204 CHAPTER 9. WEIERSTRASS APPROXIMATION THEOREM

Definition 9.3.1 Let p > 1. Then f ∈C1/p ([0,1]) means that f ∈C ([0,1]) and also

ρ p ( f )≡ sup

{| f (x)− f (y)||x− y|1/p : x,y ∈ X , x ̸= y

}< ∞

Then the norm is defined as ∥ f∥C([0,1])+ρ p ( f )≡ ∥ f∥1/p.

It is an exercise to verify that C1/p ([0,1]) is a complete normed linear space.Let p > 1. Then C1/p ([0,1]) is not separable. Define uncountably many functions, one

for each ε where ε is a sequence of −1 and 1. Thus εk ∈ {−1,1}. Thus ε ̸= ε′ if the two

sequences differ in at least one slot, one giving 1 and the other equaling −1. Now define

fε (t)≡∞

∑k=1

εk2−k/p sin(

2kπt)

Then this is 1/p Holder. Let s < t.

| fε (t)− fε (s)| ≤ ∑k≤|log2(t−s)|

∣∣∣2−k/p sin(

2kπt)−2−k/p sin

(2k

πs)∣∣∣

+ ∑k>|log2(t−s)|

∣∣∣2−k/p sin(

2kπt)−2−k/p sin

(2k

πs)∣∣∣

If t = 1 and s= 0, there is really nothing to show because then the difference equals 0. Thereis also nothing to show if t = s. From now on, 0 < t− s < 1. Let k0 be the largest integerwhich is less than or equal to |log2 (t− s)| = − log2 (t− s). Note that − log(t− s) > 0because 0 < t− s < 1. Then

| fε (t)− fε (s)| ≤ ∑k≤k0

∣∣∣2−k/p sin(

2kπt)−2−k/p sin

(2k

πs)∣∣∣

+ ∑k>k0

∣∣∣2−k/p sin(

2kπt)−2−k/p sin

(2k

πs)∣∣∣

≤ ∑k≤k0

2−k/p2kπ |t− s|+ ∑

k>k0

2−k/p2

Now k0 ≤ − log2 (t− s) < k0 + 1 and so −k0 ≥ log2 (t− s) ≥ −(k0 +1). Hence 2−k0 ≥|t− s| ≥ 2−k02−1 and so 2−k0/p ≥ |t− s|1/p ≥ 2−k0/p2−1/p. Using this in the sums,

| fε (t)− fε (s)| ≤ |t− s|Cp + ∑k>k0

2−k/p2k0/p2−k0/p2

≤ |t− s|Cp + ∑k>k0

2−k/p2k0/p(

21/p |t− s|1/p)

2