Let $E_i$ be a $\mathbb R$-Banach space, $x,y:[0,T]\to E_1$ be continuous, $a\in E_2$, $\alpha:[0,T]\to E_2$ be bounded and Borel measurable and $f:E_1\times E_2\to\mathbb R$ be Fréchet differentiable with $$x(t)-y(t)=\int_{s-\varepsilon}^tf(x(r),a)-f(y(r),a(r))\:{\rm d}r$$ for all $t\in[s-\varepsilon,s]$.
Why are we able to conclude that $$x(s)-y(s)=(f(x(s),a)-f(x(s),\alpha(s)))\varepsilon+o(\varepsilon)\tag1?$$
The claim is made in the last line on p. 114 of this lecture notes. I'm trying to understand this for a while now, but I'm not able to deduce this. Is this an application of Taylor's theorem? Or simply the definition of the Fréchet derivative? Or do we need to assume that $a$ is continuous at $s$?
Too long to be a comment:
I wrote this quite hastily, do not trust it! I'll come back to it. Anybody should feel free to take elements of this draft and turn it into a real answer, then I'll delete this.
We have $$ \Vert x(s)-y(s)-\varepsilon \left( f(x(s),a)-f(x(s),\alpha(s)) \right) \Vert \leq \int_{s-\varepsilon}^s \Vert f(x(t),a)-f(x(s),a)\Vert dt + \int_{s-\varepsilon}^s \Vert f(y(t),\alpha(t))-f(x(s),\alpha(s))\Vert dt.$$ Using continuity the first term is bounded by $$\varepsilon \sup_{t\in [s-\varepsilon;s]} \Vert f(x(t),a)-f(x(s),a)\Vert =o(\varepsilon).$$ For the second term we use $$ \Vert f(y(t),\alpha(t))-f(x(s),\alpha(s))\Vert \leq \Vert f(y(t),\alpha(t))-f(y(s), \alpha(s))\Vert+\Vert f(y(s),\alpha(s))-f(x(s), \alpha(s))\Vert.$$ We only need to show that the expression above is $o(1)$. For the first term this follows from continuity and for the second term we use that $f$ is differentiable to get $$ \Vert f(y(s),\alpha(s))-f(x(s), \alpha(s))\Vert \leq C \Vert x(s)-y(s)\Vert=o(1).$$