Consider the (enlarged) cube $A$ in $\mathbb R^n$ spanned by the $2^n$ vertices $v$ of the form $|v^{(i)}|=1$, $1\le i\le n$.
In it consider, for $1\le k\le n$, an $(n-1)$ dimensional cube $B_k$ such that all $2^{n-1}$ vertices $w$ of $B_k$ have the properties
- $|w^{(i)}|<1$ for $1\le i\le k$
- $|w^{(i)}|\le 1$ for $k< i\le n$
The cube $B_1$ can be obtained by projecting $A$ into the orthogonal complement of the first standard base vector so that we even have $w^{(0)}=0$.
Assume we have $B_k$ for some $k k$. (Thus we can take $\Psi_1=1$).
For $t\in\mathbb R$, we can consider the rotation $\Phi_t$ that maps $e_1\mapsto \cos( t) e_1+\sin(t)e_{k+1}$, $e_{k+1}\mapsto\cos(t)e_{k+1}-\sin(t)e_1$ and $e_i\mapsto e_i$ for $i\notin\{1,k+1\}$.
Let $w$ be a vertex of $B_1$ and let $u=\Psi_k\Phi_t w$.
Then $u$ is a vertex of $B_k$ if $t=0$.
By continuity of $t\mapsto\Phi_t$, we have $|u^{(i)}|<1$ for $1\le i\le k$, provided $|t|$ is small enough.
Also, $u^{(i)}=w^{(i)}$ if $i>k+1$.
We have $u=\Psi_k w + \Psi_k(\Phi_1-1)w\approx \Psi_k w -tw^{(k+1)} \Psi_k e_{k+1}$, hence
$u^{(k+1)}\approx (1-t)w^{(k+1)} $, that is for small positive $t$, we will have $|u^{(k+1)}|\approx1-t$ so that $B_{k+1}:= \Psi_{k+1}B_1$ with $\Psi_{k+1}:=\Psi_k\Phi_t$ has the desired properties.
Then $B_n$ is an $(n-1)$-dimensional hypercube that fits strictly inside the $n$-dimensional hypercube $A$. In other words, $B_n$ can be stretched by an amount $q>1$ and still fits.
This shows that
$$ f(n,n+1)>1\quad\text{for all }n\in\mathbb N.$$
Since trivially
$$f(m,n)\ge f(m,k)f(k,n)\quad\text{if }m\le k\le n,$$
we arrive at
$$ f(m+1,n) 0$).
We also find
$$f(m,n)< f(m,n)f(n,n+1)\le f(m,n+1)$$
if $m\le n$ (so that $f(m,n)>0$). Since trivially $0=f(m+1,n)
$$ f(m+1,n)