Show that the nonlinear integral equation $$f(x) = \int_0^1 e^{-sx}\cos{(\alpha f(s))}ds,$$ $0\leq x\leq 1$, $0<\alpha<1$, has a unique solution.
I originally thought was some form of Fredholm integral equation, then I could just use a theorem from my textbook, but it's not. The $f(s)$ is inside another function $\cos{(\alpha f(s))}$. I know I have to separate out the kernel of this is equation and do stuff with it but I'm not sure how exactly with this equation. And help/hints would be greatly appreciated.
If we let $Tf(x):=\int_0^1 e^{-sx} (\cos(\alpha f(s))ds$ then \begin{eqnarray} |Tf(x)-Tg(x)|&=&|\int_0^1 e^{-sx} (\cos(\alpha f(s))- \cos(\alpha g(s)))ds| \\ &\le&\int_0^1e^{-sx} |\int_0^1\sin(\alpha(tf(s)+(1-t)g(s))\,\alpha(f(s)-g(s))dt|ds \\ &\le&\alpha \int_0^1e^{-sx} |f(s)-g(s)|ds \end{eqnarray} Depending on which space you intent to work in you can now estimate the right hand side by $$\le \alpha ||f-g||$$ where $||.||$ is, e.g., the $C^0$ norm. Similar estimates for other norms should not be too difficult. So, since $\alpha < 1$, the Banach Fixed point theorem applies.