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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

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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.

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    $\begingroup$ Please provide some context. Are you working in an arbitrary Banach space? $\endgroup$
    – YCor
    Commented Mar 25, 2020 at 11:49
  • $\begingroup$ @YCor $\mathbb R^n$ $\endgroup$
    – wlad
    Commented Mar 25, 2020 at 11:49
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    $\begingroup$ a typo, $x$ in the numerator on the right-hand-side should read $y$ $\endgroup$
    – Carlo Beenakker
    Commented Mar 25, 2020 at 12:22
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    $\begingroup$ 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 $\endgroup$
    – HenrikRüping
    Commented Mar 25, 2020 at 12:39
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    $\begingroup$ @HenrikRüping It could be an inequality instead. Basically, the generalised gradient already satisfies only a weak version of the chain rule $\endgroup$
    – wlad
    Commented Mar 25, 2020 at 16:05

4 Answers 4

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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.

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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.

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    $\begingroup$ I fail to see how this addresses the question in the OP. Can you give more details? $\endgroup$
    – Willie Wong
    Commented Mar 26, 2020 at 4:21
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    $\begingroup$ 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. $\endgroup$
    – Bazin
    Commented Mar 27, 2020 at 22:48
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Even for finite-dimension, you have only an inclusion instead of equality in the chain rule. Basically, naively performing the chain rule by multiplying Clarke subdifferentials can result in a set that strictly contains the Clarke subdifferential of the composition. This is made clear in the book of Clarke.

If you want something like the Clarke subdifferential which admits a chain rule by design, I invite you to read about conservative fields.

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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 $g(x)=y$.

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    $\begingroup$ except I don't think that's true... $\endgroup$
    – wlad
    Commented Mar 25, 2020 at 12:46
  • $\begingroup$ 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 $\endgroup$
    – wlad
    Commented Mar 26, 2020 at 12:59
  • $\begingroup$ I may edit the question to include these examples $\endgroup$
    – wlad
    Commented Mar 26, 2020 at 13:06
  • $\begingroup$ I've included a simple example in the question $\endgroup$
    – wlad
    Commented Mar 26, 2020 at 13:22

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