# When does strict inclusion holds for the domain of subdifferential?

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.

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

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.

• You seem to have posted this twice. Please remove one of the copies. Jan 25, 2021 at 22:51

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, $$f(y)=-\sqrt y for each real $$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.
• I wonder if it is simpler to just make the observation that $f(x) = -\sqrt(x) + \delta_{x \geq 0}(x)$, so $\partial f(x) = -\frac{1}{2} x^{-1/2} +N_{x \geq 0}(x)$ and the first term is not defined at $x = 0$. Jan 26, 2021 at 21:53
• We can always write $\{-\frac{1}{2} x^{-1/2}\}$ as a singleton set. This is what I meant. The value in this set is not defined as $x \to 0$. Is this a simpler way to see that this inclusion is strict? Jan 26, 2021 at 22:24