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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?

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    $\begingroup$ 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 '19 at 14:51
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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.

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  • $\begingroup$ If I understood correctly this is an argument about the injectivity not surjectivity though. $\endgroup$ – Kal S. Feb 18 '17 at 10:22
  • $\begingroup$ Does this proves injectivity? I need proof for injectivity... $\endgroup$ – Pranita Gupta Jan 18 '19 at 14:47
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By Mean value theorem the condition $|f'(x)|<M$ implies $\frac{f(x)}{x}$ is bounded. So you know that:

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

and

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

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

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  • $\begingroup$ 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
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Choose a positive $a<{1\over M}$. Then $m:=1-aM>0$. Since $$g'(x)=1+af'(x)>1-a M=m\qquad(-\infty<x<\infty)$$ the function $g$ is monotonically increasing, hence injective on ${\mathbb R}$. Furthermore $$g(x)=g(0)+\int_0^x g'(t)\>dt>g(0)+mx\to\infty\qquad(x\to\infty)\ ,$$ and similarly $\lim_{x\to-\infty}g(x)=-\infty$. It follows that $g$ maps ${\mathbb R}$ bijectively onto ${\mathbb R}$. As $g'(x)>0$ for all $x$ the inverse map $\>g^{-1}\!\!:\>{\mathbb R}\to{\mathbb R}$ is $C^1$ again.

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