Trying to show that equation has a single solution using Banach space Theorems

Question

How do I show that $f(x) = \int_0^1 e^{-sx}\cos(\alpha f(s))~ds, $ $0\leq x\leq1$, $0\le\alpha\le1$ has a single solution. Using Banach space Theorems like Contraction mapping theorem?

Thanks for your help.

Answer 1

Let $T\colon C^0([0,1])\to C^0([0,1])$. $$ T(f)(x) = \int_0^1 e^{-sx} \cos(\alpha f(s))\, ds. $$ We have $$ \lvert T(f_1)(x)-T(f_2)(x)\rvert \le \int_0^1 e^{-sx}\lvert\cos(\alpha f_1(s))-\cos(\alpha f_2(s))\rvert\, ds $$ Notice that $\cos$ on $[0,1]$ is $L$ lipschitz with $L=\sin(1)$. So $$ \dots \le L\alpha\int_0^1 e^{-sx}\lvert f_1(s)-f_2(s)\rvert\, ds \le L\alpha\lVert f_1-f_2\rVert. $$ Hence $T$ is $L$ lipschitz with $L<1$ hence it is a contraction on a Banach space.