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.