I have a complete metric space $(S,||_{\infty})$ where $S$ is a space of bounded and Lipschitz continuous function that is defined in $x \in [0,1]$.
Suppose I have $s \in S$ satisfying the following equation :
$$ s(x) = \alpha \cdot (F\circ s) (x)$$ for all $x\in[0,1]$ where $\alpha \in (0,1)$ and $F:S\rightarrow S$.
One thing that I know is $ T(\alpha,s) \equiv \alpha \cdot (F\circ s) : (0,1) \times S \rightarrow S$ is a contraction mapping satisfying
$$ ||T(\alpha,u) - T(\alpha,v) ||_{\infty} < \beta || u-v||_{\infty} $$
for any $u,v \in S$ and $\alpha \in (0,1)$ where $\beta \in (0,1)$.
I am trying to get sufficient conditions that $s(x)$ and $s'(x)$, which is defined almost everywhere by Lipschitz continuity, are both differentiable at least at some $x$ with respect to $\alpha$.
My initial approach was to use contraction mapping theorem with parameter, but I still think that it's overshooting.
I could find some references, but, unlike my problem where the parameter is just simply multiplied in front of $F$, it discusses in any $T(\alpha,s)$. Having very scant knowledge about functional analysis and topological space, I was wondering if there would be a more straightforward way to show the differentiability.
Assuming Fréchet derivative $A=DF(s)$, it follows from $s= \alpha F(s)$ that we need \begin{equation*} \frac{ds}{d\alpha} = F(s)+ \alpha A \frac{ds}{d\alpha} \qquad\qquad (*). \end{equation*} Now for any $u$ $$\lVert F(s+\lambda u) - F(s) \rVert \le \lVert F(s+\lambda u)- F(s)-\lambda A u\rVert + \lVert \lambda A u \rVert $$ and as a result \begin{equation*} \frac{\lVert F(s+\lambda u) - F(s) \rVert}{\lambda \lVert u\rVert} -\frac{\lVert Au \rVert}{\lVert u\rVert} \le \frac{\lVert F(s+\lambda u)- F(s)-\lambda A u\rVert}{\lambda \lVert u\rVert}. \end{equation*} The RHS tends to $0$ as $\lambda\to 0$ (definition of a Fréchet derivative) so \begin{equation*} \frac{\lVert Au \rVert}{\lVert u\rVert} = \lim_{\lambda\to 0} \frac{\lVert F(s+\lambda u) - F(s) \rVert}{\lambda \lVert u\rVert} \le \frac{\beta \lVert s+\lambda u - s\rVert}{\lambda \lVert u\rVert} \le 1. \end{equation*} Therefore \begin{equation*} \lVert A \rVert = \sup_{u} \frac{\lVert A u \rVert}{\lVert u \rVert}\le 1. \end{equation*} Now let \begin{equation*} \phi(v)=F(s)+\alpha A v \end{equation*} which is itself a contraction mapping since \begin{equation*} \lVert \phi(v_{1}) - \phi(v_{2}) \rVert =\lVert \alpha A(v_{1}-v_{2})\rVert \le \alpha \lVert A \rVert \lVert v_{1}-v_{2} \rVert \le \alpha \lVert v_{1}-v_{2} \rVert. \end{equation*} Its fixed point is the unique fn to satisfy $(*)$ so we conclude \begin{equation*} \frac{ds}{d\alpha}=\text{f.p. }\phi. \end{equation*}