Let $(f_n)$ be a sequence of functions defined by $f_0(x)=\sin x $ and
$$f_n(x)=\int_{0} ^{x}F(x,f_{n-1}(t))dt$$ where $F\in C^1(\mathbb{R}\times\mathbb{R})$ and $|\partial_yF(x,y)|\leq k<1$ for all $x,y\in\mathbb{R}$.
How can I show $(f_n)$ converges uniformly on $[0,1]$?
I tried to differentiate both sides with respect to $x$,
$$f'_n(x)=F(x,f_{n-1}(x))+\int_{0}^{x}\partial_xF(x,f_{n-1}(t))dt$$
Actually, I'm not sure that above equation holds.
Even if it is correct, I can't use the given partial condition $\partial_yF(x,y)$
Any help would be appreciated.
Best Answer
Hint: Use the bound on $\partial_y F$ to prove that the map $$\Phi:(C^0[0,1],\lVert \bullet\rVert_\infty)\to (C^0[0,1],\lVert \bullet\rVert_\infty)\\ \Phi(g)=\int_0^x F(x,g(t))\,dt$$ is a contraction.