Return to Answer

Taken by a sort of generalisation frenzy I produced the following; I tried to make it as readable as I could.

Taken by a sort of generalisation frenzy I produced the following; I tried to make it as readable as I could. Recall that the upper and lower Dini derivatives are respectively : $$D^*f(x):=\limsup_{y\to x}\frac{f(y)-f(x)}{y-x}$$ $$D_*f(x):=\liminf_{y\to x}\frac{f(y)-f(x)}{y-x}.$$

Lemma. For every $\eta>0$, and for every pluri-interval $K\subset[a,b]$ there exists a pluriinterval $L \subset K $ such that $\big|K^f \setminus {L}^f\big|\le\eta$, and $\frac{f(\beta)-f(\alpha )}{\beta -\alpha }\ge\frac1\eta$ for every component $[\alpha,\beta]$ of $L$.

Lemma. For every $\eta>0$, and for every pluriinterval $K\subset[a,b]$ there exists a pluriinterval $L \subset K $ such that $\big|K^f \setminus {L}^f\big|\le\eta$, and $\frac{f(\beta)-f(\alpha )}{\beta -\alpha }\ge\frac1\eta$ for every component $[\alpha,\beta]$ of $L$.

Note that each component interval of $K_n$ has length at most $ 2^{-n+1}\epsilon\|f\|_\infty $. So if $x\in K$, there is a sequence $x_n\in K_n$ of endpoints of components, such that $x_n\to x$ with $\frac{f(x)-f(x_n)}{x-x_n}\ge {2^n}{\epsilon}$, that is $$K\subset \{D^*f=+\infty\}.$$
Since $\epsilon>0$ is arbitrary, the thesis follows.

Note that each component interval of $K_n$ has length at most $ 2^{-n+1}\epsilon\|f\|_\infty $. So if $x\in K$, there is a sequence $x_n\in K_n$ of endpoints of components, such that $x_n\to x$ with $\frac{f(x)-f(x_n)}{x-x_n}\ge \frac{2^n}{\epsilon}$, that is $$K\subset \{D^*f=+\infty\}.$$
Since $\epsilon>0$ is arbitrary, the thesis follows.