When does strict inclusion holds for the domain of subdifferential?

Question

Recall that, given an extended real-valued function $f: \mathbb{R}^n \to (-\infty, \infty]$

Its effective domain is, $$\text{dom}(f) = \{x \in \mathbb{R}^n : f(x) < +\infty\}$$

The subdifferential is, $$\partial f(x) = \{v \in \mathbb{R}^n: f(x^\prime) \geq f(x) + v^\top (x^\prime - x), \forall x^\prime \in \mathbb{R}^n\}$$

and its effective domain is all the vectors where it is subdifferentiable,

$$\text{dom}(\partial f) = \{x \in \mathbb{R}^n | \partial f(x) \neq \varnothing\}.$$

Now by a theorem of Rockafellar (top of page 227, "Convex Analysis", 1970), it states,

Let $f: \mathbb{R}^n \to (-\infty, \infty]$ be a closed, proper, convex function, then $$\text{rint}(\text{dom}(f))) \subseteq \text{dom}(\partial f) \subseteq \text{dom}(f).$$ where $\text{rint}$ is the relative interior.

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My question is, when does strict inclusion holds, especially for $\text{dom}(\partial f) \subseteq \text{dom}(f)$.

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For any set I can think of that is closed, you can make a point on the boundary of that set, and there will exist some subgradient at that point. There is the possibility that this strict inequality is an equality for most cases.

Answer 1

Define $f\colon\mathbb R\to\mathbb R$ by letting $f(x):=-\sqrt x$ if $x\ge0$ and $f(x):=\infty$ if $x<0$. Then $f$ is a closed proper convex function.

However, $f(0)=0<\infty$, so that $0\in\text{dom}(f)$. On the other hand, $f(y)=-\sqrt y<vy=f(0)+v(y-0)$ for each real $v$ and any small enough $y>0$. So, $0\notin\text{dom}(\partial f)$.

Thus, here the set inclusion $\text{dom}(\partial f)\subset\text{dom}(f)$ is strict.