Please let me post a question although this is possibly known already because I did not find the statement.
Let $f:(0,1)\to\mathbb{R}$ be a continuous function. Let denote by $D^+ f$ and $D^- f$ the superdifferential and the subdifferential of $f$, respectively. Also let define their domain as $$ \mathcal{D}^{\pm}f:=\{x\in(0,1) \mid D^{\pm}f(x)\neq\emptyset\}. $$ Then does that $\mathcal{D}^+ f\cup\mathcal{D}^- f=(0,1)$ hold?
I have checked that several fundamental functions satisfy it, but I could not proof generally or find counterexamples.
Thank you in advance.
I use the following definition of semidifferential: $$ D^+ f(x):=\{\phi'(x)\in\mathbb{R} \mid \text{$f-\phi$ attains a (local) maximum at $x$ for $\phi\in C^{1}(0,1)$}\} $$ and $$ D^- f(x):=\{\phi'(x)\in\mathbb{R} \mid \text{$f-\phi$ attains a (local) minimum at $x$ for $\phi\in C^1(0,1)$}\}. $$