>>344

こんなのがあったので引用する(^^
https://answers.yahoo.com/question/index?qid=20111114081521AAQ7U4A
(抜粋)
Determine whether the following functions are lipschitz. Explain? Yahoo answers 20/11/2011
(a) f(x) = 1/x, x belong to (0,1)
(b) f(x) = (x^2) sin(1/x) with x belong to (0,1]
(c) f(x) = (x^(3/2)) sin (1/x) with x belong to (0,1)
(d) f(x) = (x^2) sin [exp(1/x)] with x belong to (0,1)
(e) f: R-->R, where f is a polynomial of degree larger than 1.

Best Answer:
(a) This is not Lipschitz on (0, 1). To show this:

Define x(n) = 1/n and y(n) = 1/2 for n = 1, 2, ..., and suppose that f is Lipschitz on (0, 1).
(Note that {1/n} and {1/2} are subsets of (0, 1).)

Then, there must exist M > 0 such that
M ? |[f(x(n)) - f(y(n))] / (x(n) - y(n))|
....= |(n - 2) / (1/n - 1/2)|
....= |(n - 2) / [(1/(2n)) (2 - n)]|
....= 2n → ∞ as n→∞.

Hence, no such M can exist, and so f(x) = 1/x is not Lipschitz on (0, 1).

(b) Note that f '(x) = 2x sin(1/x) - cos(1/x).
==> |f '(x)| ? 2|x| |sin(1/x)| + |cos(1/x)| ? 2|x| * 1 + 1 ? 2 * 1 + 1 = 3 for all x in [0, 1].

Since f has bounded derivative on (0, 1], we see that f is Lipschitz.

つづく