TITLE: How to show that $\mathbb R^n$ with the $1$-norm is not isometric to $\mathbb R^n$ with the infinity norm for $n>2$?
QUESTION [17 upvotes]: Could you please give me a hint to prove that
$\mathbb{R}^n$ with the 1-norm $\lvert x\rvert_1=\lvert x_1\rvert+\cdots+\lvert x_n\rvert$ is not isometric to $\mathbb{R}^n$ with the infinity-norm $\lvert x\rvert_\infty = \max_{i=1,\ldots,n}(\lvert x_i\rvert)$ for $n\gt2$.
I do not see how the critical case $n=2$ enters the picture.
Thank you!

REPLY [20 votes]: But one can do without the Mazur-Ulam theorem just as easily. I find it convenient to put the exponent $p$ in subscript, freeing the superscript for dimension: $\ell_\infty^n$ and $\ell_1^n$ 
Suppose $F:\ell_\infty^n\to \ell_1^n$ is an isometry (which I do not assume to be surjective or linear). By translation we can achieve $F(0)=0$. Let $V\subset \ell_\infty^n$ be the set of vertices of the unit cube, i.e., the set of all vectors with entries $\pm1$. This set has three interesting properties:  

all points are at distance $1$ from the origin
all points are at distance $2$ from one another 
the cardinality of the set is $2^n$. 

Let $W=F(V)$. The set $W\subset \ell_1^n$ also has the properties 1,2,3 listed above. For each vector $v\in W$ consider its positive and negative supports $\mathrm{supp}^+v=\{1\le j\le n : v_j> 0\}$ and  $\mathrm{supp}^-v=\{1\le j\le n : v_j< 0\}$. If $n>2$, then $2^n>2n$. Therefore,  we can assume that at least $n+1$ vectors in $W$ have nonempty positive supports. But then at least two of them, say $v'$ and $v''$, have overlapping positive supports. This creates cancellation when we compute distance between them: $\|v'-v''\|_1<\|v'\|_1+\|v''\|_1=2$ , a contradiction. QED
The proof actually shows that $\ell_{\infty}^n$ does not admit an isometric embedding (linear or not) into $\ell_1^N$ for $N<2^{n-1}$.