Abstract Nonsense

Crushing one theorem at a time

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.

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October 28, 2010 - Posted by | Analysis, Fun Problems | , ,

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