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This is an old pre-lim question I've been working on

if $f(x)$ is continuously differentiable on $(-a,a), f(0)=0$ and $|f'(y)| \leq k < 1$ on $(-a,a)$, Then $\exists \epsilon > 0$ and a unique diff. $g$ on $(-\epsilon, \epsilon)$ s.t. $x=g(x)+f(g(x))$

the only thing I can think of is that x+f(x) should have an inverse by CH 5 exercise 3 of Rudin.

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1 Answer 1

up vote 5 down vote accepted

Define $\phi(x)=x+f(x)$. Then $\phi'(x)=1+f'(x)\geq 1-k>0$. This means that $f$ is increasing, continuous, and therefore invertible with inverse $\phi^{-1}: (c,d)\to(-a,a)$. Since the derivative of $\phi$ is never zero, it follows that the inverse of $\phi$ is also differentiable. The relation is $x=\phi(\phi^{-1}(x))=\phi^{-1}(x)+f(\phi^{-1}(x))$. Therefore, your function $g$ is in fact $\phi^{-1}$, which is a diffeomorphism.

Since $0\in (c,d)$ you can find $\varepsilon$ as you want, such that $(-\varepsilon,\varepsilon)\subset (c,d)$.

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