Suppose that $0\le a \le 1$ and $0<f(x), g(ax)<\infty$ for $x\ge 0$ and $f(x)\to g(x)$ when $x\to \infty$. And also $g(a x)<g(b x)$ if $a<b$.
Question: Can we use $g(a_k x)$ $(0\le a_1\lt a_2 \lt a_3<\cdots\lt 1)$ as the basis functions to expand $f(x)$?
Here are my thoughts:
Let $$\displaystyle f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} x^k$$ $$\displaystyle g(x) = \sum_{k=0}^\infty \frac{g^{(k)}(0)}{k!} x^k$$
Define $$f_1(x):=g(x)+\sum^\infty_{m=0} b_m g(a_m x)$$ Then $$f_1(x)-g(x)=\sum^\infty_{m=0} \sum_{k=0}^\infty b_m \frac{g^{(k)}(0)}{k!} a^k_m x^k$$
From $f(x)=f_1(x)$ we get: $$\sum^\infty_{m=0} A_{k,m} b_m:=\sum^\infty_{m=0} g^{(k)}(0) a^k_m b_m=f^{(k)}(0)-g^{(k)}(0)$$
If $\det A \not = 0$, then we have: $$b_m=\sum^\infty_{k=0}\left(A^{-1}\right)_{m,k} \left(f^{(k)}(0)-g^{(k)}(0)\right)$$
I am seeking a rigorous proof (or references) of what I just outlined.
What else needs to be included to make a rigorous proof?
best regards- mike
EDIT: It is shown in this question that the determinant of matrix $$ A_{k,m}:= \alpha_k \beta_m$$ is zero. So matrix $A$ has no inverse.
