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I have a function $f(x)$ defined for $x \in [0,\, 1]$ which is such that $$ f'(x) = \beta \,f(x /\alpha) \qquad\text{for all } x \text{ with } 0 \leq x \leq \alpha, $$ where $\beta$ and $\alpha$ are constants with $\beta > 0$ and $0 < \alpha < 1$. I know that $f(0) = 0$ and that $f(x)$ is positive for $x \in (0,\, 1)$ and infinitely differentiable on $[0,\, \alpha]$, so that $f^{(n)}(0)=0$ for all $n \geq 0$.

I conjecture that that $$ \lim_{x \to 0} \;\frac{f(x) \,f(x/\alpha^2)}{f(x / \alpha)^2} = \alpha, $$ or equivalently, that the function defined by $g(x) := f(x)/f(x /\alpha)$ is such that $g'(x)$ tends to $0$ for $x \to 0$. How can I prove/disprove?

Note: as a special case, we can consider the nice dilatation equation $f'(x) = f(2 x)$.

The motivation is as follows. Consider a first order autoregressive process $Y_t = \alpha Y_{t-1} + (1 - \alpha) \varepsilon_t$ where $0 < \alpha < 1$ and $\varepsilon_t$ is a white noise with uniform distribution on $(0, \,1)$. Then if $F$ is the stationary distribution function and $f(x) := 1- F(1 -x)$, the equation above holds. My purpose is to prove that the famous Von Mises conditions in Extreme Value Analysis hold, so that $F$ is in the Gumbel domain of attraction.

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    $\begingroup$ Suppose that $x \in (\alpha^2, \alpha)$, then $x/\alpha > \alpha$, so how can you say that $x \mapsto f(x/\alpha)$ is again differentiable for all $x \in [0,\alpha)$? $\endgroup$
    – Tom
    Oct 22, 2016 at 17:34
  • $\begingroup$ Oh yes, "it is clear" is not true. I should replace by "assume that" for the statement on differentiability, and then $f^{(n)}(0) = 0$ remains "clear"? $\endgroup$
    – Yves
    Oct 22, 2016 at 17:50
  • $\begingroup$ Unfortunately I noticed this question only now and it is a bit later to produce the analytical details, but however have you tried to attack your problem by using the Mellin transform $$\mathcal{M}[f,s]=\int\limits_{0}^\infty x^{s-1}f(x)\mathrm{d}x\;?$$ This tool is sometimes very effective when dealing with the multiplicative group $\mathbb{R}_+=\{\alpha\in\mathbb{R}|\alpha>0\}$. $\endgroup$ Nov 9, 2018 at 13:07
  • $\begingroup$ Will this be better than using the Laplace transform, say $\psi$, of $f$? A functional equation can be found for $\psi$, allowing it to be expressed as an infinite product. But I can not see that studying the behaviour of $\psi$ near $\infty$ will be simpler than that of $f$ near $0$. I am not a mathematician; For now, I can just show two easy things, namely that $g(x)= O(x)$ for $x \to 0$ and that the only finite limit possible for $g(x) / x$ for $x \to 0$ is $0$. $\endgroup$
    – Yves
    Nov 9, 2018 at 13:22

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