$f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, on $[-1, 1]$

Terms

Term	This example	Why
continuous	✓
Lipschitz continuous	✓	The function $f(x) = x^2 \sin(1/x)$ is Lipschitz continuous on $[-1, 1]$ because it is differentiable everywhere with $\abs{f^{\prime}(x)} \leq 3$.
uniformly continuous	✓
absolutely continuous	✓
continuously differentiable	✗	The function $f$ is not continuously differentiable, because its derivative is not continuous at $0$. At $0$, we have $f^{\prime}(0) = 0$ because \[\lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} h \sin(1/h) = 0.\] Away from $0$, the product and chain rules give \[f^{\prime}(x) = 2x \sin(1/x) - \cos(1/x),\] The points $x_n = 1 / (2 \pi n)$ tend to $0$, but $f^{\prime}(x_n) = 0 - \cos(2 \pi n) = -1$ for every $n$, so $f^{\prime}(x_n)$ does not tend to $f^{\prime}(0) = 0$. That is, \[f^{\prime}(0) \neq \lim_{x \to 0} f^{\prime}(x),\] because the limit does not exist. The discontinuity of $f^{\prime}$ at $0$ is not a jump discontinuity, at which the limits from the left and from the right exist but differ. Instead, $f^{\prime}$ oscillates: as $x \to 0$, the term $2x \sin(1/x)$ tends to $0$, but the term $-\cos(1/x)$ oscillates between $-1$ and $1$ faster and faster, so $f^{\prime}(x)$ has no limit from either side. This oscillation is the only way that a derivative can fail to be continuous, by Darboux’s theorem: every derivative has the intermediate value property, so a derivative cannot have a jump discontinuity, such as the jump of a step function.