# APMO 89 Functional Equation

Determine all functions $f$ from the reals to the reals for which

(1) $f(x)$ is strictly increasing and
(2) $f(x) + g(x) = 2x$ for all real $x$,

where $g(x)$ is the composition inverse function to $f(x)$. (Note: $f$ and $g$ are said to be composition inverses if $f(g(x)) = x$ and $g(f(x)) = x$ for all real $x$.)

Let us define a sequence $\{a_n\}_{0}^{\infty}$,where $a_0=x$ and $a_{n+1}=f(a_n)$.
The problem now translates to $$a_{n+1}+a_{n-1}=2a_n$$
The characteristic polynomial is
$$x^2-2x+1$$
whose roots are $1$ and $1$.Thus,
$$a_n=P\cdot n(1^n)+Q\cdot(1^n)$$Since $a_0=x$, then $Q=x$. $\color{red} {Hence, f(x)=a_1=x+c, for \;some \;constant \;c.}$

EDIT: As pointed out in the comments, $P$ cannot be be concluded to be a constant and the line in red color above is wrong.How can I proceed from here?

• Thanks for pointing out,I will edit it – user362405 Apr 17 '17 at 12:53
• Is $f^{-1}$ a function from $\Bbb R$? – DHMO Apr 17 '17 at 12:56
• Since $f$ is from$\Bbb R \to$$\Bbb R$,$f^{-1}$ should also be function from $\Bbb R$ – user362405 Apr 17 '17 at 13:06
• There was another wording of the same problem..I have changed to it for better clarity.PLease let me know if there are any ambiguities – user362405 Apr 17 '17 at 13:09
• You never used the fact that $f$ is increasing. There is a simple subtle issue with your solution: since you fix $x$ for the problem, your constant $c$ at the end of the problem depends on $x$, so it is not a constant. – N. S. Apr 17 '17 at 13:15

Suppose there exists a $y$ with $f(y)\ne y$. Then, since $f$ satisfies the condition in question, $f-c$ also satisfies it, we can suppose that $y=0$.
Let $f(0)=x$. Then one can see $f(x)=2x,f(2x)=3x$, and generally $f(mx)=(m+1)x$ for any positive or negative integer $m$. WLOG, suppose $x$ is positive.
Now suppose there exists $a$ such that $0<a<x$ and $f(a)\ne a+x$. $f(a)>f(0)=x>a$, so $f(a)-a$ is positive. By the property of $f$, $f(a+m(f(a)-a))=a+(m+1)(f(a)-a)$ for all integer $m$. Also, since $f(a)-a \ne x$, there exists smallest integer $N$ such that $a+N(f(a)-a)>(N+1)x$ or $a+(N+1)(f(a)-a)<Nx$. In both cases, we get contradiction on strictly increasing property of $f$ because $a+(N-1)(f(a)-a)<Nx$ and $f(a+(N-1)(f(a)-a))>f(Nx)$, or $a+N(f(a)-a)>(N-1)x$ and $f(a+N(f(a)-a))<f((N-1)x)$.
Therefore, every $0<a<x$ satisfies $f(a)=a+x$ and it is easy to see that $f(a)=a+x$ for all real number $a$.
• What if $f(0)=0$? You ruled out the case where $f=id$, but not the case where $f(0)=0$... – DHMO Apr 17 '17 at 14:18
• I think first paragraph covers the case $f\ne id$ and $f(0)=0$... – didgogns Apr 17 '17 at 14:21