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

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 $$n$$ $$$$\alpha>\frac{f(a_{n+1})}{f(a_n)}.$$$$

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

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

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?