2
$\begingroup$

Consider $f: \mathbb{R}^n \to \overline{\mathbb{R}}$ a lower-semicontinuous, proper, closed and convex. My question is, can the subdifferential of $\partial f$ be unbounded in the interior of $\text{dom}(f)$ (all points where $f(x) < \infty$?

I was pretty convinced that it can only be unbounded on the boundary and not the interior until I found the following counter-example (unfortunately I've lost track of where I encountered it, possibly in some online notes by Dimitri Bertsekas):

Suppose $f(x) = \|x\|^2, x = (x_1, x_2, x_3)$ is defined on an affine hyperplane of $\mathbb{R}^3$, then the subdifferential necessarily includes a ray that is perpendular to that hyperplane at any point in the interior of the domain, hence the subdifferential of this function at any point (interior or not) is unbounded.

Can someone please verify or provide guidance on this matter?

enter image description here

Improve this question
$\endgroup$
2
  • $\begingroup$ If a point is in the interior of the domain, then the function is locally bounded there and there is a neighborhood in which the function is also Lipschitz continuous. This implies bounded subgradients. Does you counterexample actually have interior points in its domain? $\endgroup$
    – Dirk
    Commented Jun 22, 2021 at 11:04
  • $\begingroup$ @Dirk Ok, I see. At relative interior points there is ray of the form $(a,\ldots, a), a \in [0, \infty)$ that makes the subgradient unbounded. $\endgroup$
    – Olórin
    Commented Jun 22, 2021 at 19:00

1 Answer 1

Reset to default
1
$\begingroup$

If the question is about the interior of the domain, it seems to me that the subgradients should be bounded.

If the question is about the relative interior, then your counterexample seems to work.

A perhaps simpler example: Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a convex function and let $g(x_1, x_2) = f(x_1)$ if $x_2 = 0$ and $g(x_1, x_2) = + \infty$ otherwise. Then, at a point $(x_1, 0)$ we have $\partial g(x_1, 0) = \partial f(x_1) \times \mathbb{R}$, which is clearly unbounded.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks. I was a bit unsure of my counterexample because there are many references that says the subdifferential is always nonempty, convex and compact. See: en.wikipedia.org/wiki/Subderivative#The_subgradient $\endgroup$
    – Olórin
    Commented Jun 23, 2021 at 6:35
  • $\begingroup$ Funnily, the wikipedia page says that the subdifferential can be empty… (but it always is closed and compact though). $\endgroup$
    – Dirk
    Commented Jun 23, 2021 at 6:41
  • $\begingroup$ On the interior of the domain the subgradient is always nonempty, convex and compact. This is to what wikipedia refers since in their example f is defined on an open set. However, the subgradient is always empty outside of the domain (by definition). And as both counterexamples show it can be unbounded. But the reason for that is that the functions from the examples are naturally defined on a set of lower dimension, hyperplane in your case, $\mathbb{R} \times \{ 0\}$ in mine. So yes it may be unbounded, but in some sense just because we are not looking at it in the right way. $\endgroup$
    – J. Doe
    Commented Jun 23, 2021 at 18:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .