How to show the slope of converged function of the Contraction Mapping Theorem is bigger in this case?

Suppose I have the following recursive equation that satisfies in every $x\in(0,1)$ : $$v(x) = g(\alpha)\cdot F(x,v(x))$$ where $v(x)$ is a function of $x$, and $g(\alpha)$ is a strictly increasing function with $\alpha \in (0,1)$.

Suppose that, regardless of $\alpha$, the RHS satisfies all the sufficient conditions to apply the Contraction Mapping Theorem in sup norm and that $F$ transforms a strictly increasing function to another.

My guess is that both $v'(x)$ and $v(x)$ are increasing with $\alpha$, but I am struggling to prove it.

Here is how I understood Ali's proof for the case of $v(x)$ increases with $\alpha$ in every $x$, but I think I missed to specify that $F$ is a contraction mapping with sup norm in functional space. Please let me know if I'm the one wrong.

$F$ is a contraction mapping for function to function. Therefore the last step does not necessarily satisfy the definition of contraction mapping on sup norm.

(1) What I want to show is that $v(x)>\tilde{v}(x)$ for any $x$.

(2) For $\alpha < \alpha'$, I get $v(x)= g(\alpha)F(x,v(x)) < g(\tilde{\alpha})F(x,v(x)) $ for every $x$

(3) I define $ v=$ f.p. $T_{\alpha}$ and $\tilde{v}=$ f.p. $T_{\tilde{\alpha}}$.

(3-1) Dropping out $x$ for simplicity, $v =T_{\alpha}(v)=g(\alpha)F(v) < g(\tilde{\alpha})F(v)=T_{\tilde{\alpha}}(v)$ for any $x$

(4) Suppose, on the contrary, $v(x) \geq \tilde{v}(x)$ for some $x$.

(5) $|v(x)-v'(x)| < |T_{\tilde{\alpha}}(v(x)) - T_{\tilde{\alpha}}(\tilde{v}(x)) | $

Up to my knowledge, (5) does not necessarily contradicts the definition of contraction mapping defined on functional space with sup norm : $ sup|T_{\tilde{\alpha}}(v)-T_{\tilde{\alpha}}(\tilde{v})| \leq sup|v-\tilde{v}| $. It is because the contradiction is only for some $x$.

Assume $x$ fixed and write \begin{equation*} T_{\alpha}(u)=g\!\left(\alpha\right)F\!\left(x, u\right). \end{equation*}

To show that $v=\text{f.p.}\,T_{\alpha}$ is a strictly increasing function of $\alpha$ it will be enough that \begin{equation*} \alpha < \alpha' \implies T_{\alpha}(u) < T_{\alpha'}(u) \quad\text{for all }u \end{equation*} because then \begin{equation*} \alpha < \alpha' \implies \text{f.p.}\,T_{\alpha}= T_{\alpha}\left(\text{f.p.}\,T_{\alpha}\right) < T_{\alpha'}\left(\text{f.p.}\,T_{\alpha}\right) \end{equation*} so if for some $\alpha < \alpha'$ it were the case that $ \text{f.p.}\,T_{\alpha'} \le \text{f.p.}\,T_{\alpha}$ then \begin{equation*} \left\vert \text{f.p.}\, T_{\alpha}-\text{f.p.}\,T_{\alpha'}\right\vert < \left\vert T_{\alpha'}\left(\text{f.p.}\, T_{\alpha}\right) -\text{f.p.}\,T_{\alpha'}\right\vert =\left\vert T_{\alpha'}\left(\text{f.p.}\, T_{\alpha}\right) -T_{\alpha'}\left(\text{f.p.}\,T_{\alpha'}\right)\right\vert \end{equation*} which is impossible since $T_{\alpha'}$ is a contraction.

The condition is met with your assumptions plus one additional—that $F\!\left(x, u\right)$ takes only positive values.