Net change expression that smells like gradient theorem

Question

I have run across the following in a proof.

Given a continuous function $\;f : \mathbb{R}^n \times \mathcal{D}_f \times \mathcal{I}_f \rightarrow \mathbb{R}^m$ where $\mathcal{D}_f \subseteq \mathbb{R}^m$ is an open domain, $\mathcal{I}_f \subseteq \mathbb{R}$ an interval, $R(t) \in L(\mathbb{R}^n)$ a linear mapping, and $\eta := (I - R(t))y$

$f(y,x,t) - f(R(t)y,x,t) = \int_0^1 f_y(sy + (1-s)R(t)y, x, t)\eta \, ds$

I have not been able to figure out how they came up with this expression. It smells like the Gradient theorem, but how to get from the LHS to the RHS given the standard definition of the theorem? Maybe I am barking up the wrong tree here?

Answer 1

Let $y\mapsto g(y)$ be a scalar function of the vector variable $y$, and assume that the segment $[a,b]$ is contained in the domain of $g$. We want to compute the quantity $Q:=g(b)-g(a)$. To this end introduce the auxiliary function $$\phi(s):=g\bigl(s b+(1-s)a\bigr)\qquad(0\leq s\leq1)\ .$$ Then $$Q=\phi(1)-\phi(0)=\int_0^1\phi'(s)\>ds\ .$$ Now by the chain rule $$\phi'(s)=\nabla g\bigl(s b+(1-s)a\bigr)\cdot (b-a)\ ,$$ so that we obtain $$Q=\int_0^1\nabla g\bigl(s b+(1-s)a\bigr)\cdot (b-a)\>ds\ .\tag{1}$$ Now your problem is just an uploaded version of this: Instead of a scalar function $g$ you have a vector-valued function $f$. There are additional irrelevant variables $x$ and $t$, hence the author writes $f_y$ to idicate that the gradient with respect to $y$ is meant. Then $b:=y$ and $a:=R(t)y$. Finally $f_y$ is interpreted as a linear map operating on $\eta=b-a$, whereas in $(1)$ we had a scalar product.