Suppose $f(x)$ is a continuously differentiable function on $\mathbb{R}$ such that its derivative is unbounded.
Claim: $f(x)$ is not uniformly continuous.
Assume f(x) is uniformly continuous, then given $\epsilon$ we have a suitable $\delta$. Since $f'(x)$ is not bounded, their exists an element $c$ such that $f'(c) > k \frac{\epsilon}{\delta}$, $k \gg 0$. Since $f'(x)$ is continuous, if we choose a $\delta$ neighbourhood of $c$, we would then by mean value theorem have
$$f(b) – f(a) = f'(c')(b-a), c' \in \Big(c- \frac{\delta}{2},\; c + \frac{\delta}{2}\Big).$$ or
$$ f'(c') = \frac{\epsilon}{\delta}$$ – a contradiction since $f'(c')$ has value in around some small neighborhood of $k \frac{\epsilon}{\delta}$.
So finally I have this conclusion:
If $f(x)$ is continuously differential function, then $f(x)$ is uniformly continuous iff $f'(x)$ is bounded?
Is my conclusion right?
New details:
I realize that that my argument $f'(c) \to f'(c')$ is slippery and might not yield the desired conclusion every time, but suppose if I have a specific $f(x)$ and I can somehow show that $m \leq \mid f'(x) \mid \leq M$ for some interval $I$ where $f'(c) = k\frac{\epsilon}{\delta}$ and $m, M$ are suitable close enough (say $\epsilon^{1000000})$, then does my argument holds?
Best Answer
Take any unbounded continuous integrable function $g$ and define $f(x)=\int_0^{x} g(y)\, dy$. Then $f$ is uniformly continuous but its derivative is not bounded. To construct such a function $g$ draw triangles with bases $(n-\frac 1 {n^{3}}, n+\frac 1 {n^{3}})$ and height $n$. Think of a function $g$ whose graph is made up of these triangles. $f$ is uniformly continuous because it is continuous and it has finite limits at $\pm \infty$.