# Prove that $f(\mathbb R) = \mathbb R$. Partially solved.

Let $$f:\mathbb R\to \mathbb R$$ and $$f(f(x)) = x + f(x)$$, $$x\in\mathbb R$$

Prove that: a) $$f$$ is injective

b) $$f(0) = 0$$

c) $$f(\mathbb R) = \mathbb R$$

My solution:

a) Let $$x_1, x_2 \in \mathbb R$$ and $$f(x_1) = f(x_2)$$ then

$$f(f(x_1)) = f(f(x_2))$$

$$x_1 + f(x_1) = x_2 + f(x_2)$$

$$x_1 = x_2$$ therefore $$f$$ is injective.

b) Let $$x=0$$ then $$f(f(x)) = x + f(x)$$

$$f(f(0)) = 0 + f(0)$$

$$f(f(0)) = f(0)$$ but $$f$$ is injective therefore

$$f(0) = 0$$

c) Here is the point where I get stuck. How do I solve this?

I think I have to begin saying, let $$y\in\mathbb R$$ and prove that there is a $$x\in\mathbb R$$ for which $$f(x) = y$$.

• I am not giving the answer, it is just a doubt clearing.If I consider an identity mapping i.e. $f(x)=x$, then we have $f(x)=x+x\implies x=2x\implies x=0$....., does this solution hold for the second question?
– user249332
Sep 8 '15 at 21:41
• take an arbitrary $y \in \mathbb{R}$ and show that there exists an $x \in \mathbb{R}$ so that $f(x)=y$ so the image of $f$ is the whole $\mathbb{R}$. Probably some reduction ad absurdum (mprr...) will be needed Sep 8 '15 at 21:41
• If we could show that $f$ is additive, then $x=f(f(x))-f(x)=f(f(x)-x)$. -- And you didn't forget an important condition such as continuity, did you? (That would trvialize part c as $f$ is unbounded) Sep 8 '15 at 21:41
• Two solutions for $f$ are $f(x)=kx$ where $k^2 =k+1$. These are the only polynomial solutions for $f$. Are there others? Sep 8 '15 at 23:24
• A remark: By induction, $f^{\circ n}(x)=F_nf(x)+F_{n-1}x$ for $n\ge 0$ ($F_n$ the Fibonacci numbers) Sep 9 '15 at 6:04

It works if you assume $f(x)$ is continuous.

Claim: Suppose $f(x)$ is continuous and $f(f(x)) = x + f(x)$ for all $x \in \mathbb{R}$. Then $f(\mathbb{R})=\mathbb{R}$.

Proof: We already know that $f(0)=0$ and $f(x)$ is injective. It follows that $f(1)\neq 0$.

-Case 1: Suppose $f(1)>0$. Then (by the intermediate value theorem) we must have: \begin{align} &f(x) > 0 \: \: \mbox{ if $x >0$} \\ &f(x) < 0 \: \: \mbox{ if $x<0$} \end{align}

For all positive integers $n$ we have: \begin{align*} &f(f(n)) = n + f(n) \geq n\\ &f(f(-n)) = -n + f(-n) \leq -n \end{align*} and so $f$ takes arbitrarily large values and arbitrarily small values. By the intermediate value theorem, it must take all values in $\mathbb{R}$.

-Case 2: Suppose $f(1)<0$. Then we must have:

\begin{align} &f(x) < 0 \quad \mbox{ whenever $x >0$} \\ &f(x) > 0 \quad \mbox{ whenever $x<0$} \end{align}

Suppose there is a finite constant $-M$ such that $f(x) \in [-M,0]$ for all $x\geq 0$ (we reach a contradiction). Then the infinite sequence $\{f(n)\}_{n=1}^{\infty}$ is in the compact interval $[-M,0]$, so the Bolzano-Wierstrass theorem ensures there is a subsequence $n_k$ such that $n_k\rightarrow\infty$ and $f(n_k) \rightarrow x^*$ for some $x^* \in [-M,0]$. But for all $k$ we have: $$f(f(n_k)) = n_k + f(n_k)$$ and taking a limit as $k\rightarrow \infty$ gives $f(x^*) = \infty + x^*$, a contradiction. Thus, $f(x)$ takes arbitrarily small values over $x\geq 0$.

A similar argument shows $f(x)$ takes arbitrarily large values over $x \leq 0$. Hence, by continuity, $f(\mathbb{R}) = \mathbb{R}$.

• Case 2 can only occur if $f(x)=-\frac1\phi x$: By induction $f^{\circ n}(x_0)=F_nf(x_0)+F_{n-1}x_0$, so $\frac1{F_{n-1}}f^{\circ n}(x_0)=\frac{F_n}{F_{n-1}}f(x_0)+x_0\to \phi f(x_0)+x_0\ne 0$ if we pick $x_0$ with $f(x_0)\ne -\frac1\phi x_0$. Hence for large $n$, the numbers $x:=f^{\circ n}(x_0)$ and $f(x)$ have the same sign. For continuous injective $f$ with $f(0)=0$ this means $\operatorname{sgn}f(x)=\operatorname{sgn}x$ for all $x$. Sep 10 '15 at 5:49
• To continue my previous comment: Assume $\operatorname{sgn}f(x)=\operatorname{sgn} x$ (e.g., $f$ continiuous and not $x\mapsto -\frac1\phi x$). Then $\lim \frac 1{F_{n-1}}f^{\circ n}(x)=\phi f(x)+x$has the same sign as $x$, so that $f$ is not bounded from below and not bounded from above. Sep 10 '15 at 6:01
• @HagenvonEitzen : Nice observation about Fibonacci. I assume you mean $\phi = \frac{1+\sqrt{5}}{2}$. So if $f(1)<0$ and $f$ is continuous, then $f(x) = (-1/\phi)x$. Sep 11 '15 at 17:26

The solution (given by my teacher).

Let $y\in \mathbb R$ I'll prove that there is a $x \in \mathbb R$ for which $f(x) = y$

$f(x) = y$

$f(f(x)) = f(y)$

$x + f(x) = f(y)$

$x + y = f(y)$

$x = f(y) - y$

therefore $f(\mathbb R) = \mathbb R$

• Can you do these steps in reverse, or did you actually need to start with $f(x)=y$? Because starting with $f(x)=y$ seems to be assuming what you want to prove.
– Ian
Sep 9 '15 at 15:25
• You want to prove the existence of $x$ for which $f(x) = y$ is valid. Therefore you take $f(x) = y$ for granted and prove that it actually has a solution, which is $f(y) - y$. I think that this is the logic, I remind you, it isn't my solution but my teacher's. Sep 9 '15 at 15:30
• As we were discussing in the comments yesterday, this is really "scratch work", rather than a proof. By that I mean that you started with what you want and you reverse engineered a setup in which it should work. But then you need to start from your setup and prove that it does what you want. So this amounts to running through your steps in reverse, but not using any step in which you replace $f(x)$ by $y$ or vice versa. And doing that I run into trouble (but maybe I did some algebra wrong).
– Ian
Sep 9 '15 at 15:36
• That just shows that if there is an $x$ such that $f(x) = y$, then $x=f(y) -y$. The same working may work in reverse though, to prove if $x = f(y) - y$, then $f(x) = y$. Sep 9 '15 at 15:36
• My algebra: $x=f(y)-y$ so $x+f(x)=f(y)-y+f(x)$ so $f(f(x))=f(y)-y+f(x)$. But now what? This is still consistent with $f(x)=y$ (if you plug that in you get $f(y)=f(y)$), but it does not show why that must occur.
– Ian
Sep 9 '15 at 15:40