Let $x,y\in\mathbb{R}^n$, $n>0$. Suppose we have an arbitrary nonlinear mapping $T(x):\mathbb{R}^n\to\mathbb{R}^n$, that basically performs a nonlinear coordinate transformation for which we assume that $\frac{\partial T}{\partial x}$ has full rank for all $x$ (so it is a 'valid' transformation, so to speak). My question is; Does there always exists a $Q:\mathbb{R}^n\times \mathbb{R}^n\to\mathbb{R}^{n\times n}$, such that for all $x,y$ $$T(x)-T(y)=Q(x,y)(x-y).$$ And if not, what are the conditions on $T$, such that the above holds?
For $n=1$, it is quite easy, i.e. $Q(x,y)=\frac{T(x)-T(y)}{x-y}$, but for higher dimensions, this is a bit more tricky...
Yes, if $T:\mathbb{U}\rightarrow R^n$ is continuously differentiable on $\mathbb{U}$ and $\mathbb{U}$ is open, it holds that
$$ T(x)-T(y)= \underbrace{\left(\int_0^1 \frac{\partial T}{\partial x}(y+t(x-y))\,d t\right)}_{Q(x,y)}(x-y)$$ for $x,y\in\mathbb{U}$, see Lemma 1 here.