Show that $(x\in\mathopen]0,1\mathclose[\mapsto \frac{1}{1-ax})_{a\in\mathopen]0,1\mathclose[}$ is free I would like to show that $(x\in\mathopen]0,1\mathclose[\mapsto \frac{1}{1-ax})_{a\in ]0,1[}$ is free in $\mathbb R^\mathbb R$.
I have this solution: let $a_1,\dots, a_n$ be distincts elements of $\mathopen]0,1\mathclose[$, $\lambda_1, \dots, \lambda_n$ be real numbers such that:
$$\forall x\in\mathopen]0,1\mathclose[,\sum_{i=1}^n \frac{\lambda_i}{1-a_ix} = 0.$$
Thus:
$$\forall x\in \mathopen]0,1\mathclose[, \sum_{i=1}^n \lambda_i\prod_{j=1, j\neq i}^n(1-a_jx)=0$$
And so, the polynomial $P=\sum_{i=1}^n \lambda_i\prod_{j=1, j\neq i}^n(1-a_jX)$ is the null polynomial.
Moreover, $\forall i \in [[1, n]], P(\frac{1}{a_i}) = \lambda_i \prod_{j=1,j\neq i}^n(1-\frac{a_j}{a_i}) = \lambda_i \mu_i$, where $\mu_i\neq 0$ since the $a_i$ are
distincts. So $\forall i\in[[0,n]], \lambda_i=0$.
Question: do you have a simpler proof? For example, which doesn't use the polynomials?
 A: Here is another solution (I don't know if it is simpler for you) that still uses polynomials.
Let $\phi: x \mapsto \sum_{k=1}^n \frac{\alpha_k}{1-xa_k}$ and we suppose that: $\phi$ is the null function, then: 
$$\forall p\in \mathbb{N},\qquad \phi^{(p)}(0)= p! \sum_{k=1}^n \alpha_ka_k^p=0,$$ 
and then for every polynomial $Q$: 
$$\sum_{k=1}^n \alpha_kQ(a_k)=0.$$ 
We conclude thanks to Lagrange interpolating polynomial (or use Vandermonde system).
A: First proof 
We will prove it by induction on $n$ (by reducing the number of $a_i$)
Assume without loss of generality that $a_1=\max(a_k)$ and $I=\big]\frac{-1}{a_1},\frac{1}{a_1}\big[$. Define the function $f$ by :
 $$\forall x\in I , f(x)=\sum_{i=1}^n \frac{\lambda_i}{1-a_ix}$$
 we know that $\forall i \forall x\in I \,\, |a_ix|<1$, hence  the function $f$ is developable into a power series in $0$ over $I$ and because the restriction of $f$ to $]0,1[$ is null$f_{]0,1[}=0$ we conclude than $f$ is null over $I$ in particular $\lim_{x\to \frac{1}{a_1}^-} f(x)=0$, using the definition of $f$ we conclude that $\lambda_1=0$.
This process can be repeated until $\lambda_1=\lambda_2=\cdots=\lambda_n=0$
Second proof
Let's take the same function as in the first proof, we have :
$$\forall x\in]0,1[ f(x)=\sum_{i=0}^{+\infty} (\lambda_1 a_1^i+\lambda_2a_2^i+\cdots+\lambda_n a_n^i)
x^i=0$$
we conclude that all the coefficient of $f$ are null, hence :
$$ \forall i\in \mathbb{N} \lambda_1 a_1^i+\lambda_2a_2^i+\cdots+\lambda_n a_n^i=0$$
and this is a vendermonde system (when $i\leq n-1$) which has the only solution $\lambda_1=\lambda_2=\cdots=\lambda_n=0$
