I have a complete metric space $(,||_{\infty})$ where $$ is a space of bounded and continuously differentiable function that is defined in $x∈[0,1]$ .
Suppose I have $v∈$ satisfying the following equation :
$$ v(x) = F(x) ⋅v(f())$$ for all $∈[0,1]$
One thing I know is that the mapping $T$ of $( v)(x) \equiv F(x) ⋅v(f())$ is a contraction mapping in the space.
What I'm interested in is to prove if the derivative of $v(x)$ also satisfies the contraction mapping. Assuming $F(x)$ is continuously differentiable, if I take the derivative on both sides :
$$v'(x) = F'(x)\cdot v(f(x)) + F(x) v'(f(x)) f'(x) $$
The way I'm trying to approach the problem is as follows :
- Define $g(x) \equiv v'(x)$ and the fixed-point function of the original problem as $v^{*}(x)$. Then, the equation can be written by :
$$g(x) = F'(x) v^{*}(f(x)) + F(x) g(x)f'(x)$$
- If I prove that the mapping $\widetilde{T}$ of $(\widetilde{T} g) \equiv F'(x) v^{*}(f(x)) + F(x) g(x)f'(x)$ satisfies all the conditions of the contraction mapping theorem, then the fixed-point function $g^{*}(x)$ is $v'(x)$.
I wondered if this sounds plausible. One of my concerns is that if I can really plug in $v^{*}(x)$ in the first step. Should it be replaced by $\int g(x)dx$ so that it changes while doing iteration convergence?
Any comments or suggestions would be really helpful.