Its all about functions

Question

Let $f: \Bbb{R} \to \Bbb{R}$ be a continuous function such that for all $x \in \Bbb{R}$ and for all $t \ge 0$, $$f(x) = f(xe^t).$$

Show that $f$ is a constant function.

I tried finding out its derivative but it didn't lead me anywhere.

Any help will be appreciated.

Thank you.

Answer 1

Let $x \in \mathbb R$. Let $s >0$ and let $y = xe^{-s}$. We have $f(y) = f(ye^{r})$ for every $r > 0$. So, $f(xe^{-s}) = f(xe^{r-s})$ for every $r > 0$. In particular, $f(x) = f(xe^{-s})$. This is true for every $s>0$. Hence,

$$f(x) = \lim_{s \to \infty} f(xe^{-s}) = f(x\times 0) = f(0)$$

where in the second equality we used the continuity of $f$.

This shows that $f$ is constant.

Answer 2

$$\forall x \in (\mathbb R^+\cup\mathbb R^-), \forall u\gt1, f(xu)=f(x)$$ $\implies \forall x,y\in\mathbb R^+, f(x)=f(y)$ and $\forall x,y\in\mathbb R^-, f(x)=f(y)$

By continuity, $\forall x,y\in\mathbb R, f(x)=f(y)$