4
$\begingroup$

I have two functions $g(x)$ and $h(z)$ where $h:\mathbb R^n\to \mathbb R$ and $g: \mathbb R\to \mathbb R$. Both are convex, neither are smooth. How can I apply the chain rule to find $\partial (g\circ h)$? (in terms of $\partial g$ and $\partial h$).

What I have so far: the definition for subdifferential (that's all) $$ \partial(g\circ h)(x) = \{ z : g(h(y)) \geq g(h(x)) + z^T(y-x), \forall y\} $$

If in fact $h$ was smooth, we do have this result: $$ \partial(g\circ h)(x) = \nabla h(x)^T \partial g(h(x)) $$

My guess is that the answer will be something like $\partial(g\circ h)(x)=S$ where $$ S = \{b \cdot a: a\in \partial h(x), b\in \partial g(h(x))\} $$ though I'm not sure...

Also does the problem become easier if I restrict the domain of $g$ to nonnegative scalars and claim that $g$ is monotonic?

Edit: Ok one of the special cases I am thinking of is that $h$ is a norm, and is nonsmooth only at 0, with $h(0)=0$. So, we only need to consider $\partial(g\circ h)(0)$; the rest follows the chain rule for smooth $h$. If, in addition, $g$ is monotonic, then $$ a\in \partial h(0) \iff a^Ty \leq h(y) \forall y. $$ $$ b\in \partial g(0) \Rightarrow b(a^Ty) \leq g(a^Ty) \leq g(h(y)) $$ which gives $S \subseteq \partial (g\circ h)$.

The general case is still open!

$\endgroup$
  • 1
    $\begingroup$ Assuming ${\rm dom}(g) \subseteq [0,+\infty[$ seems to help. Fix $x_0 \in \Bbb R^n$. If $t \in \partial g(h(x_0))$ and $v \in \partial h(x_0)$, then $t\geq 0$ and $tv \in \partial(g\circ h)(x_0)$: $$\langle tv, x-x_0\rangle = t\langle v,x-x_0\rangle \leq t(h(x_0)-h(x)) \leq g(h(x)) - g(h(x_0)).$$this way, $\partial g(h(x_0)) \cdot \partial h(x_0) \subseteq \partial (g\circ h)(x_0)$. I'm not sure about the converse. I have never studied this properly, so this'll stay just a comment. $\endgroup$ – Ivo Terek Nov 22 '17 at 0:55
  • 1
    $\begingroup$ Niiice. I got stuck because your last step, I tried to use monotonicity which doesn't carry over with subtraction, but of course if $t\in \partial g$ then it doesn't matter! Thanks! $\endgroup$ – Y. S. Nov 22 '17 at 1:00
2
$\begingroup$

Check out Corollary 16.72 in the book by Bauschke and Combettes (second edition), which states:

Let $f\colon H\to\mathbb{R}$ be continuous and convex, and let $\phi$ be lower semicontinuous, convex, and increasing on the range of $f$. Suppose that (the relative interior of the range of $f$ + the positive reals) intersected with the relative interior of the domain of $\phi$ is nonempty. Let $\bar{x}$ be in $H$ such that $f(\bar{x})$ is in the domain of $\phi$. Then $$ \partial (\phi\circ f)(\bar{x}) = \left\{ \alpha u \mid| (\alpha,u)\in\partial\phi(f(\bar{x}))\times\partial f(\bar{x})\right\}.$$

So your conjecture is true, with some assumptions. The proof is nontrivial and makes use of coderivatives.

$\endgroup$
  • $\begingroup$ thanks for the pointer! This is actually a pretty general condition, and definitely seems to encapsulate the monotonic and positive domain $\phi$ case. Very nice, thanks! $\endgroup$ – Y. S. Nov 26 '17 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.