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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.

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

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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.

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  • $\begingroup$ The example has now been replaced by a slightly simpler one. A few details have been added. $\endgroup$ Jan 26, 2021 at 20:54
  • $\begingroup$ Thanks. So it seems if it goes towards the boundary too "sharp" then there won't be a subdifferential at the boundary $\endgroup$ Jan 26, 2021 at 21:28
  • $\begingroup$ 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 $x = 0$. $\endgroup$ Jan 26, 2021 at 21:53
  • $\begingroup$ @ShamisenExpert : Your latter expression for the subdifferential will not work -- for one, the subdifferential of a function at a point is, not a number, but a set. $\endgroup$ Jan 26, 2021 at 22:21
  • $\begingroup$ 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? $\endgroup$ Jan 26, 2021 at 22:24

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