TITLE: Continuous functions $f$ with $f(A)$ linearly independent when $A$ is independent
QUESTION [27 upvotes]: Is there any characterization of continuous functions $f : \Bbb{R}\longrightarrow \Bbb{R}$ such that for any linearly independent set $A$ (over the rationals) $f(A)$ is also linearly independent ?

REPLY [9 votes]: Trevor Wilson has reduced the possibilities to $\alpha x^\beta$ with $\alpha\ne 0$ and $\beta\ge 0$.  The following argument should reduce the possibilities to $\beta=0$ or $\beta=1$, modulo a lemma that I can't quite seem to prove. Perhaps some other MO reader can plug the gap.
The lemma I need is:

For any $q_1, q_2, q_3\in\mathbb{Q}$, not all zero, and any positive $\beta\ne1$, the equation $$q_1+q_2x^\beta+q_3(1+x^{\beta^2})^{1/\beta} = 0$$ has at most countably many real solutions.

This lemma implies that for any positive $\beta\ne1$, there exists a constant $c>0$ such that the three numbers $x_1:=1$, $x_2:=c$, and $x_3:=(1+c^{\beta})^{1/\beta}$ are linearly independent over $\mathbb{Q}$.  On the other hand, $\alpha x_1^\beta + \alpha x_2^\beta = \alpha x_3^\beta$, so their images under the function are linearly dependent.