9.3. THE Lp SPACES 225

Each subsequence{

f( j,r)}∞

j=1 is a subsequence of the one above it{

f( j,r−1)}∞

j=1 and con-

verges at d j for j ≤ r. Consider the subsequence{

f(r,r)}∞

r=1 the diagonal subsequence.Then

{f(r,r)

}∞

r= j is a subsequence of{

f(i, j)}∞

i=1 and so{

f(r,r) (di)}∞

r= j converges for each

i≤ j. Since j is arbitrary, this shows that{

f(r,r)}∞

r=1converges at every point of D as r→∞.From now on, denote this subsequence of the original sequence as {gk}∞

k=1. It has theproperty that it converges at every point of D.

Claim: {gk}∞

k=1 converges at every x ∈ K.Proof of claim: Let ε > 0 be given. Let δ go with ε/4 in the definition of equiconti-

nuity. Then pick d ∈ D such that |d−x|< δ . Then

|gk (x)−gl (x)| ≤ |gk (x)−gk (d)|+ |gk (d)−gl (d)|+ |gl (d)−gl (x)|

<ε

4+ |gk (d)−gl (d)|+

ε

4

There exists N such that if k, l ≥ N, then |gk (d)−gl (d)|< ε

3 . Thus, if k, l ≥ N,

|gk (x)−gl (x)|<ε

4+

ε

3+

ε

4< ε

which shows that, since ε is arbitrary, {gk (x)}∞

k=1 is a Cauchy sequence and so it convergesto some g(x). This shows the claim. Now from the Lemma 9.2.3, this g is in C (K,Rn) and∥gk−g∥

∞→ 0. ■

9.3 The Lp SpacesLet (Ω,F ,µ) be a measure space. Recall that the space L1 (Ω) consists of functions f :Ω→ F such that f is measurable and

∫Ω| f (ω)|dµ < ∞. The Lp spaces are defined as

follows.

Definition 9.3.1 Let (Ω,F,µ) be a measure space. Then Lp (Ω) consists of thosemeasurable functions f such that

∫Ω| f |p dµ < ∞. Here it is assumed that p > 1. Also

define the conjugate exponent q as satisfying 1p +

1q = 1 In case p = 1, we let q = ∞ and

give a special meaning to L∞ (Ω) discussed later.

Here we assume p > 1. There is an essential inequality which makes possible the studyof Lp (Ω) .

Proposition 9.3.2 Let 0≤ a,b. Then for p > 1

ap

p+

bq

q≥ ab (9.5)

Proof: Let b ≥ 0 be fixed and let f (a) ≡ ap

p + bq

q − ab. Then f (0) = bq

q ≥ 0 andf ′ (a) = ap−1 − b. If b = 0 the desired inequality is obvious. If b > 0, then f ′ (a) < 0for a close to 0 and f ′ (a) > 0 if ap−1 > b. Thus f has a minimum at the point whereap−1 = b. But p− 1 = p/q and so, at this point ap = bq. Therefore, at this point, f (a) =ap

p + ap

q − aap−1 = ap− ap = 0. Therefore, f (a) ≥ 0 for all a ≥ 0 and it equals 0 exactlywhen ap = bq. ■

This implies the following major result, Holder’s inequality.