Question

Suppose \[-\infty \le a < c < b \le∞\] and f:(a,b)rightarrow Real is continuous on (a,b). (a) If f is uniformly continuous on both (a,c) and (c,b), prove that f is uniformly continuous on (a,b). (b) Give an example to show that the conclusion in part (a) may be false if f is not continuous on (a,b).

Answer 1

First the counter example
f(x) =1 if x is in the interval [0, 1{
f(x) = -1 if x is in the interval [1, 2]
Can you show that this f is a counter example?

Answer 2

Let
\[
\epsilon > 0 \text { and choose } \delta > 0 \text { so that }
\\
| x - c| < \delta \text { implies } \, |f(x) - f(c) \le \frac {\epsilon}2 \\
| x - y| < \delta \text { implies } \, |f(x) - f(y) \le \frac {\epsilon}2 \\
\text { if both } x,y \text { are } in (a, c)\\

| x - y| < \delta \text { implies } \, |f(x) - f(y) \le \frac {\epsilon}2 \\
\text { if both } x,y \text { are } in (c, b)\\
\]
now let
\[
x \in (a, c)\\
y \in (c,b)\\
|x - y | < \delta\\ \text { implies }\\
|x-c|\le \delta\\
|y-c| \le \delta\\ \text{ which implies }\\
|f(x) - f(y)| \le |f(x) - f(c)| + | f(c) - f(y)| \le \le \frac {\epsilon}2 + \le \frac {\epsilon}2=\epsilon
\]

Answer 3

If x and y are in the same sub-interval, there is nothing to prove.