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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

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  • $\begingroup$ Is $f$ allowed to depend on the sequence? $\endgroup$
    – user56202
    Oct 29, 2021 at 0:55
  • $\begingroup$ doesn't $f(x)=x$ defined over $(0,\infty)$ get the job done with $\alpha=1$? or am I blind? $\endgroup$
    – Schach21
    Oct 29, 2021 at 0:56
  • $\begingroup$ @Schach21 It seems like it. This question seems trivial yet it has 3 upvotes, so I think I'm missing something. $\endgroup$
    – user56202
    Oct 29, 2021 at 1:02
  • $\begingroup$ Is the question supposed to read "for all $\alpha$"? The question is trivial if it were "there exists" an $\alpha$ $\endgroup$
    – okzoomer
    Oct 29, 2021 at 1:27
  • $\begingroup$ @okzoomer This question is either trivial or very poorly written. $\endgroup$
    – Ovi
    Oct 29, 2021 at 1:50

1 Answer 1

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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?

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