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Prince Rupert's problem asks how large an m dimensional cube can be inscribed in an n dimensional unit cube.

For $m=1$ and $m=2, n=3$, this is pretty easy, where $f(m,n)$ is the edge of the cube.

$$ f(1,n) = \sqrt n $$

$$ f(2,3)=\frac 3 4 \sqrt 2 $$

From there it's supposed to be straightforward to show:

$$ f(m+1,n) < f(m,n) < f(m,n+1) $$

While this is easy for the case where m=1 I can't get any further than that.

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With $\le $ instead of $<$ the bounds are trivial. It seems that strict inequality (as long as $m<n$) is equivalent to $f(k,k+1)>1$ for all $k$. –  Hagen von Eitzen Oct 20 '12 at 20:49

1 Answer 1

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 $k<n$. Also assume that we have an orthogonal map $\Psi_k$ such that $B_k=\Psi_k B_1$ and that $\Psi_k e_i=e_i$ for $i> k$. (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 $$ f(m+1,n)<f(m,m+1)f(m+1,n)\le f(m,n)$$ at least if $m+1\le n$ (i.e. if $f(m+1,n)> 0$). 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 $0=f(m+1,n)<f(m,n)=1$ if $m=n$, we finally obtain

$$ f(m+1,n)<f(m,n)<f(m,n+1)\quad\text{if }1\le m\le n.$$

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