9.4. EXERCISES 205

≤ |t− s|Cp + ∑k>k0

2−(k−k0)/p(

21/p |t− s|1/p)

2

≤ Cp |t− s|+(

21+1/p) ∞

∑k=1

2−k/p |t− s|1/p

= Cp |t− s|+Dp |t− s|1/p ≤Cp |t− s|1/p +Dp |t− s|1/p

Thus fε is indeed 1/p Holder continuous.Now consider ε ̸= ε

′. Suppose the first discrepancy in the two sequences occurs withε j. Thus one is 1 and the other is −1. Let t = i+1

2 j+1 ,s =i

2 j+1

| fε (t)− fε (s)− ( fε ′ (t)− fε ′ (s))|=∣∣∣∣∣ ∑∞k= j εk2−k/p sin

(2kπt

)−∑

∞k= j εk2−k/p sin

(2kπs

)−(

∑∞k= j ε ′k2−k/p sin

(2kπt

)−∑

∞k= j ε ′k2−k/p sin

(2kπs

)) ∣∣∣∣∣Now consider what happens for k > j. Then sin

(2kπ

i2 j+1

)= sin(mπ) = 0for some integer

m. Thus the whole mess reduces to∣∣∣∣(ε j− ε′j)

2− j/p sin(

2 jπ (i+1)2 j+1

)−(ε j− ε

′j)

2− j/p sin(

2 jπi2 j+1

)∣∣∣∣=

∣∣∣∣(ε j− ε′j)

2− j/p sin(

π (i+1)2

)−(ε j− ε

′j)

2− j/p sin(

πi2

)∣∣∣∣= 2

(2− j/p

)In particular, |t− s|= 1

2 j+1 so 21/p |t− s|1/p = 2− j/p

| fε (t)− fε (s)− ( fε ′ (t)− fε ′ (s))|= 2(

21/p)|t− s|1/p

which shows that

sup0≤s<t≤1

| fε (t)− fε ′ (t)− ( fε (s)− fε ′ (s))||t− s|1/p ≥ 21/p (2)

Thus there exists a set of uncountably many functions in C1/p ([0,T ]) and for any two ofthem f ,g, you get

∥ f −g∥C1/p([0,1]) > 2

so C1/p ([0,1]) is not separable.

9.4 Exercises1. Let (X ,τ) ,(Y,η) be topological spaces and let A⊆ X be compact. Then if f : X→Y

is continuous, show that f (A) is also compact.