568 CHAPTER 21. BANACH SPACES
+ ∑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
≤ |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 with
ε j. Thus one is 1 and the other is −1. Let t = i+12 j+1 ,s =
i2 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
)