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?

  • $\begingroup$ please provide solution for injectivity...I tried solving but I did the same as Kim Jong un ...I don't know how it proves injectivity...it doesn't show x=y . $\endgroup$ – Pranita Gupta Jan 18 at 14:51

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.

  • $\begingroup$ If I understood correctly this is an argument about the injectivity not surjectivity though. $\endgroup$ – Test123 Feb 18 '17 at 10:22
  • $\begingroup$ Does this proves injectivity? I need proof for injectivity... $\endgroup$ – Pranita Gupta Jan 18 at 14:47

By Mean value theorem the condition $|f'(x)|<M$ implies $\frac{f(x)}{x}$ is bounded. So you know that:

$$\lim_{x \to \infty}g(x)=\lim_{x \to \infty}x+af(x)=\lim_{x \to \infty}x+ax\frac{f(x)}{x}=\lim_{x \to \infty}x(a+\frac{f(x)}{x})=\mp \infty$$


$$\lim_{x \to -\infty}g(x)=\pm \infty$$

But $g$ is contionous, so by Intermediate value theorem $g$ is surjective.

  • $\begingroup$ But $f$ is not necessarily bounded. However, we know that $g'(x)\ge 1-\frac aM$ is bounded away from $0$ ... $\endgroup$ – Hagen von Eitzen Aug 26 '14 at 19:26
  • $\begingroup$ @ Hagen: completely agree with you. For example, $f(x)=x$ has $|f'|=1$ but $f$ isn't bounded. $\endgroup$ – Kim Jong Un Aug 26 '14 at 19:29
  • $\begingroup$ Ok, I fixed it. $\endgroup$ – agha Aug 26 '14 at 19:33
  • $\begingroup$ @ agha: Do you use $a<\frac{1}{M}$ somewhere? $\endgroup$ – Kim Jong Un Aug 26 '14 at 19:59
  • $\begingroup$ No, I prove only that $f$ is surjection, not bijection. $\endgroup$ – agha Aug 26 '14 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.