Problems and Solutions in - Mathematics - University of Hawaii

1.6 2004 April 19 1 REAL ANALYSIS
2. (a) State Egorov’s theorem.
(b) State Fatou’s lemma.
(c) Let {fn} ⊂ L p [0, 1], where 1 ≤ p < ∞. Suppose that fn → f a.e., where f ∈ L p [0, 1]. Prove that
fn − fp → 0 if and only if fnp → fp.
Solution:
(a) See theorem A.8.
(b) See theorem A.6.
(c) (⇒) By the Minkowsky inequality, fnp = fn − f + fp ≤ fn − fp + fp. Similarly, fp ≤
fn − fp + fnp. Together, the two inequalities yield |fnp − fp| ≤ fn − fp. Therefore, fn − fp → 0
implies |fnp − fp| → 0. This proves necessity.
(⇐) I know of three proofs of sufficiency. The second is similar to the first, only much shorter as it exploits the full
power of the general version of Lebesgue’s dominated convergence theorem, whereas the first proof merely relies
on Fatou’s lemma. 29 The third proof uses both Fatou’s lemma and Egoroff’s theorem, so, judging from parts (a)
and (b), this may be closer to what the examiners had in mind. Note that none of the proofs use the assumption that
the measure space is finite, so we may as well work in the more general space L p (X, M, µ).
Both proofs 1 and 2 make use of the following:
Lemma 1.4 If α, β ∈ [0, ∞) and 1 ≤ p < ∞, then (α + β) p ≤ 2 p−1 (α p + β p ).
Proof: When p ≥ 1, φ(x) = x p is convex on [0, ∞). Thus, for all α, β ∈ [0, ∞),
p
α + β α + β
= φ ≤
2
2
1
1
[φ(α) + φ(β)] =
2 2 (αp + β p ).
When α, β ∈ R, the triangle inequality followed by the lemma yields
Proof 1: By (19),
|α − β| p ≤ ||α| + |β|| p ≤ 2 p−1 (|α| p + |β| p ). (19)
|fn − f| p ≤ 2 p−1 (|fn| p + |f| p ).
In particular, fn − f ∈ L p , for each n ∈ N. Moreover, the functions
gn = 2 p−1 (|fn| p + |f| p ) − |fn − f| p . (20)
are non-negative. Now notice that lim gn = 2p |f| p . Applying Fatou’s lemma to (20), then,
2 p |f| p
=
lim gn ≤ lim
2p−1 gn = lim (|fn| p + |f| p ) − |fn − f| p .
Since fnp → fp, this implies
2 p
|f| p ≤ 2 p
|f| p
− lim
|fn − f| p .
Equivalently 0 ≤ − lim |fn − f| p . This proves fn − f p → 0. ⊓⊔
29 Disclaimer: I made up the first proof, so you should check it carefully for yourself and decide whether you believe me.
28