Let $\Omega$ be a bounded open subset with $C^2$ boundary, $f:\mathbb{R}^d\to \mathbb{R}$ be a Lipschitz function, and $u,v:\Omega\to \mathbb{R}^d$ be measurable functions such that $u_i,v_i\in L^2(\Omega)$ for all $i=1,\ldots, d$.
I would like to know whether it makes sense to write: $$ f(u(x))-f(v(x))=\sum_iB_i(x)(u_i-v_i)(x),\qquad \textrm{for a.e. $x\in \Omega$,} \tag{1} $$ with some measurable functions $B_i\in L^\infty(\Omega)$ for all $i$.
One natural candidate of $B_i: \Omega \to\mathbb{R} $ seems to be $$B_i(x)=\int_0^1 \frac{\partial f}{\partial z_i}\big(s(u(x)-v(x))+v(x)\big)\,ds,\qquad\textrm{for a.e. $x\in \Omega$,}\tag{2}$$ with some function $\frac{\partial f}{\partial z_i}\in L^\infty(\Omega)$ for all $i$.
My intuition is that, if we fix $x\in \Omega$, (1) is similar to apply fundamental theorem of calculus to the one-dimensional Lipschitz function $g(s)=f(s(u(x)-v(x))+v(x))$, $s\in [0,1]$. Since $f$ is Lipschitz, we know the weak derivative of $g:[0,1]\to \mathbb{R}$ is defined a.e. and $\frac{\partial g}{\partial s}\in L^\infty(0,1)$: $$ f(u(x))-f(v(x))=\int_0^1 \frac{\partial g}{\partial s}(s(u(x)-v(x))+v(x))\,ds. $$ If we can apply the chain rule and show $B_i$ given in (2) is measurable, then we are done. However I don't know whether the procedure is mathematically correct.
The reason is that in Brezis's book (see Proposition 9.5), the differentiation of a function composition requires $f$ to be $C^1$. Hence I am not sure (2) is valid or not in terms of measure theory.
But actually I don't need $\frac{\partial f}{\partial z_i}$ to be the derivative of $f$, which is only defined almost everywhere. My target is to find Lebesgue measurable $B_i\in L^\infty(\Omega)$ such that (1) is valid.
I found that actually we can define $B_i$ as the difference quotient of each components. For example, if we consider $d=3$. then we can write \begin{align} &f(u_1,u_2,u_3)-f(v_1,v_2,v_3)\\ &=f(u_1,u_2,u_3)-f(v_1,u_2,u_3)+f(v_1,u_2,u_3)-f(v_1,v_2,u_3)+f(v_1,v_2,u_3)-f(v_1,v_2,v_3)\\ &=\sum_{i=1}^d B_i(u_i-v_i), \end{align} where $B_i:\Omega\to \mathbb{R}$ is given as the difference quotients, e.g. $$ B_1(x)=\begin{cases} \frac{f(u_1,u_2,u_3)-f(v_1,u_2,u_3)}{u_1-v_1}(x), &\{x\in \Omega \mid u_1(x)-v_1(x)\not=0\}\\ 0,& \textrm{otherwise}. \end{cases} $$ One can see $B_i$ are measurable, and then Lipschitz of $f$ implies that they are essentially bounded.
Do you think the arguments are correct?