I have a problem, that equation $x(t) = t + \varepsilon x(t^{a})$, where $|\varepsilon| < 1, a > 0$, has only one solution in space $C[0,1]$.
I know, that it is about using the contraction theorem, so I checked all needed conditions of that theorem. I just have to do the rest part: prove that $||A|| < 1$, where $A$ is an operator: $A: C[0, 1] \rightarrow C[0, 1]$. I tryed some examples of $A$, like: $A(x) = \frac{x - t}{\varepsilon}$, but haven't reached sufficient results.
Can anyone give me a hint, how to solve this problem?
If you want to use contraction theorem, you want $A$ such that $x^*$ that satisfies your equation is fixed point of $A$, i.e. $(Ax^*)(t) = x^*(t)$. That only leaves you with one choice of $A$, i.e. $(Ax)(t) = t + \varepsilon x(t^a)$. Such $A$ is not linear, but it is affine, so it will work out fine because you want to calculate $$\|Ax-Ay\| = \sup_{t\in[0,1]}|(t+\varepsilon x(t^a))-(t+\varepsilon y(t^a))| = |\varepsilon|\sup_{t\in[0,1]}|x(t^a)-y(t^a)|. $$
To finish up, note that $t\mapsto t^a$ is bijection on $[0,1]$, so the second supremum is equal to...?