Measure, Integration & Real Analysis, 2021a

198 Chapter 7 L p Spaces
7.11 Example L p (E)
We adopt the common convention that if E is a Borel (or Lebesgue measurable)
subset of R and 0 < p ≤ ∞, then L p (E) means L p (λ E ), where λ E denotes
Lebesgue measure λ restricted to the Borel (or Lebesgue measurable) subsets of R
that are contained in E.
With this convention, 7.10 implies that
if 0 < p < q < ∞, then L q ([0, 1]) ⊂L p ([0, 1]) and ‖ f ‖ p ≤‖f ‖ q
for f ∈L q ([0, 1]). See Exercises 12 and 13 in this section for related results.
Minkowski’s Inequality
The next result is used as a tool to prove Minkowski’s inequality (7.14). Once again,
note the crucial role that Hölder’s inequality plays in the proof.
7.12 formula for ‖ f ‖ p
Suppose (X, S, μ) is a measure space, 1 ≤ p < ∞, and f ∈L p (μ). Then
{∣∫
∣∣
‖ f ‖ p = sup
}
fhdμ∣ : h ∈L p′ (μ) and ‖h‖ p ′ ≤ 1 .
Proof If ‖ f ‖ p = 0, then both sides of the equation in the conclusion of this result
equal 0. Thus we assume that ‖ f ‖ p ̸= 0.
Hölder’s inequality (7.9) implies that if h ∈L p′ (μ) and ‖h‖ p ′ ≤ 1, then
∫
∫
∣ fhdμ∣ ≤ | fh| dμ ≤‖f ‖ p ‖h‖ p ′ ≤‖f ‖ p .
Thus sup {∣ ∣ ∫ fhdμ ∣ : h ∈L p′ (μ) and ‖h‖ p ′ ≤ 1 } ≤‖f ‖ p .
To prove the inequality in the other direction, define h : X → F by
f (x) | f (x)|p−2
h(x) =
‖ f ‖ p/p′
p
(set h(x) =0 when f (x) =0).
Then ∫ fhdμ = ‖ f ‖ p and ‖h‖ p ′ = 1, as you should verify (use p − p p ′ = 1). Thus
‖ f ‖ p ≤ sup {∣ ∣ ∫ fhdμ ∣ ∣ : h ∈L p′ (μ) and ‖h‖ p ′ ≤ 1 } , as desired.
7.13 Example a point with infinite measure
Suppose X is a set with exactly one element b and μ is the measure such that
μ(∅) =0 and μ({b}) =∞. Then L 1 (μ) consists only of the 0 function. Thus if
p = ∞ and f is the function whose value at b equals 1, then ‖ f ‖ ∞ = 1 but the right
side of the equation in 7.12 equals 0. Thus 7.12 can fail when p = ∞.
Example 7.13 shows that we cannot take p = ∞ in 7.12. However, if μ is a
σ-finite measure, then 7.12 holds even when p = ∞ (see Exercise 9).
Measure, Integration & Real Analysis, by Sheldon Axler