$Let f=o(g)$ stand for $\lim\_{x\to\infty} f(x)/g(x)=0$.
It is simple to prove the following fact.


**Proposition.** Let $f\_0,f\_1:\mathbb{R}\to(0,\infty)$ be such that $f\_0=o(f\_1)$. Then there is a family of functions $(f\_t)\_{t\in[0,1]}$ such that $f\_s=o(f\_t)$ if $0\leq s<t\leq1$.


*Proof.* Consider $f\_t=f\_0^{1-t}f\_1^t$.


I wonder if the following conjecture is also true.


**Conjecture.** Let $f\_0,f\_1:\mathbb{R}\to(0,\infty)$ be such that $f\_0=o(f\_1)$. Then there is a bounded totally ordered set $I$ and a family of 
functions $(f\_i)\_{i\in I}$ such that


1. $|I|>\mathfrak{c}$,
2. for every $i,j\in I$ with $i<j$ one has $|\{k\in I: i<k<j\}|=|I|$,
3. $f\_{\min I} = f\_0$, $f\_{\max I}=f\_1$,
4. for every $i,j\in I$ with $i<j$ one has $f\_i=o(f\_j)$.



Cardinality of growth rates
Let me rather define $f=o(g)$ as $\forall \varepsilon > 0, \exists x\_0, \forall x \geq x\_0, |f(x)| \leq \varepsilon |g(x)|$. 


Let $f\_0,f\_1$ be such that $f\_0 = o(f\_1)$. Let us assume that we have a family $(f\_i)\_{i \in I}$ satisfying the properties $1,3,4$.


I am going to show that $\exists x\_0, \forall x \geq x\_0, f\_0(x)= 0$ (and conversely, if the latter holds, just take $f\_i = f\_0$ for $i < \mathrm{max} \ I$). If this is not the case, then there exists a sequence $(x\_n)\_n$ tending to $+\infty$ such that for eqch $n$, $f\_0(x\_n) \neq 0$. 


Let us consider the map $I \rightarrow \mathbb{R}^{\mathbb{N}}$ sending $i$ to $(f\_i(x\_n))\_n$. Since $\mathbb{R}^{\mathbb{N}}$ has the cardinality of the continuum, this map is not injective, so we can find $i < j$ such that $\forall n, f\_i(x\_n) = f\_j(x\_n)$. Since $f\_i = o(f\_j)$, this implies $f\_i(x\_n) = 0$ for $n$ large enough. Since $f\_0 = o(f\_i)$, this implies $f\_0(x\_n)= 0$ for $n$ large enough, a contradiction.


