Question 6.T.9: A function f : D → R is not uniformly continuous if, and onl...
A function f : D → \mathbb{R} is not uniformly continuous if, and only if, there exist two sequences (x_{n}) and (t_{n}) in D such that
(i) |x_{n} − t_{n}| → 0
(ii) |f(x_{n}) − f(t_{n})| \nrightarrow 0.
If f is not uniformly continuous, then there is an ε > 0 such that, for all δ > 0, we can find two points x, t ∈ D that satisfy |x − t| < δ and |f(x) − f (t)| ≥ ε. By taking
δ = 1,\frac{1}{2},\frac{1}{3}, · · · ,
we can form two sequences (x_{n}) and (t_{n}) in D such that
|x_{n} − t_{n}| < \frac{1}{n}, |f(x_{n}) − f (t_{n})| ≥ ε.
Conversely, suppose ƒ is uniformly continuous and (x_{n}), (t_{n}) are two sequences in D which satisfy (i). If ε is any positive number, the uniform continuity of f implies that there is a δ > 0 such that
x, t ∈ D, |x − t| < δ ⇒ |ƒ(x) − ƒ(t)| < ε.
Since |x_{n} − t_{n}| → 0, there is an integer N such that
n ≥ N ⇒ |x_{n} − t_{n}| < δ ⇒ |f(x_{n}) − f(t_{n})| < ε,
and therefore |f(x_{n}) − f(t_{n})| → 0.