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A lot of results are available for the following chain-rule problem:

(CRP1) Let $f\colon \mathbb R \to \mathbb R$ be a $C^1$/Lipschitz function and let $g \colon \mathbb R^d \to \mathbb R$ be a weakly differentiable function (e.g. $W_{\rm loc}^{1,p}$ or $BV_{\rm loc}$). Then the function $f \circ g$ is weakly differentiable as well and explicit chain rule formulas hold, like for instance in the Sobolev setting $$ (f \circ g)'(x) = f'(g(x)) g'(x) $$ a.e. with respect to Lebesgue measure (with some standards caveat when $f$ is Lipschitz).

I am wondering for the other way round, i.e.

(CRP2) Let $f\colon \mathbb R \to \mathbb R^d$ be a $C^1$/Lipschitz function and let $g \colon \mathbb R^d \to \mathbb R$ be a weakly differentiable function (e.g. $W_{\rm loc}^{1,p}$ or $BV_{\rm loc}$). What can we say about the function $g \circ f \colon \mathbb R \to \mathbb R$? For instance in the Sobolev setting it seems to me that the formula $$ (g \circ f)'(x) = \nabla g(f(x)) \cdot f'(x) $$ (a.e. with respect to Lebesgue measure) makes sense, doesn't it? Are there any references about this topic?

Thanks.

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Your formula can be wrong even if $f$ and $g$ are both Lipschitz. For criteria when such a result holds (and related results) see e.g. Leoni, Giovanni, Morini, Massimiliano: Necessary and sufficient conditions for the chain rule in $W^{1,1}_{loc}(ℝ^N;ℝ^d)$ and $BV_{loc}(ℝ^N;ℝ^d)$. J. Eur. Math. Soc. (JEMS) 9 (2007). https://mathscinet.ams.org/mathscinet-getitem?mr=2293955

They give the counterexample $g(y_1,y_2)=\max(y_1,y_2)$, $f(x)=(x,x)$, where the right hand side is nowhere defined as $f'(x)=(1,1)$ everywhere but $g$ is not differentiable at $y_1=y_2$.

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Here is a related example from:

P. Hajlasz, Sobolev mappings: Lipschitz density is not a bi-Lipschitz invariant of the target. Geom. Funct. Anal. 17 (2007), 435-467.

Theorem. There is a Lipschitz function $\varphi\in {\rm Lip}\, (\mathbb{R}^2)$ with compact support such that the bounded operator $\Phi:W^{1,p}([0,1],\mathbb{R}^2)\to W^{1,p}([0,1])$ defined as composition $\Phi(u)=\varphi\circ u$ is not continuous for any $1\leq p<\infty$.

The operator is bounded in the sense that the Sobolev norm of $\varphi\circ u$ is bounded by constant times that of $u$. However, the operator is not continuous as a mapping between Banach spaces $W^{1,p}([0,1],\mathbb{R}^2)$ and $W^{1,p}([0,1])$.

However as was proved in

M. Marcus,V. J. Mizel, Every superposition operator mapping one Sobolev space into another is continuous. J. Funct. Anal. 33 (1979), 217-229,

the composition operator in the case in which $\varphi$ is a Lipschitz function on $\mathbb{R}$ is continuous.

Another reference for understanding the chain rule for Sobolev and BV functions is:

Ambrosio, L.; Dal Maso, G. A general chain rule for distributional derivatives. Proc. Amer. Math. Soc. 108 (1990), no. 3, 691-702.

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