Uniform asymptotic expansion

Question

Let the function $f(x,\nu)$, with $x>0$ and $\nu\in[0,+\infty)$, have the asymptotic expansion \begin{equation} f(x,\nu)\sim\sum_{n=0}^{\infty}a_{n}(\nu)x^{n}\;, \end{equation} as $x\to 0$. Assume also that $|a_{n}(\nu)|\leq M_{n}$ for some constant $M_{n}$ for $\nu\in[0,+\infty)$. Can one conclude that the above asymptotic expansion is uniform in the variable $\nu$? namely can one conclude that \begin{equation} f(x,\nu)=\sum_{n=0}^{N}a_{n}(\nu)x^{n}+O(x^{N+1})\;, \end{equation} with $O(x^{N+1})$ independent of $\nu$?

Answer 1

In general no. For example, $e^{-\nu^2/x^2} = O(x^n)$ as $x \to 0$ for all $n \geq 0$ and $\nu \neq 0$ fixed, so we have

$$ \frac{1}{1-x} + e^{-\nu^2/x^2} \sim \sum_{n=0}^{\infty} x^n$$ as $x \to 0$ for all fixed $\nu \neq 0$. This asymptotic is not uniform with respect to $\nu$ when, say, $\nu = x$.