Uniform continuity of a bounded function on an unbounded interval Are all bounded continuous functions on an unbounded interval uniformly continuous ?
 A: A counterexample is given by $f(x)=\sin(x^2)$.
Let $x_n=\sqrt{2\pi n}$ and $y_n=\sqrt{2\pi n+\pi/4}$. Then $$|f(x_n)-f(y_n)|=1.$$
But $|x_n-y_n|\to0$. So $f$ is not uniformly continuous.
(One way to see that $|x_n-y_n|\to0$ is to apply the Mean Value Theorem to the function $\phi(t)=\sqrt t$.)
A: The answer to your question is not necessarily. The counter example provided by @DavidC.Ullrich demonstrates this.
A common reason a function is not uniformly continuous is that its derivative is actually unbounded (but a uniformly continuous function can actually have an unbounded derivative). For instance, if $g(x)=\sin(x^2)$ then $g'(x) = 2x \cos(x^2)$ which can take arbitrarily large positive and negative values.
If a differentiable function has a derivative that is bounded over the whole real line, then it would be uniformly continuous. This follows from the mean value theorem.
For instance suppose that $f:\mathbb{R} \to \mathbb{R}$ is differentiable and that $|f'(x)| < M$ for all $x \in \mathbb{R}$. When we consider $x < y$ and $$|f(x) - f(y)| = |f'(\xi)||x-y|$$ where $\xi \in (x,y)$, then we can see that $$|f(x) - f(y)| \le M |x-y|.$$
Now suppose we let $\epsilon > 0$. If we fix $x_0$, and consider all $y \in \mathbb{R}$ such that $|x_0 - y| < \epsilon/M$ we see that $$|f(x_0) - f(y)| < \epsilon.$$ Finally, since $\epsilon/M$ is independent of $x_0$, $f$ is uniformly continuous.
