Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm having a bit of difficulty getting my head around the different types of subgradients we're currently covering in a nonsmooth optimisation class I'm taking.

These subgradients are (assume $x \in$ dom$f$):

  • Proximal subgradient: $\partial_p(f(x)) = \{v\ |\ \exists \delta>0,\rho > 0\ s.t.\ f(y) \geq f(x) + \langle v, y-x\rangle - \frac{1}{2}\rho ||y - x||^2\ \forall\ y \in B_\delta(x) \}$
  • regular subgradient: $\{v\ |\ \exists\ \delta > 0\ s.t. f(y) \geq f(x) + \langle v, y-x\rangle + o(||y-x||)\ \forall\ y\in B_\delta(x) \}$
  • limiting subgradient, defined by $\partial^\infty(f(x)) = \{v\ |\ (v, 0) \in N_{epi f}(x, f(x))\}$, where $N_C(x)$ is the limiting normal cone $\limsup_{x'\rightarrow _C x}\hat{N}_C(x)$ and $\hat{N}_C(x)$ is the normal cone to $C$ at $x$, with $v \in \hat{N}_C(x)$ iff $\langle v, y - x\rangle \leq o(||y-x||)\ \forall\ y\in C$

However, I'm having trouble conceptualising these; for instance, playing with the relatively straightforward function

$f(x) = -|x|$

it seems like all three of these subgradients coincide ($x = 0$ has no subgradient, everything else has $\partial f(x) = \nabla f(x)$ since the limiting normal cone coincides with the normal cone and there exists no $\delta$ for which $B_\delta(0)$ does not intersect $f(x)$), but playing with the less straightfoward

$f(x) = \begin{cases}x^2 \sin(1/x) &\text{if }x \not= 0\\0 &\text{if } x = 0\end{cases}$

just leaves me completely utterly lost; I can't see how the difference between a limiting normal cone and a normal cone makes a difference, I'm struggling to derive the proximal subgradient, etc.

If anyone could provide some intuition for these three concepts, preferably while working through the two functions I've been playing with, that would be fantastic!

share|improve this question
    
If you're going to -1 me, it would be fantastic if you could tell me what the -1 is for so I can correct my mistakes! –  Ben Stott Aug 11 '13 at 11:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.