Let $f:\mathbb{R}\rightarrow \mathbb{R}$ a $C^1$ function with $|f'(x)|<M$ for all $x$. Then I want to show that there exists an $a>0$ such that $g:\mathbb{R}\rightarrow \mathbb{R}$, $g(x)=x+af(x)$ is a bijection with differentiable inverse.
For $a<1/M$ I have shown that it is strictly monotone, which implies it's injective and that it has a differentiable inverse. How can I show that is surjective?
Best Answer
Another look at surjectivity: suppose $y>x$ is such that $g(x)=g(y)$. Then $$ y+af(y)=x+af(x)\implies(y-x)=a(f(x)-f(y))=a(x-y)f'(z) $$ for some $z\in(x,y)$. Taking absolute values of both sides: $$ |y-x|=a|y-x||f'(z)|\implies 1=a|f'(z)|. $$ But $a|f'(z)|<\frac{1}{M}M=1$ so you have the desired contradiction.