Measure, Integration & Real Analysis, 2021a

Section 7A L p (μ) 197
Proof Suppose 1 < p < ∞, leaving the cases p = 1 and p = ∞ as exercises for
the reader.
First consider the special case where ‖ f ‖ p = ‖h‖ p ′ = 1. Young’s inequality (7.8)
tells us that
| f (x)h(x)| ≤ | f (x)|p + |h(x)|p′
p p ′
for all x ∈ X. Integrating both sides of the inequality above with respect to μ shows
that ‖ fh‖ 1 ≤ 1 = ‖ f ‖ p ‖h‖ p ′, completing the proof in this special case.
If ‖ f ‖ p = 0 or ‖h‖ p ′ = 0, then
Hölder’s inequality was proved in
‖ fh‖ 1 = 0 and the desired inequality
holds. Similarly, if ‖ f ‖ p = ∞ or
1889 by Otto Hölder (1859–1937).
‖h‖ p ′ = ∞, then the desired inequality clearly holds. Thus we assume that
0 < ‖ f ‖ p < ∞ and 0 < ‖h‖ p ′ < ∞.
Now define S-measurable functions f 1 , h 1 : X → F by
f 1 =
f and h
‖ f ‖ 1 = h .
p ‖h‖ p ′
Then ‖ f 1 ‖ p = 1 and ‖h 1 ‖ p ′ = 1. By the result for our special case, we have
‖ f 1 h 1 ‖ 1 ≤ 1, which implies that ‖ fh‖ 1 ≤‖f ‖ p ‖h‖ p ′.
The next result gives a key containment among Lebesgue spaces with respect to a
finite measure. Note the crucial role that Hölder’s inequality plays in the proof.
7.10 L q (μ) ⊂L p (μ) if p < q and μ(X) < ∞
Suppose (X, S, μ) is a finite measure space and 0 < p < q < ∞. Then
‖ f ‖ p ≤ μ(X) (q−p)/(pq) ‖ f ‖ q
for all f ∈L q (μ). Furthermore, L q (μ) ⊂L p (μ).
Proof Fix f ∈L q (μ). Let r = q p
. Thus r > 1. A short calculation shows that
r ′ =
q
q−p
. Now Hölder’s inequality (7.9) with p replaced by r and f replaced by | f |p
and h replaced by the constant function 1 gives
∫ (∫ ) 1/r (∫ ) 1/r
| f | p dμ ≤ (| f | p ) r ′
dμ 1 r′ dμ
= μ(X) (q−p)/q( ∫
| f | q dμ) p/q.
Now raise both sides of the inequality above to the power 1 p , getting
(∫ ) 1/p ( ∫
| f | p dμ ≤ μ(X)
(q−p)/(pq) 1/q,
| f | dμ) q
which is the desired inequality.
The inequality above shows that f ∈L p (μ). Thus L q (μ) ⊂L p (μ).
Measure, Integration & Real Analysis, by Sheldon Axler