Finding all continuous function which maps any sequence in geometric progression to another geometric progression

Question

Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for any geometric progression $x_n$ the sequence $f(x_n)$ is also a geometric progression.

I tried first by taking constant sequences. But it does not helps much.

Answer 1

Keeping as general as possible: $x_n = ar^n$

Require $f(ar^n) = bs^n$. (Yes the $n$ can be different but you can adjust $s$ to make the $n$ equal).

We also ask $f(ar^{n+1}) = bs^{n+1}$. For any $x \in \mathbb{R}$, we can find $a$ such that $x = ar^{n}$.

Then what we ask is equivalent to asking that for any $x \in \mathbb{R}, \exists s :f(ar^{n+1}) = f(rx) = bs^{n+1} = sf(x)$. So this is like "almost linear" what are asking for.

Solving this functional equation, we can see $f(0) = 0$ (or $s=1$).

But if I take $x=1$, $f(r) = sf(1) \iff \frac{f(r)}{f(1)} = s$. This gives us clues. $f(r^k) = s^k f(1)$. The function that maps $ar^n$ to $bs^n$ is thus characterised by $\frac{f(r)}{f(1)} = s$. Next step: $f(ar^n) = b(\frac{f(r)}{f(1)})^{n}$. Of course also $f(a)=b$.

$f(ar^n) = f(a)(\frac{f(r)}{f(1)})^{n}$.

Key identity is $f(rx) = \frac{f(r)}{f(1)}f(x)$

Some examples of solutions: $f(x) = x$, $f(x) = cx, c \in \mathbb{R}$, $f(x) = cx^k, k \in \mathbb{Z}$.

Answer 2

If $f(1)=0$ and $x\neq 0$ then we have that $f(x)^2=f(1)f(x^2)=0$, so $f\equiv 0$. Assume from now that $f(1)\neq 0$.

Take $r>0$. There exists some $s\in\Bbb R$ such that $$f(r^n)=s^nf(1),\,\forall n\in\Bbb Z$$ If $q\in\Bbb Z_+$, $$f((r^{1/q})^q)=sf(1)$$ Then, if $f(r^{1/q})=tf(1)$, then $sf(1)=t^qf(1)$, so $t=s^{1/q}$. Note that this implies $s>0$. This shows that $$f(r^\alpha)=s^\alpha f(1)\,\forall \alpha\in\Bbb Q$$ and, by continuity, for all $\alpha\in\Bbb R$.

Take now $x>0$. Then $$f(x)=f(r^{\log_r x})=s^{\log_r x}f(1)=f(1)x^k$$ where $k=\log_rs$.

If $x<0$, consider the geometric progression $(-1)^nx$. Then $f(x),f(-x),f(x),\ldots$ is also a geometric progression, so $|f(-x)|=|f(x)|$. There exists a function $g$ of the form $g(x)=ax^k$ such that $|f(x)|=|g(x)|$. Again by continuity, this means that $f$ is $g$, $-g$, $|g|$ or $-|g|$.

Remark: Since the domain of $f$ is $\Bbb R$, $k$ must be an integer.