Chain rule for subdifferential

Question

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!

Answer 1

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.

Answer 2

You may be interested in Theorem 4.19 and Theorem 13.23 (and theorems on various subdifferentials later on) in Introduction to Nonsmooth Analysis and Optimization by Christian Clason and Tuomo Valkonen

Theorem 4.19. Let $F : \to ℝ$ be convex and $\phi : ℝ \to ℝ$ be convex, increasing, and differentiable. Then $\phi \circ F$ is convex, and for all $x \in $,

$$\partial\left[\phi \circ F\right](x) = \phi'(F(x))\partial F(x) = \{\phi'(F(x))x^* | x^* \in \partial F(x)\}$$

Remark 4.20. The differentiability assumption on $\phi$ in Theorem 4.19 is not necessary, but the proof is otherwise much more involved and demands the support functional machinery of Section 13.3.

Theorem 13.23. Let $$ be a separable Banach space, $ : \rightarrow $ be continuously Fréchet differentiable at $ \in $ , and $ : \rightarrow \mathbb{ℝ}$ be locally Lipschitz continuous near $()$. Then,

$$\partial_( \circ )() \subset ′()^*\partial_(()) := \{′()^*^* | ^* \in \partial_(())\}$$ If $$ is regular at $()$, then $ \circ $ is regular at $$, and equality holds.