Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying $f(x)= f\left(\frac{x}{1-x}\right) \ \forall x \neq 1$. If $f$ is continuous at $0$...

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying $f(x)= f\left(\frac{x}{1-x}\right) \ \forall x \neq 1$. If $f$ is continuous at $0$ then find all possible $f$.

My approach :

If $f(x)= f\left(\frac{x}{1-x}\right)$ then $$\displaystyle f\left(\frac{x}{1-x}\right) = f\left(\frac{x}{1-x} \left(1-\frac{x}{1-x} \right)^{-1}\right)=f \left(\frac{x}{1-2x} \right)$$

On successively doing this, we will get $f(x) = f\left(\frac{x}{1-nx}\right)$.

Taking limit both sides as $n \to \infty$, we will get $f(x)=f(0)$ .This implies $f(x)=f(0)$ for all $x$. Thus $f(x)$ is a constant function. Lot of people pointed out about $f(1)$, can someone help me with this? Can I define $f(1) = a$ for any $a \in \mathbb R$ and then $f(x) = f(0)$ for all $x \in \mathbb{R} \setminus \left\{ 1 \right\}$? Is this correct?

The limit does not necessarily equal to $f(0)$. You will have to use the continuity argument to establish that. — sudeep5221, Jul 08 '21 at 13:03
lim (f(x/1-nx)) as n tends to infinity will be zero right since n is in denominator? — Nidhi Rai, Jul 08 '21 at 13:06
@NidhiRai How do you know you can move the limit operation inside $f$? — Gary, Jul 08 '21 at 13:07
Yes, but you have to phrase this properly. Also in the established equality, you need to assume $x \neq 1/n$. — Gary, Jul 08 '21 at 13:09

score 1 · Accepted Answer · answered Jul 08 '21 at 13:31

1

As remarked in the comments, $x=\frac1n$ for $n\in\Bbb N_+$ demands special consideration. As $f(\frac1n)=f(\frac1{n-k})=f(1)$, $k=1,...,n-1$, then again by continuity, $f(1)=f(0)$.

answered Jul 08 '21 at 13:31

Lutz Lehmann

126,666

$f$ is not necessarily continous at $1$. – Hagen von Eitzen Jul 08 '21 at 14:15
@HagenvonEitzen But it is continuous at $x=0$. $f(1)=f(\frac1n)=\lim_{n\to\infty}f(\frac1n)=f(0)$. – Lutz Lehmann Jul 08 '21 at 14:17
The functional equation is not applicaple to $x=1$ – Hagen von Eitzen Jul 08 '21 at 14:20
1

@HagenvonEitzen But it is applicable at $x=\frac12$. And at $x=\frac13$, etc. – Lutz Lehmann Jul 08 '21 at 14:24
Oh, now I see – Hagen von Eitzen Jul 08 '21 at 14:24

score 1 · Answer 2 · answered Jul 08 '21 at 14:48

A different-looking approach:

If $y=\frac x{1-x}$, then $x=\frac y{y+1}$. We conclude

If $y\ne -1$ then $f(y)=f(\frac y{y+1})$.

Consider $a>0$. Define the sequence $(x_n)$ by setting $x_0=a$ and recursively $x_{n+1}=\frac {x_n}{1+x_n}$. By induction, all $x_n$ are positive (in particular, $1+x_n\ne0$), the sequence is strictly decreasing, and $f(x_n)=f(a)$ for all $n$. In particular, the sequence converges to some limit $x^*$. For it, we have $x^*=\frac {x^*}{1+x^*}$, i.e., $x^*=0$. Then by continuity at $0$, $f(a)=f(0)$.

Now consider $a<0$. This time, let $x_0=a$ and $x_{n+1}=\frac{x_n}{1-x_n}$. Similar to above, $x_n$ converges to $0$ and we conclude $f(a)=f(0)$.

Hence $f$ is constant.

Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying $f(x)= f\left(\frac{x}{1-x}\right) \ \forall x \neq 1$. If $f$ is continuous at $0$...

My approach :

2 Answers2