How discontinuous can a derivative be?
Classroom notes - On a discontinuity of a derivative - J Klippert (2000)
2. Discontinuous derivatives and fundamental essential discontinuities
Theorem 2.1. Let $f :[a,b]\to\Bbb R$ be such that $f$ is differentiable on $(a,b)$. Let $x_0\in(a,b)$. If $f'$ is discontinuous at $x_0$, then $x_0$ is a fundamental essential discontinuity of $f'$.
证明.
(a) $x_0$ 不是可去间断点(即,极限存在,但不等于函数值)假设$x_0$是可去间断点,则存在$L∈\Bbb R$使得$\lim _{x \rightarrow x_{0}} f^{\prime}(x)=L \neq f^{\prime}\left(x_{0}\right)$,不妨设$L<f'(x_0)$,设$$\varepsilon=\frac{f^{\prime}\left(x_{0}\right)-L}{2}$$存在$\delta>0$使$\left|f^{\prime}(x)-L\right|<\varepsilon$ 对任意 $x \in[a, b]$ 使得 $x_0<x<x_0+\delta$. 注意$L+\varepsilon<f^{\prime}\left(x_0\right)$, 令 $\gamma \in\left(L+\varepsilon, f^{\prime}\left(x_0\right)\right)$, 取 $t \in\left(x_0, x_0+\delta\right)$. 则 $f$ 在 $\left[x_0, t\right]$ 可导. 因为$f^{\prime}(t)<L+\varepsilon$, 有$f^{\prime}(t)<\gamma<f^{\prime}\left(x_0\right)$. 由Darboux定理, 存在$c \in\left(x_0, t\right)$ 使 $f^{\prime}(c)=\gamma$. 但 $f^{\prime}(c)<L+\varepsilon$, 因为 $\gamma>L+\varepsilon$, 推出 $f^{\prime}(c) \neq \gamma$矛盾. $\square$
(b) $x_0$ 不是跳跃间断点(即,左右极限都存在且不等)假设$x_0$是$f^{\prime}$的跳跃间断点. 则$f^{\prime}\left(x_0^{-}\right)$与$f^{\prime}\left(x_0^{+}\right)$均存在且有限,且$f^{\prime}\left(x_0^{-}\right) \neq f^{\prime}\left(x_0^{+}\right)$. 不妨设$f^{\prime}\left(x_0^{-}\right)<f^{\prime}\left(x_0^{+}\right)$. 设
\[
\varepsilon=\frac{f^{\prime}\left(x_0^{+}\right)-f^{\prime}\left(x_0^{-}\right)}{3}
\]Note that $f^{\prime}\left(x_0^{-}\right)+\varepsilon<f^{\prime}\left(x_0^{+}\right)-\varepsilon$. Then there exists $\delta_1>0$ such that for $x \in[a, b]$ for which $x_0-\delta_1<x<x_0$,
\[\tag1
\left|f^{\prime}(x)-f^{\prime}\left(x_0^{-}\right)\right|<\varepsilon
\]
and there exists $\delta_2>0$ such that if $x \in[a, b]$ and $x_0<x<x_0+\delta_2$,
\[\tag2
\left|f^{\prime}(x)-f^{\prime}\left(x_0^{+}\right)\right|<\varepsilon
\]
Letting $\delta=\min \left\{\delta_1, \delta_2\right\}$, choose $\alpha, \beta \in[a, b]$ such that
\[
x_0-\delta<\alpha<x_0<\beta<x_0+\delta
\]
and let $\gamma \neq f^{\prime}\left(x_0\right)$ be such that
\[
f^{\prime}(\alpha)<f^{\prime}\left(x_0^{-}\right)+\varepsilon<\gamma<f^{\prime}\left(x_0^{+}\right)-\varepsilon<f^{\prime}(\beta)
\]
由Darboux定理, 存在$c \in(\alpha, \beta)$使$f^{\prime}(c)=\gamma$. Since $\gamma \neq f^{\prime}\left(x_0\right), c \neq x_0$ and thus either $x_0-\delta<c<x_0$ or $x_0<c<x_0+\delta$. In the former case, we have $x_0-\delta_1 \leqslant x_0-\delta<c<x_0$ and thus
\[\tag3
\left|f^{\prime}(c)-f^{\prime}\left(x_0^{-}\right)\right|=\gamma-f^{\prime}\left(x_0^{-}\right)>\varepsilon
\]
In the latter, we have that $x_0<c<x_0+\delta \leqslant x_0+\delta_2$ and so
\[\tag4
\left|f^{\prime}(c)-f^{\prime}\left(x_0^{+}\right)\right|=f^{\prime}\left(x_0^{+}\right)-\gamma>\varepsilon
\]
Since inequality (3) contradicts (1) and inequality (4) contradicts (2), we conclude that (b) holds.
(c) $f'(x_0^+)\ne\pm\infty,f'(x_0^-)\ne\pm\infty$我们来证明$f\left(x_0^{+}\right) \neq-\infty$(用相同方法可以证明 $f\left(x_0^{+}\right) \neq-\infty$ 和 $f^{\prime}\left(x_0^{-}\right) \neq \pm \infty$.)
假设 $f^{\prime}\left(x_0^{+}\right)=+\infty$. 取$M>f^{\prime}\left(x_0\right)$. 则存在$\delta>0$使对任何$x \in[a, b]$满足$x_0<x<x_0+\delta$, 我们有$f^{\prime}(x)>M$. 取$t \in\left(x_0, x_0+\delta\right)$. 则$f$在$\left[x_0, t\right]$可导且$f^{\prime}\left(x_0\right)<f^{\prime}(t)$. 取$\gamma \in \Bbb R$使$f^{\prime}\left(x_0\right)<\gamma<M<f^{\prime}(t)$. 则 $∀c \in\left(x_0, t\right), f^{\prime}(c)>M>\gamma$与Darboux定理矛盾. $\square$ | 
