Does the generalised directional derivative satisfy any version of the chain rule?

Is there a chain rule of any kind for the generalised directional derivative (of the Clarke type)? There is certainly a chain rule for the generalised gradient.

The generalised directional derivative is: $$f^\circ(x;v)=\limsup_{y \to x, t \downarrow 0} \frac{f(y+tv) - f(y)}{t},$$ where $$x,v \in \mathbb R^n$$ for some $$n$$, and $$f:\mathbb R^n \to \mathbb R^m$$. Albeit, the definition is valid over any Banach space (but I'd like to keep things to finite dimensions for simplicity's sake).

Some information about it can be found here: https://www.encyclopediaofmath.org/index.php/Clarke_generalized_derivative

The naive version of the chain rule is false: Consider $$f(x) = |x|, g(x) = -x$$. We have that $$(f\circ g)^\circ(0;1) = 1$$ while $$f^\circ(g(0); g^\circ(0;1)) = f^\circ(0;-1) = -1$$. What I'm looking for must therefore be an inequality.

• Please provide some context. Are you working in an arbitrary Banach space? – YCor Mar 25 at 11:49
• @YCor $\mathbb R^n$ – ogogmad Mar 25 at 11:49
• a typo, $x$ in the numerator on the right-hand-side should read $y$ – Carlo Beenakker Mar 25 at 12:22
• So if $m=n=1$ and $f$ is a function whose slope alters between $+1$ and $-1$ faster and faster as $x$ goes to zero, we get that $f^\circ(0)=1$ since close to one, there is always a point with slope $+1$ and we get $(-f)^\circ(0)=1$, since the same is true for $-f$. Doesnt that contradict the chai rule – HenrikRüping Mar 25 at 12:39
• @HenrikRüping It could be an inequality instead. Basically, the generalised gradient already satisfies only a weak version of the chain rule – ogogmad Mar 25 at 16:05

A nice and non-trivial extension of the chain-rule occurs from the DiPerna-Lions theory of rough vector fields: take on an open subset $$\Omega$$ of $$\mathbb R^n$$ a vector field $$X$$ with $$L^\infty_{loc}(\Omega)$$ coefficients and null divergence such that $$X\in W^{1,1}_{loc}(\Omega)$$. Let $$u$$ be an $$L^\infty_{loc}(\Omega)$$ function such that $$Xu\in L^1_{loc}(\Omega)$$. Then $$X(u^2)=2u Xu.$$ The previous chain-rule formula is the main point to prove uniqueness of weak solutions for these vector fields. There are generalizations in several directions: instead of looking at $$u^2$$, you may check $$X(F(u))=F'(u) Xu, \quad F\in C^1.$$ Also, you can relax the regularity $$W^{1,1}$$ to $$BV$$, play a bit with regularity $$W^{1,p}$$ and relax as well the condition on the divergence by requiring only absolute continuity wrt Lebesgue measure.

• I fail to see how this addresses the question in the OP. Can you give more details? – Willie Wong Mar 26 at 4:21
• Well the chain rule is about finding the derivative of $F\circ u$. My remark above is highlighting the fact that when you have a limited amount of regularity, say when the function $u$ is not better than $L^\infty$, if you have the weak information that $Xu$ belongs to $L^1$, you can still apply the chain rule and this is the core argument for the DiPerna-Lions theory for first-order linear equations with $W^{1,1}$ coefficients. – Bazin Mar 27 at 22:48

According to theorem 8.14 of https://arxiv.org/pdf/1708.04180.pdf, we have that for locally Lipschitz $$g:Y\to \mathbb R$$ and Frechet-differentiable $$f: X \to Y$$, that

$$(g\circ f)^\circ(x;v) \leq g^\circ(f(x);f'(x)v).$$

Equality holds if $$g$$ is regular.

We can furthermore say that:

$$(g\circ f)^\circ(x;v) \geq -g^\circ(f(x);-f'(x)v)$$ under the same conditions.

The straightforward generalization of the usual chain rule would give $$(f\circ g)^\circ(x,v)=v\cdot\bigl(Dg(x)\bigr)^{\rm T}\cdot\bigl(\nabla f(y)\bigr),$$ with $$Dg$$ the Jacobian matrix and $$g(x)=y$$.

• except I don't think that's true... – ogogmad Mar 25 at 12:46
• consider $f(x) = |x|$, $g(x) = -|x|$, $h(x) = f(x) + g(x)$. We have $f^\circ(0;1) = 1$, $g^\circ(0;1) = 1$, while $h^\circ(0;1) = 0$. This contradicts the addition rule, and therefore your naive version of the chain rule – ogogmad Mar 26 at 12:59
• I may edit the question to include these examples – ogogmad Mar 26 at 13:06
• I've included a simple example in the question – ogogmad Mar 26 at 13:22