f |pdµ

)1/p

= I ( f )

and let (∫|g|pdµ)1/q = I (g). Then using the lemma,∫ | f |

I ( f )|g|

I (g)dµ ≤ 1

p

∫ | f |p

I ( f )p dµ +1q

∫ |g|q

I (g)q dµ = 1.

Hence, ∫| f | |g| dµ ≤ I ( f ) I (g) =

(∫| f |pdµ

)1/p(∫|g|qdµ

)1/q

.

This proves Holder’s inequality.The following lemma will be needed.