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For a convex function $\varphi: \mathbb{R}^n \to \mathbb{R},$ the subdifferential is defined as $$\partial \varphi (x) = \{z \in \mathbb{R}^n \mid \forall y \in \mathbb{R}^n, \varphi(y) \geq \varphi(x) + \langle z, y - x\rangle\}.$$ My question is under what conditions there exists a measurable function $f : \mathbb{R}^n \to \mathbb{R}^n$ such that $f(x) \in \partial \varphi(x)$ for all $x \in \mathbb{R}^n.$ I am aware of Kuratowski and Ryll-Nardzewski measurable selection theorem, but I am not sure how to apply this to the subdifferential correspondence. Any reference will be helpful!

Thanks!

Paruru
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    By "measurable function", do you mean Borel or Lebesgue measurable? – Akira Sep 02 '23 at 22:31
  • I originally meant the Borel measurability, but I'm also interested in the Lebesgue one if this difference affects the result! – Paruru Sep 02 '23 at 22:35
  • Because convex function on $\mathbb R^d$ is differentiable a.e. and the Lebesgue $\sigma$-algebra is complete, such Lebesgue measurable selection exists. – Akira Sep 02 '23 at 22:36
  • This question may be of your interest. – Akira Sep 02 '23 at 22:41
  • Also, this question... – Akira Sep 02 '23 at 22:57
  • @Akira Thank you for your comments! I don't get the point perfectly yet. I understand that the set of non-differentiable points is Lebesgue measurable and has zero measure. But values on the set matters for the measurabilty of the selection, right? How should I select values from the subdifferential on non-differentiable points to make the selection Lebesgue measurable? – Paruru Sep 03 '23 at 00:30
  • Please see my below answer. – Akira Sep 03 '23 at 07:42

1 Answers1

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Let $D$ be the set on which $\varphi$ is Fréchet differentiable. By this thread, $D$ is a Borel set. On the other hand, convex functions on $\mathbb R^n$ is differentiable a.e., so $D^c := \mathbb R^n \setminus D$ is a Lebesgue null set. By this thread, the restriction $\nabla \varphi:D \to \mathbb R^n$ is Borel measurable. We define $f:\mathbb R^n \to \mathbb R^n$ as follows:

  • If $x \in D$ then $f(x) := \nabla \varphi (x)$.
  • If $x \notin D$ then we pick any $y \in \partial f(x)$ and let $f(x) := y$.

Because the Lebesgue $\sigma$-algebra is complete, a subset of a Lebesgue null set is Lebesgue measurable. This implies arbitrary modification of a Lebesgue measurable function on a Lebesgue null set remains Lebesgue measurable. Hence $f$ is Lebesgue measurable.

Akira
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  • Why are you talking about Fréchet differentiability in finite dimension? What does it add by using this level of abstraction? – Jürgen Sukumaran Sep 03 '23 at 07:49
  • @mordecaiiwazuki The usual derivative in finite dimensions is also Fréchet derivative. I explicitly use the term "Fréchet" because the quoted theorem in the book uses it. I don't think using the term "Fréchet" appeals to any high level of abstraction. – Akira Sep 03 '23 at 07:51
  • What's wrong with my answer that I got a downvote? – Akira Sep 03 '23 at 07:52
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    Well, behold, lot's of people here and MO get triggered for all kinds of reasons. It might be the Frechet / over-abstraction thing, it could just be due to not enough sunlight, anything really. Anyways, I've upvoted the answer because it fits the bill perfectly (and I learned something new). – dohmatob Sep 03 '23 at 13:41
  • @dohmatob Thank you so much for your supportive comment :) – Akira Sep 03 '23 at 13:42
  • @Akira Thanks! This makes sense! Please let me know if you have something you can say about the Borel case. – Paruru Sep 03 '23 at 21:02
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    I'm also curious by the same question, so I have just posted it on MO... – Akira Sep 03 '23 at 21:42