Do there exist continuous functions $c_n\colon\mathbb R^+\to\mathbb R$ and $\gamma\colon\mathbb R^+\to\mathbb R$ such that $\lim_{x\to\infty}\gamma\left(x\right)=0$ and the following equation is true for all $\alpha,x>0$ $$ e^{-\alpha x}=\sum_{n=1}^{\infty}c_n\left(\alpha\right)\gamma\left(x\right)^n $$ If yes, how regular can we make $\gamma$, can it be smooth? Analytic? What about the $c_n$?

Clarifying edit: all the $x$-dependence is in $\gamma$ and all the $\alpha$ dependence is in the $c_n$


1 Answer 1


The answer is no.

Suppose that $c_n(\alpha)$ and $\gamma(x)$ with the desired properties exist. Necessarily $\gamma$ is one-to-one, and hence $$e^{-\alpha \gamma^{-1}(x)} = \sum_{n = 0}^\infty c_n(\alpha) x^n.$$ Since the right-hand side converges for some $x > 0$, it defines a function $$f_\alpha(x) = \sum_{n = 0}^\infty c_n(\alpha) x^n$$ which is real-analytic near $0$ and not identically equal to zero. In particular, $f_\alpha(x) \sim c_{k(\alpha)}(\alpha) x^{k(\alpha)}$ as $x \to 0$, where $k(\alpha)$ is the index of the first non-zero coefficient $c_n(\alpha)$. This means that $$e^{-\gamma^{-1}(x)} \sim c_{k(\alpha)}(\alpha) x^{k(\alpha)/\alpha},$$ and therefore $k(\alpha) / \alpha$ is constant. Since $k(\alpha)$ is integer-valued, this is not possible, unless $k(\alpha) = 0$. But this means that $e^{-\gamma^{-1}(x)}$ has a positive right limit at $0$, contrary to the assumption that $\gamma^{-1}(x) \to \infty$ as $x \to 0^+$.

  • $\begingroup$ Why is \gamma one-to-one? $\endgroup$ Jan 31, 2022 at 12:52
  • 1
    $\begingroup$ If $\gamma(x) = \gamma(y)$, then $e^{-\alpha x} = e^{-\alpha y}$, and hence $x = y$. $\endgroup$ Jan 31, 2022 at 12:56

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