Prove or disprove that for $\alpha>0$ and a positive, strictly decreasing sequence $a_n$, there exists $f$ such that $\alpha>f(a_{n+1})/f(a_n)$

Question

Let $\{a_n\}_{n=1}^\infty \subset(0,\infty)$ be strictly decreasing, and let $\alpha>0$. Prove or disprove that there exists a function $f\colon (0,\infty) \to (0,\infty)$ such that for all positive integer $n$ \begin{equation} \alpha>\frac{f(a_{n+1})}{f(a_n)}. \end{equation}

Here is my attempt: As an example, let $\alpha>0$, $a_1=1$, and for all $n\ge 1,$ let $a_{n+1}=\frac{1}{2}a_n$. Then, for all $b> (\ln \alpha)/\ln (1/2)$, $f$ defined by $f(x)=x^b$ satisfy the claim. So I think, similar to this example, in the case where the sequence $a_n$ is exponentially convergent, then we can come up with $f$. However, I am not sure how to do the general case. Any help is appreciated

Answer 1

I don't know if I'm missing something, but can't you define $f$ by $f(a_0) := 1, f(a_{n+1}) = \frac {\alpha}{2}f(a_n)$, and $f(x) :=$ whatever everywhere else?