I am currently struggling with calculating the sub-derivative for $x = 0$ of the function $f(x) = - \sqrt{x}$, with $f:[0, 1] \rightarrow \mathbb{R}$.
Our lecture book does not state anything about sub-derivatives for functions in a closed interval, and I haven't been able to find any useful examples online. I am leaning toward $\partial f(0) = \emptyset$, but I am missing the argumentation.
Any kind of help would be appreciated.
The function is convex and differentiable on $(0,1]$ so on this interval $\partial f(x)$ coincides with $f'(x) = -\frac{1}{2\sqrt{x}}$. I see that you struggle with the remaining value $\partial f(0)$, for which it's best to use the definition: \begin{align} g\in \partial f(0) &\iff f(x) \ge f(0) + g(x-0) \quad \forall x \in [0,1] \\ & \iff -\sqrt{x} \ge gx \quad \forall x \in [0,1] \\ & \iff -\sqrt{x} \ge gx \quad \forall x \in (0,1] \\ & \iff -\frac{1}{\sqrt{x}} \ge g \quad \forall x \in (0,1] \\ & \iff \inf_{x \in (0,1]}\left(-\frac{1}{\sqrt{x}}\right) \ge g. \end{align} The LHS of the last inequality is $-\infty$ (when $x\to 0)$ so such $g$ does not exist. Therefore your result is correct: $\partial f(0) = \emptyset$.