About tight rational bounds

Question

Suppose we have rational functions $f$, $g$, and $h$ defined for all natural numbers $n$ such that $f \leq g \leq h$ for all $n \in \mathbb{N}$.

How can we prove that there is no rational functions $f_1$ and $g_1$ such that $f \leq f_1 \leq g \leq g_1\leq h$ ?

Knowing that the field $\mathbb{Q}$ is dense, that is for any $q_1$ and $q_2$ in $\mathbb{Q}$, there always exists $q \in \mathbb{Q}$ such that $q_1< q< q_2$. Is this already sufficient to show that my previous question's implication is false; that is there are always rational functions $f_1$ and $g_1$.

Now suppose a function $g$ not defined by any elementary operation but the range $R$ of $g$ is a subset of $\mathbb{Q}$, did my previous question now possible?

Thanks

Answer 1

Perhaps I am misinterpreting what you mean by rational function (usually this refers to a function that is the ratio of two polynomials).

Consider $f(n) = n-2, g(n) = n,$ and $h(n) = n+2$. These are defined for all natural numbers $n \in \mathbb{N}$ and the desired inequality holds.

Now use $f_1(n) = n-1$ and $g_1(n) = n+1$ to obtain the inequalities you were trying to avoid. (!)

Moreover, from the phrasing of your question: why not just take $f_1 = f$ and $g_1 = g$?

Answer 2

Let $f$, $g$, and $h$ be rational functions of $n \in \mathbb{N}$, such that

$$f(n) \leq g(n) \leq h(n).$$

In general, there is always a family of rational functions $f_1, g_1$ that satisfies $$f(n) \leq f_1(n) \leq g(n) \leq g_1(n) \leq h(n);$$ we can simply take $$f_1(n) = \frac{f(n) + g(n)}{2},$$ and $$g_1(n) = \frac{g(n) + h(n)}{2}.$$