Let $f$ be a differentiable function from a closed interval to $\mathbb{R}$. Then, is $f'$ bounded?

Question

Let $f$ be a differentiable function from a closed interval to $\mathbb{R}$. Then, $f'$ is bounded.

Can somebody give a proof or a counterexample for this?

Moreover, if $f$ is differentiable, then is $f'$ continuous?

Answer 1

No. Let $f\colon[-1,1]\to\mathbb{R}$ be the function $$ f(x) \;=\; \begin{cases}x^2 \sin (1/x^2) & \text{if }x\ne 0, \\[3pt] 0 & \text{if } x=0.\end{cases} $$ Here is a plot of this function, with the parabolas $y = \pm x^2$ shown in red:

Note that $f$ is differentiable at $0$, with $f'(0) = 0$. However, the oscillations become steeper and steeper as $x\to 0$. In particular, for $x\ne 0$ we have $$ f'(x) \;=\; 2x \sin(1/x^2) - \frac{2\cos(1/x^2)}{x} $$ which is clearly unbounded.

Answer 2

Standard counterexample: $f(x) = x^2 \sin(1/x^2)$ on, say, $[0,1]$ (with $f(0)$ defined as $0$).

Answer 3

No; if $f$ is differentiable, $f'$ is not necessarily continuous; the functions that do satisfy this property are called $C^1$ functions. Basically, take any function that is not continuous but integrable and you can find a function that is not $C^1$.

A standard example is that of $f(x)= x^2 sin(1/x) ; x\neq 0 ; f(0)=0$, whose derivative $2xsin(1/x)-cos(1/x)$ EDIT exists at $x=0$, since $Lim _{h\rightarrow 0} {h}{Sin(1/h)=0}$ , but the derivative is not continuous at $x=0$; if $f'(x)$ was continuous at $0$, then $Lim_{h\rightarrow 0} cos(1/x)$ would exist, but it does not, as we can choose values $x_i, x_j$ that are indefinitely-close to each other, so that $|cos(1/x_i)-cos(1/x_j)|=2$ , by periodicity of $cos(1/x)$, so $cos(1/x)$ would fail a $\delta-\epsilon$ test for any $\epsilon <2$ .