Problems and Solutions in - Mathematics - University of Hawaii

1.6 2004 April 19 1 REAL ANALYSIS
3. (a) Let S = [0, 1] and let {fn} ⊂ Lp (S), where 1 < p < ∞. Suppose that fn → f a.e. on S, where f ∈ Lp (S). If
there is a constant M such that fnp ≤ M for all n, prove that for each g ∈ Lq (S), 1 1
p + q = 1, we have
lim
n→∞
fng =
S
fg.
S
(b) Show by means of an example that this result is false for p = 1.
Solution:
(a) Since g ∈ Lq (S), for all ɛ > 0 there exists δ > 0 such that if A is a measurable set with µA < δ then
|g| q dµ < ɛ.
A
Let B0 ⊂ S denote the set on which fn does not converge to f. Let B ⊂ S be such that fn → f uniformly in
S \ B, and such that µB < δ. (Such a set exists by Egoroff’s theorem since µS < ∞.) Now throw the set B0 in
with B (i.e. redefine B to be B ∪ B0). Then,
Dn := |fng − fg| dµ = |fn − f||g| dµ
S
S
=
|fn − f||g| dµ +
S\B
|fn − f||g| dµ
B
≤ (fn − f)χ S\Bpgq + fn − fpgχBq.
The final inequality holds because fn, f ∈ L p implies |fn − f| ∈ L p and by Hölder’s inequality. Now, by
Minkowski’s inequality, fn − fp ≤ fnp + fp, so
Dn ≤ sup{|fn(x) − f(x)| : x /∈ B}µ(S \ B) 1/p
gq + (fnp + fp) |g|
B
q 1/q dµ .
Now we are free to choose the δ > 0 above so that
|g|
B
q dµ
1/q
<
ɛ
2(M + fp)
holds whenever µB < δ. Also, fn → f uniformly on S \ B, so let N be such that
ɛ
sup{|fN(x) − f(x)| : x /∈ B} <
2µ(S \ B) 1/p .
gq
Then DN < ɛ/2 + ɛ/2. ⊓⊔
(b) The result is not true for p = 1 as the following example shows: Let fn = nχ [0,1/n], n = 1, 2, . . . . First note
that fn → 0 a.e. For if x ∈ (0, 1], there exists N > 0 such that 1/N < x and thus fn(x) = 0 for all n ≥ N.
Therefore, {x ∈ [0, 1] : fn(x) 0} = {0}. Now, if g is the constant function g ≡ 1, then fng dµ = 1 for all
n = 1, 2, . . . . Therefore, fng dµ → 1, while fg dµ = 0g dµ = 0. Finally, note that fn1 = nµ([0, 1
n ]) = 1
for n = 1, 2, . . . , so {fn} satisfies the hypothesis fn1 ≤ some constant M. ⊓⊔
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