258 CHAPTER 10. REGULAR MEASURES

them as ∥ f∥p and ∥g∥q respectively, consider the above inequality in∫ f∥ f∥p

g∥g∥q

dµ .This is Holder’s inequality. When does equality hold? See previous problem todetermine this.

7. ↑ Lp (Ω,µ) consists of those measurable functions f such that | f |p is integrable.Show using Holder’s inequality that if µ (Ω) < ∞ that if 1 < p < q, then Lq (Ω) ⊆Lp (Ω) . Give an example which shows that sometimes the opposite inclusion holdsif µ (Ω) = ∞.

8. ↑ Show that if ∥ f∥p ,∥g∥p < ∞, then ∥ f +g∥p ≤ ∥ f∥p +∥g∥p . Hint: Show∫| f +g|p dµ ≤

∫| f +g|p/q | f |dµ +

∫| f +g|p/q |g|dµ

Now apply Holder’s inequality.

9. ↑If we regard f = g when the two differ only on a set of measure zero, explain why∥ f∥p is a norm. The collection of these functions with this convention for ∥·∥p iscalled Lp (Ω,µ)

10. ↑Now suppose { fn} is a Cauchy sequence in Lp. That is: For every ε > 0 thereexists nε such that if m,n > nε then ∥ fm− fn∥p < ε . Show there exists f ∈ Lp

such that limn→∞ ∥ f − fn∥p = 0. Hint: Recall from Theorem 3.2.2 on Page 65 thatwe only need to obtain a subsequence which converges. Here is how you can getsuch a subsequence. Pick a subsequence

{fnk

}denoted as {gk} for short such that

∥gk−gk+1∥pp < 4−k. Now let Ek ≡

{ω : |gk (ω)−gk+1 (ω)|p > 2−k

}. Explain why

4−k ≥∫

Ek

|gk (ω)−gk+1 (ω)|p dµ ≥ 2−kµ (Ek) , µ (Ek)< 2−k

Now recall the Borel Cantelli lemma, Lemma 8.2.5, there is a set of measure zeroN such that if ω /∈ N then ω is in only finitely many of the Ek. Thus for suchω {gk (ω)} is a Cauchy sequence which converges to some f (ω) . Now explainusing Fatou’s lemma how ∥gk− f∥p → 0 and that f ∈ Lp. First argue that ∥ f∥p ≤liminfk→∞ ∥gk∥p < ∞ because the ∥gk∥p are bounded due to the fact that {gk} is aCauchy sequence. Next pick m such that if k, l > m, then ∥gk−gl∥< ε. Use Fatou’slemma again to obtain that for k > m,∥gk− f∥p < ε which was to be shown.

11. ↑Show that the simple functions are dense in Lp. Hint: Consider positive and nega-tive parts of f ∈ Lp and use Theorem 8.1.6 about pointwise limits of simple functionsalong with the dominated convergence theorem or some such thing.

12. ↑In case the measure space is (X ,F ,µ) where µ is regular and Borel and X is aPolish space, show that Cc (X) is dense in Lp (X).

13. ↑In the situation of (Rn,Fp,mp) , Lebesgue measure, define fy (x) ≡ f (x−y) .Show limy→0 ∥ fy− f∥p = 0. This is very important and is called continuity of trans-lation in Lp. Hint: Let g ∈Cc (Rn) . Then

∥ fy− f∥p ≤ ∥ fy−gy∥p +∥gy−g∥p +∥g− f∥p

Now from the above problem, pick g∈Cc (Rn) such that ∥ f −g∥p = ∥ fy−gy∥p < ε.This comes from a change of variables exercise of using translation invariance of themeasure. Now if y is small enough, the right side is no more than 3ε .