If $f(x)$ is a continously differentiable function such that $f'(x)$ is unbounded, then $f(x)$ is not unifromly continuous?

Question

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?

Answer 1

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 $\pm \infty$.

Answer 2

No, you are not right. Consider for example the $C^{\infty}$ function $$f(x)=\frac{\cos(e^x)}{1+x^2}.$$ $f$ is uniformly continuous in $\mathbb{R}$ because it is continuous and its limit at $\pm \infty$ is zero.

On the other hand, its derivative is unbounded in $\mathbb{R}$: $$f'(x)=-\frac{e^x\sin(e^x)}{1+x^2}-\frac{2x\cos(e^x)}{(1+x^2)^2}$$ and $\lim_{n\to +\infty} f'(x_n)=+\infty$ where $x_n=\ln(3\pi/2+2\pi n)$.

Answer 3

The error in OP's argument is that they fix $\epsilon, \delta$ beforehand for the uniformity claim, but later use the fact that the derivative $f'$ is above a certain value in an interval of length $\delta$, which is not guaranteed, as Kavi explained.

Intuitively, if the interval where $f'$ is high is very small, and becomes smaller with higher $f'$, for a given $\epsilon$ the 'spikes' in $f'$ (as described by Kavi) are not able to increase $f$ by much (more than $\epsilon$) except for a finite number at the beginning.