Why if $f'$ is unbounded, then $f$ isn't uniformly continuous?

Solution 1:

As @Kannappan suggested, here is my previous comment, expanded into an answer.

Assertion c) is indeed not correct, not even for bounded intervals. That is, every function that is differentiable on a closed and bounded interval is continuous and hence automatically uniformly continuous there, even if its derivative is unbounded (which is of course possible). An example for the latter case (differentiable everywhere, unbounded derivative on a bounded and closed interval) is $$ f(x) = \begin{cases} x^{3/2} \sin \frac{1}{x} (0 < x \le 1) \\ 0 (x = 0) \end{cases} $$ For the case $I = [0,\infty)$, a counterexample is given by $g(x) = f(\frac{x}{1+x})$ where $f$ is defined as above.

Solution 2:

This is a wrong claim as the answer by Professor Aguirre suggests. I give an example in the case of a neither closed nor open set. It is known to be false even in general as the other answer suggests.

You seem to claim that:

If a function $f$ has unbounded derivative, this does not mean that $f$ is not uniformly continuous.

This is a wrong claim. Here is a counter example.

Consider $f:(0,1] \to \mathbb R$ defined by $f(x)=\sqrt x$. I leave it to you to prove that $f'$, the derivative of $f$ is unbounded on $(0,1]$ but $f$ is uniformly continuous on $(0,1]$.

Conclusion:

What is true is, there are functions with unbounded derivatives nevertheless uniformly continuous.