# Abstract Nonsense

## Functional Equation Problem

Point of post: In this post I will answer the question

” What functions are continuous at $0$ and satisfy

$\displaystyle f(x)=f\left(\frac{x}{1-x}\right),\quad x\ne -1$

Problem: Find all functions $f:\mathbb{R}\to\mathbb{R}$ which are continuous at $0$ and satisfy

$\displaystyle f(x)=f\left(\frac{x}{x-1}\right)$

Proof: Note first that if

$\displaystyle g(x)=\frac{x}{1-x}$

that

$\displaystyle g\left(\frac{1}{n}\right)=\frac{1}{n-1}\implies g^{n-1}\left(\frac{1}{n}\right)=1$

And so in particular, for any $\frac{1}{n}$ we have that

$\displaystyle f\left(\frac{1}{n}\right)=f\left(g\left(\frac{1}{n}\right)\right)=\cdots=f\left(g^{n-1}\left(\frac{1}{n}\right)\right)=f(1)$

and thus by continuity at $0$ we see that

$f(0)=f\left(\lim \frac{1}{n}\right)=\lim f\left(\frac{1}{n}\right)=\lim f(1)=f(1)$

so that

$\displaystyle f(1)=f\left(\frac{1}{n}\right)=f(0),\quad \forall n\in\mathbb{N}$

Otherwise, we note that $1-xn\ne 0$ for any $x\notin\{\frac{1}{1},\frac{1}{2},\cdots\}$ and so we notice that

$\displaystyle f(x)=f\left(g(x)\right)=f\left(g^{2}(x)\right)=\cdots=f\left(g^n(x)\right)=f\left(\frac{x}{1-nx}\right)$

and thus by continuity

$\displaystyle f(0)=f\left(\lim \frac{x}{1-nx}\right)=\lim f\left(\frac{x}{1-nx}\right)=\lim f(x)=f(x)$

from where it follows that $f$ is a constant function.

References:

1. Kaczor, Wiesĺawa J., and Maria T. Nowak. Problems in Mathematical Analysis. Providence (R.I.): American Mathematical Society, 2001. Print.