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Lipschitz Functions
Lorianne Ricco
February 4, 2004
Definition 1 Let f (x) be defined on an interval I and suppose we can find two positive constants M and α such that
|f (x 1 ) − f (x 2 )| ≤ M |x 1 − x 2 |α^ for all x 1 , x 2 ∈ I.
Then f is said to satisfy a Lipschitz Condition of order α and we say that f ∈ Lip(α).
Example 1 Take f (x) = x on the interval [a, b]. Then
|f (x 1 ) − f (x 2 )| = |x 1 − x 2 |
That implies that f ∈ Lip(1). Now take f (x) = x^2 on the interval [a, b]. Then
|f (x 1 ) − f (x 2 )| = |x 12 − x 22 | = |x 1 − x 2 ||x 1 + x 2 | ≤ M |x 1 − x 2 |
with M = 2max(|a|, |b|). Hence, again f ∈ Lip(1). The function f (x) = 1/x on (0, 1). Is it Lip(1)? How about Lip(1/2)? How about Lip(α)?
Theorem 1 Lip(α) is a linear space.
Proof We will look at a part of this proof. Let f, g ∈ Lip(α).
(f + g)(x) ∈ Lip(α)
Then,
|f (x) + g(x) − f (y) + g(y)| ≤ M |x − y|α
If f ∈ Lip(α) it implies that
|f (x) − f (y)| ≤ M 1 |x − y|α
If g ∈ Lip(α) it implies that
|g(x) − g(y)| ≤ M 2 |x − y|α If (f + g) ∈ Lip(α) it imples that
|(f + g)(x) − (f + g)(y)| ≤ M 3 |x − y|α
|f (x) + g(x) − f (y) − g(y)| =
|f (x) − f (y) + g(x) − g(y)| =
|f (x) − f (y)| + |g(x) − g(y)| =
By the triangle inequality,
|f (x) − f (y)| + |g(x) − g(y)| ≤
M 1 |x − y|α^ + M 2 |x − y|α^ = |M 1 + M 2 ||x − y|α
Theorem 2 If f ∈ Lip(α) with α > 1 then f = constant.
Proof Left as homework for everyone.
1 Lipschitz and Continuity
Theorem 3 If f ∈ Lip(α) on I, then f is continous; indeed, uniformly contiu- ous on I.
Last time we did continuity with and δ. An alternative definition of con- tinuity familar from calculus is: f is continuous at x = c if:
- f (c) exists
- limx→cf (x) exists
- limx→cf (x) = f (c)
In order to be continuous, if |x − x 0 | < δ, then |f (x) − f (x 0 )| < . Proof |f (x) − f (c)| ≤ M |x − c|α
limx→c|f (x) − f (c)| ≤ M limx→c|x − c|α^ = 0
This implies
limx→cf (x) = f (c)
How about continuous implies Lip(α)?
