4
$\begingroup$

Let $\gamma\in\left]0,+\infty\right[$, let $f$ be a proper, convex, lower semicontinuous function from a real Hilbert space $\mathcal{X}$ to $\left]-\infty,+\infty\right]$, and set $g=f-\frac{\gamma}{2}\|\cdot\|^2$. Then $g$ is weakly convex.

I'm looking for a reference characterizing for which $x\in\mathcal{X}$ the following holds

\begin{equation} \partial_{Clarke} g(x) = \partial_{convex} f(x) - \gamma x. \tag{*} \end{equation}

where $\partial_{Clarke}$ is the Clarke subdifferential,

$$\partial_{convex} f(x) = \left\{ u \in \mathcal{X} \mid (\forall y \in \mathcal{X}) \quad \langle y - x \mid u \rangle + f(x)\leq f(y) \right\},$$

and the righthand side in (*) denotes Minkowski subtraction. I know that $(\nabla \frac{\gamma}{2}\|\cdot\|^2) (x) = \gamma x$ and that $\partial_{convex}$ coincides with $\partial_{Clarke}$ on convex functions. However, I am only working with a weakly convex function. I've perused Rockafellar/Wets but not found much.

I'm actually not entirely positive that (*) is true everywhere, e.g. it may fail on the boundary of the domain of $g$. Any relevant info is greatly appreciated!

EDIT: I think it would suffice to find a reference for when the sum rule holds for Clarke subdifferentials. I believe that $\partial_{Clarke}g(x)\supset\partial_{convex}f(x)-\gamma x$, so it would suffice to show the reverse inclusion.

$\endgroup$
6
  • $\begingroup$ Rockafellar is the closest to this topic than other references. Maybe you can try this reference jstor.org/stable/2589712?seq=1 $\endgroup$ Commented Jul 30, 2020 at 20:54
  • $\begingroup$ My comment might be obvious or useless, but I think (*) holds at least at the points where $f$ is locally Lipschitz. I used Corollary 5.4 of link.springer.com/article/10.1007/s11228-007-0043-y since you're in a reflexive space and $g$ is differentiable... I'm not sure, though. $\endgroup$
    – Koto
    Commented Jul 31, 2020 at 1:19
  • $\begingroup$ Clarke’s book Nonlinear Analysis has the sum rule for the generalized gradient that can be used here $\endgroup$ Commented Jul 31, 2020 at 2:47
  • $\begingroup$ @NeutralElement thanks for the suggestion, I think the sum rule is exactly what I'm looking for! I'm not sure if I've got the right book, but I'm looking at Theorem 8.10 from Nonlinear Analysis, Diff. Eq's and Control and it looks like only of an approximate sum-rule. It does not show the equality (*) for the same $x$, it shows approximate equality for something in a neighborhood of $x$. Is this the result you were referring to? $\endgroup$
    – Zim
    Commented Jul 31, 2020 at 16:25
  • 1
    $\begingroup$ @Zim Optimization and Nonsmooth Analysis Thm 2.9.8, p. 102 $\endgroup$ Commented Jul 31, 2020 at 17:00

1 Answer 1

2
$\begingroup$

@Zim Optimization and Nonsmooth Analysis Thm 2.9.8, p. 102

$\endgroup$

You must log in to answer this question.

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