If $f(x)$ is bounded and differentiable on a compact interval $I$, then $f'(x)$ is bounded on $I$.

Question

Prove if $f(x)$ is bounded and differentiable on a compact interval $I$, then $f'(x)$ is bounded on $I$, or give a counterexample.

I think it's true, but don't know how to prove it.

Answer 1

False: $$ f(x) = \begin{cases} x^2 \sin \frac{1}{x^2} & x \ne 0 \\ 0 & x = 0 \end{cases} $$ on the interval $[-1, 1]$. The derivative is $0$ at $x = 0$ (by the squeeze lemma) and is \begin{align} f'(x) &= 2x \sin \frac{1}{x^2} + x^2 \cos (\frac{1}{x^2} )(-2)x^{-3} \\ &= 2x \sin \frac{1}{x^2} -2\frac{1}{x}\cos (\frac{1}{x^2} ) \end{align}

That's unbounded on the interval $[-1, 1]$.

Answer 2

For a counterexample, set:

$$f(x)=x^2\sin\frac{1}{x^2}$$

for $x\ne 0$, and $f(0)=0$: it is bounded (by $1$) and differentiable on $[0,1]$ as it can be proven that $f'(0)=0$:

$$\lim_{h\to 0}\frac{h^2\sin\frac{1}{h^2}}{h}=\lim_{h\to 0}h\sin\frac{1}{h^2}=0$$

because $h\to 0$ and $\sin(\ldots)\le 1$. Now, calculate $f'(x)$:

$$f'(x)=\begin{cases}2x\sin\frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2}&x\ne 0\\0&x=0\end{cases}$$

is unbounded on $[0,1]$ (take $x_n=\frac{1}{\sqrt{2n\pi}}$, $n=1,2,\ldots$).