Consider the following vectors:
$v_m = (f(ma_1), f(ma_2), \dots, f(ma_n))$ for $m \in \{1,\dots, n\}$,
where $f(x) \in \mathbb{R}$ is a continuous, strictly monotonic, and nonlinear function and $a_1 \neq a_2 \neq \dots a_n$ take values in $\mathbb{R}$. Can we prove that there exist $a_1, \dots a_n$ such that $v_1, \dots, v_n$ are linearly independent? We can put additional costraints on $f$ if necessary.
I tried to show that for $n = 2$ by putting the following constraint on $f$: there exist $x$ and $y$ such that
$\frac{f(x)}{f(y)} \neq \frac{f(2x)}{f(2y)}.$
See here for the types of functions that satisfy $\frac{f(x)}{f(y)} = \frac{f(kx)}{f(ky)}.$
Now, we can do a proof by contradiction. Suppose that for all $a_1 \neq a_2$, $v_1$ and $v_2$ are linearly dependent. We can write
$v_1 = k v_2 \Rightarrow (f(a_1), f(a_2)) = (kf(2a_1), kf(2a_2))$.
Therefore, we have
$\frac{f(a_1)}{f(a_2)} = \frac{f(2a_1)}{f(2a_2)}$ for all $a_1, a_2$,
which contradicts our assumption.
Is it possible to derive something similar for $n > 2$?
I tried the following. Suppose that $v_1, \dots, v_n$ are linearly dependent. Then there exist $k_1, \dots k_n$ such that the following equations
$k_1 f(a_m) + k_2 f(2a_m) + \dots + k_{n-1} f((n-1)a_m) = f(na_m)$ for $m \in \{1, \dots, n\}$,
are satisfied for any choice of $a_1, \dots, a_n$. Can we say that the following condition for $f(x)$ is sufficient for linear independence of $v_1, \dots, v_n$? If so, what functions satisfy this property?
Condition on $f$: We cannot find any set of values for $k_1, \dots, k_{n-1}$ such that for any $x \in \mathbb{R}$
$k_1 f(x) + k_2 f(2x) + \dots + k_{n-1} f((n-1)x) = f(nx)$.