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Let $E^n$ be the real or complex space of dimension $n$. If $N$ and $M$ are two norms over $E^n$, and if $A$ is an endomorphism, then $$\|A\|^M_N:=\sup_{x\ne0}\frac{M(Ax)}{N(x)}$$ is an operator norm of $A$.

The Banach-Mazur distance between $N$ and $M$ is $$d(N,M):=\log\inf_{A\in Aut(E)}\|A\|^M_N\|A^{-1}\|^N_M.$$ It is actually a semi-distance, with $d(N,M)=0$ iff one passes from $N$ to $M$ by a change of coordinates. Thus $d$ is a distance on some quotient space, in which there is only one Euclidian norm, for instance.

Let us define the usual $\ell^p$-norm by $$\|x\|_p:=\left(\sum_j|x_j|^p\right)^{1/p}.$$ The $L^\infty$-norm is as usual. It has been known for a long time that if $1\le p,q\le2$, or if $2\le p,q\le\infty$, then $$d(\ell^p,\ell^q)=\left|\frac{1}{p}-\frac{1}{q}\right|\log n.$$ What is known if $1\le p\le2\le q\le+\infty$ ?

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I've just looked in the "Handbook of the Geometry of Banach Spaces", Chapter 1, Section 8, which gives that $$d(\ell^p,\ell^q) = \log\max(n^{|1/p-1/2|}, n^{|1/q-1/2|})$$ up to a constant (which seems hard to find!) The argument is not entirely elementary, using cotype estimates. I'd also look in Tomczak-Jaegermann's book "Banach-Mazur distances and finite-dimensional operator ideals" but I don't have that on my shelf right now.

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  • $\begingroup$ This does not seem correct, because the distance is zero when $p=q$, whereas your formula gives $|1/p-1/2|\log n$ instead. $\endgroup$
    – Denis Serre
    Sep 16, 2010 at 10:49
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    $\begingroup$ @Denis: Matt gave the formula for $p \le 2 \le q$ because that is what you asked for. BTW, it is conventional to define the B-M distance without the log, making the distance submultiplicative instead of subadditive, so the the B-M distance is the isomorphism constant between the two spaces. $\endgroup$
    – Bill Johnson
    Sep 16, 2010 at 11:31
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    $\begingroup$ Yeah, I said "up to a constant" (but, sure, I meant absolute constant). And, yes, it's only valid for $p\leq 2\leq q$. $\endgroup$
    – Matthew Daws
    Sep 16, 2010 at 12:43
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    $\begingroup$ Is this a correct reading of the above: to compare $l^p$ with $l^q$, if $p$ and $q$ are on the same side of $2$, then the map $A = I$ is optimal. If they are on opposite sides of $2$, then it's hard to find a recipe for an optimal linear map, but it's only boundedly suboptimal to compare them to the $l^2$ norm. It shouldn't be hard to find optimal comparisons numerically in low dimensions. It would be interesting to see the optimal $A$ for some of these starting with dimension 3, e.g., what linear map best fits an octahedron inside the cube? Are pictures available somewhere? $\endgroup$
    – Bill Thurston
    Sep 16, 2010 at 13:31
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    $\begingroup$ @Denis: With complex scalars, $d(\ell_1^2,\ell_\infty^2)>0$. @Bill Thurston: For the upper bound, consider the case $n=2^k$ and the mapping from $\ell_p^n$ to $L_q^n$, $p\le 2 \le q$, that takes the unit vector basis to the Walsh basis. (Here $L_q^n$ is $L_q(\mu)$ with $\mu$ the uniform probability on $\{1,...,n\}$). $\endgroup$
    – Bill Johnson
    Sep 17, 2010 at 2:31

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