I have the function $F(x) = x + f(x)$ where $f(x)$ is a contraction: $|f(x)-f(y)| \leq \alpha|x-y|$ for some $0 < \alpha < 1$ and all $x, y \in \mathbb{R}$

I want to show that $F$ is a bijection:

Proof of injection:

By contradiction suppose $F$ is not injective then $F(x_1) = F(x_2)$ but $x_1 \neq x_2$.

\begin{align*} x_1 + f(x_1) &= x_2+f(x_2) \\ |x_1 - x_2| &= |f(x_2) - f(x_1)| \\ \end{align*} But we have that $|f(x_1)-f(x_2)| \leq \alpha|x-y|$

as $x_1 \neq x_2, |x_1 - x_2| \neq 0 \implies |f(x_1)-f(x_2)| \neq 0$

as $0<\alpha < 1$ then $|f(x_1)-f(x_2)| < |x-y|$ Thus we have a contradiction.

Proof of surjection: Let $y\in\mathbb{R}$. We must show that there exists $x$ such that $F(x)=y$.

Let $y = F(x)$ \begin{align} y = x + f(x)\\ \iff x = y - f(x) \end{align}

We must show that for any $y$ and $f(x)$ we can find an $x$. I am not quite sure how to proceed from here. I think I need to use some properties of $f$ being a contraction, which will tell me that I can find this $x$ for all $f(x)$. Also, is my injective proof correct?

  • $\begingroup$ In your surjection part you must be more careful as you write "we must show there exists $\,x\,\,\,s.t.\,\,\,F(x) = y \;$ . Let $\,y=F(x)\,$..." This doesn't make much sense. $\endgroup$
    – DonAntonio
    Apr 24, 2013 at 2:17

1 Answer 1


I'm assuming that the domain of $f$ is $\mathbb{R}$, so $F$'s domain is also $\mathbb{R}$.

Injection: Your proof seems correct to me, but I think a direct one is more elegant: Suppose $F(x_1)=F(x_2)$. We must show that $x_1=x_2$. Then $x_1+f(x_1)=x_2+f(x_2)$, hence $$|x_1-x_2|=|f(x_1)-f(x_2)|\leq\alpha|x_1-x_2|\leq|x_1-x_2|$$ Therefore inequalities above are equalities, so $|x_1-x_2|=\alpha|x_1-x_2|$, that is, $(1-\alpha)|x_1-x_2|=0$. Since $\alpha<1$, then $1-\alpha\neq 0$, so $|x_1-x_2|=0$, and we conclude that $x_1=x_2$.

Surjection: You should be familiar with the Intermediate Value Theorem: if $g:\mathbb{R}\rightarrow\mathbb{R}$ is a continuous function and $x_1<x_2$ are numbers in $\mathbb{R}$, then $[g(x_1),g(x_2)]\cup [g(x_2),g(x_1)]\subseteq g\left([x_1,x_2]\right)$.

Since $f$ is a contraction, it is continuous, therefore $F$ is also continuous. It's then enough to prove, by the above theorem, that $\lim_{x\rightarrow\infty}F(x)=\infty$ and $\lim_{x\rightarrow -\infty}F(x)=-\infty$.

For every $x\in\mathbb{R}$, one has $-\alpha|x|\leq f(x)-f(0)\leq\alpha|x|$, therefore, for $x>0$, $$F(x)=f(x)+x\geq f(0)-\alpha|x|+x=f(0)+x-\alpha x=f(0)+(1-\alpha)x$$ which shows that $\lim_{x\rightarrow\infty}F(x)=\infty$ (since $1-\alpha>0$). Similarly, for $x<0$, $$F(x)=f(x)+x\leq f(0)+\alpha|x|+x=f(0)+(1-\alpha)x$$ which shows that $\lim_{x\rightarrow-\infty}F(x)=-\infty$.


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