226 CHAPTER 9. BASIC FUNCTION SPACES
Theorem 9.3.3 Let f ,g be measurable and nonnegative functions. Then
∫Ω
f gdµ ≤(∫
Ω
f pdµ
)1/p(∫Ω
gqdµ
)1/q
Proof: If either (∫
Ωf pdµ)1/p or (
∫Ω
gqdµ)1/q is 0, then there is nothing to show be-cause if (
∫Ω
f pdµ)1/p = 0, then∫
Ωf pdµ = 0 and you could let An ≡ {ω : f p (ω)≥ 1/n} .
Then0 =
∫Ω
f pdµ ≥∫
An
f pdµ ≥ (1/n)µ (An)
and so µ (An) = 0. Therefore, {ω : f (ω) ̸= 0}=∪∞n=1An and each of these sets in the union
has measure zero. It follows that {ω : f (ω) ̸= 0} has measure zero. Therefore,∫
Ωf gdµ =
0 and so indeed, there is nothing left to show. The situation is the same if (∫
Ωgqdµ)1/q = 0.
Thus assume both of the factors on the right in the inequality are nonzero. Then let A ≡(∫
Ωf pdµ)1/p ,B≡ (
∫Ω
gqdµ)1/q. Proposition 9.3.2,∫Ω
fA
gB
dµ ≤∫
Ω
f p
Ap pdµ +
∫Ω
gq
Bqqdµ
=1p
∫Ω
f pdµ
Ap +1q
∫Ω
gqdµ
Bq =1p+
1q= 1
Therefore,∫
Ωf gdµ ≤ AB = (
∫Ω
f pdµ)1/p (∫
Ωgqdµ)1/q ■
This makes it easy to prove the Minkowski inequality for the sum of two functions.
Theorem 9.3.4 Let f ,g be two measurable functions with values in F. Then(∫Ω
| f +g|p dµ
)1/p
≤(∫
Ω
| f |p dµ
)1/p
+
(∫Ω
|g|p dµ
)1/p
(9.6)
Proof: First of all,
| f +g|p = | f +g|p−1 | f +g| ≤ | f +g|p−1 (| f |+ |g|)
Recall that p−1 = p/q. Then, using Theorem 9.3.3,∫Ω
| f +g|p dµ ≤∫
Ω
| f +g|p/q | f |dµ +∫
Ω
| f +g|p/q |g|dµ
≤(∫
Ω
| f +g|p dµ
)1/q(∫| f |p dµ
)1/p
+
(∫Ω
| f +g|p dµ
)1/q(∫Ω
|g|p dµ
)1/p
=
(∫Ω
| f +g|p dµ
)1/q((∫
| f |p dµ
)1/p
+
(∫Ω
|g|p dµ
)1/p)
If∫
Ω| f +g|p dµ = 0, then 9.6 is obvious. If
∫Ω| f +g|p dµ = ∞, then
∞≤∫
Ω
2p−1 (| f |p + |g|p)dµ