Partial Differential Equations, 2nd Edition, L.C.Evans
Chapter 5 Sobolev Spaces
Shih-Hsin Chen
∗
, Yung-Hsiang Huang
†
2016/11/01
1. Proof. We prove the case k = 0 for simplicity. The general case is almost the same (except we
need the standard convergence theorem between {f
n
} and {Df
n
}).
Given a Cauchy sequence {f
n
} ⊂ C
0,β
(U), then there is a f ∈ C(U) such that f
n
→ f uniformly
and a constant M > 0 such that for each n ∈ N and s 6= t,
|f
n
(s) − f
n
(t)|
|s − t|
β
≤ M.
Then for each s 6= t, there is some N = N(s, t) ∈ N such that
|f(s) − f(t)| ≤ |f(s) − f
N
(s)| + |f
N
(s) − f
N
(t)| + |f
N
(t) − f(t)|
≤ 2|s − t|
β
+ M|s − t|
β
.
Therefore, f ∈ C
0,β
(U). It remains to show f
n
→ f in C
0,β
(U). For each s, t ∈ U and > 0,
by Cauchy’s criteria, we can find a N = N() such that for k, n ≥ N
|f
k
(s) − f
n
(s) − f
k
(t) + f
n
(t)| ≤ |s − t|
β
For each η > 0, we can find a K = K(η, s, t) > N, such that |f(x) − f
K
(x)| ≤ η for x = s, or
x = t. Therefore, for every n ≥ N
|f(s)−f
n
(s)−f(t)+f
n
(t)| ≤ |f(s)−f
K
(s)|+|f
K
(s)−f
n
(s)−f
K
(t)+f
n
(t)|+|f(t)−f
K
(t)| ≤ 2η+|s−t|
β
.
Letting η → 0, we obtain that |f(s) − f
n
(s) − f(t) + f
n
(t)| ≤ |s − t|
β
for all n ≥ N().
∗
Department of Math., National Taiwan University. Email: d03221002@ntu.edu.tw
†
Department of Math., National Taiwan University. Email: d04221001@ntu.edu.tw
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