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I have been searching for an analogue of chain rule for Sobolev functions. While in general, I could not find chain rule for composition for arbitrary functions (and neither do I expect such a result to hold), I found one result for the composition of a Lipschitz continuous function with a Sobolev function. I state it verbatim from here$^1$.

Let $f: \mathbb{R}^d \rightarrow \mathbb{R}$ be a Lipschitz continuous function. Then, for every $u \in W^{1, 1}_{loc} \left( \mathbb{R}^N, \mathbb{R}^d \right)$, the composite function $v = f \circ u$ is in $W^{1, 1}_{loc} \left( \mathbb{R}^N \right)$ and for a.e. $x \in \mathbb{R}^N$ the restriction of $f$ to the affine space $$T^u_x := \left\lbrace w \in \mathbb{R}^d | w = u (x) + \nabla u(x)z, \text{ for some } z \in \mathbb{R}^N \right\rbrace$$ is differentiable at $u(x)$ and $$\nabla (f \circ u ) (x) = \nabla_u (f|_{T^u_x}) (u(x)) \nabla u(x).$$

Here, $\nabla$ is the derivative of the function.

I cannot see how does the right hand side always make sense? If we look closely, $T^u_x$ need not be full-dimensional. How do we understand $\nabla_u(f|_{T^u_x})$? Moreover, how does one compose the two derivatives if they are not defined on appropriate dimensional spaces?

Any insights into this would be appreciated!

${}^1$: Leoni, Giovanni; Morini, Massimiliano, Necessary and sufficient conditions for the chain rule in $W^{1,1}_{loc}(\mathbb R^N;\mathbb R^d)$ and $BV_{loc}(\mathbb R^N;\mathbb R^d)$. J. Eur. Math. Soc. (JEMS) 9, No. 2, 219-252 (2007). Zbl 1135.26011, doi:10.4171/JEMS/78

Calvin Khor
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Aniruddha Deshmukh
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  • It seems that this result is from citation 2 which is a paper of Ambrosio and Dal Maso. In addition, citation 19 proves a slightly weaker result. So checking these papers may help in understanding this statement – Calvin Khor Feb 09 '23 at 10:22
  • The protoypical chain rule for Sobolev spaces can be seen in, e.g., this question https://math.stackexchange.com/questions/1110231/chain-rule-in-the-sobolev-space-w1-p - does this help? – daw Feb 09 '23 at 10:26
  • @CalvinKhor: Even in the reference, it is quite unclear how the derivative of the restriction is taken. The same question arises about the composition of the derivative operators. – Aniruddha Deshmukh Feb 12 '23 at 02:41
  • @daw: Yes, the prototypical chain rule is not new. However, the problem I posed was not about finding a chain rule in the Sobolev spaces. That was only my motivation to ask the question. The main question is about the result I have mentioned here, and about the composition of the derivative operators. – Aniruddha Deshmukh Feb 12 '23 at 02:42
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    @AniruddhaDeshmukh yes, my only point is that you are quoting a result that is not proven in the cited paper, but in a further citation, so there is some chasing to do. In any case, the dimension of $T^u_x$ is precisely the rank of $\nabla u(x)$. So I would guess that that $f|{T^u_x} $ has a derivative at the point $u(x)$ in sense that the map $ \nabla_uf|{T^u_x}(u(x))$ is a linear map from the range of $\nabla u$ (really, from the tangent space of $T^u_x$) to $\mathbb R$, and can be defined e.g. as $$\nabla_uf|{T^u_x}(u(x)) h = \lim{t\downarrow 0} \frac1t(f(u(x)+th) -f(u(x))$$ – Calvin Khor Feb 12 '23 at 06:40

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