Show $e^{f(x)} = 4x$ has a unique solution using Banach fixed point theorem

Question

I am stuck trying to prove the following:

Let $f: \mathbb{R} \to [0,1]$ be a contractive map. Use the Banach contraction principle to show that the equation $$e^{f(x)} = 4x$$ has a unique solution.

In order to use the principle, I need to restrict the domain to $[0,1]$. Then there is some unique $\overline{x} \in [0,1]$ such that $f(\overline{x}) = \overline{x}$.

The problem I am having is trying to derive the right hand side, since $$e^{f(\overline{x})} = e^\overline{x}$$ does not seem to help much.

Answer 1

You're looking for a fixed point of $e^{f(x)}/4$, what's a Lipschitz constant for that in terms of the Lipschitz constant for $f$? (Hint: you can use the derivative of the exponential itself even if $f$ isn't differentiable.)

Answer 2

You want to prove that $e^f(x)/4$ is contractive.

Hint: use the mean value theorem for the exponential function and look at the range for $f$.