BM-distances between $B(\ell_p^n)$ and $\ell^{n^2}_p$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T17:23:10Z http://mathoverflow.net/feeds/question/109008 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109008/bm-distances-between-b-ell-pn-and-elln2-p BM-distances between $B(\ell_p^n)$ and $\ell^{n^2}_p$ Tomek Kania 2012-10-06T16:19:49Z 2012-10-07T16:47:58Z <p>Let $\ell_p^n$ be the $n$-dimensional real or complex $\ell_p$-space and let $\mathscr{B}(\ell_p^n)$ be the space of matrices on $\ell_p^n$ endowed with the operator norm. I am looking for any references which give (reasonable) estimates for the Banach-Mazur distance between $\mathscr{B}(\ell_p^n)$ and $\ell_p^{n^2}$.</p> <p>I am also wondering whether there is a Banach space $E$ with a Schauder basis $(e_n)$ and corresponding basis projections $(P_n)$ which satisfies $$\sup_{n\in \mathbb{N}} \;d_{{\rm BM}}\big( \mathscr{B}(P_n(E)) ), P_{n^2}(E) \big)=\infty.$$</p> http://mathoverflow.net/questions/109008/bm-distances-between-b-ell-pn-and-elln2-p/109080#109080 Answer by Bill Johnson for BM-distances between $B(\ell_p^n)$ and $\ell^{n^2}_p$ Bill Johnson 2012-10-07T16:47:58Z 2012-10-07T16:47:58Z <p>For the first question, $n^{1/2}$ is the right order for $p=1,2,\infty$. The case $p=\infty$ is arguably the easiest, because $B(\ell_\infty^n)=\ell_\infty^n(\ell_1^n)$ isometrically and the Banach-Mazur distance between $\ell_\infty^n$ and $\ell_1^n$ is of order $n^{1/2}$. That gives the upper bound, and the lower bound is also true because $\gamma_\infty(\ell_1^n)$ (the factorization constant of the identity on $\ell_1^n$ through an $\ell_\infty$-space) is of order $n^{1/2}$.<br>  Also $B(\ell_1^n)=\ell_\infty^n(\ell_1^n)$ isometrically, and you get the lower bound from the fact that $\gamma_1(\ell_\infty^n)$ is of order $n^{1/2}$. For the upper bound assume that $n$ is a power of two (I think standard arguments from this case give the general case but did not try to think it through). In $L_1^N(\ell_1^n)$ consider the basis $w_k\otimes e_j$, $1\le j,k \le n$, where $(w_k)$ is the Walsh basis for $L_1^n$. Take the basis to basis mapping from $L_1^N(\ell_1^n)$ onto $\ell_\infty^n(\ell_1^n)$ that maps $w_k\otimes e_j$, $1\le j \le n$, onto the $k$-th copy of $\ell_1^n$. This mapping has norm at most $n^{1/2}$ because the Walsh basis has an upper $\ell_2$ estimate, and the inverse has norm one.  For $B(\ell_2^n)$, Yemon's comment gives a lower estimate of $n^{1/2}$ (in fact, for all $1\le p \le 2$ because $\gamma_p(\ell_\infty^n)$ is of order $n^{1/2}$ in this range). The upper estimate is just the fact that the norm of an operator in $\ell_2^n$ is at most $n^{1/2}$ times the Hilbert-Schmidt norm of the operator.  For all values of $p$, if you write a vector in $\mathbb{R}^{n^2}$ as an $n$ by $n$ matrix $A$, then the $\ell_p^{n^2}$ norm of $A$ is the $p'$-summing norm of $A$ considered as an operator from $\ell_p^n$ into $\ell_{p'}^{n}$. That is what is used above in the $p=2$ case, but I don't see that this helps for other values of $p$. For <code>$2&lt;p&lt;\infty$</code> you get a lower bound from the fact that $B(\ell_p^n)$ contains <code>$\ell_\infty^n$</code> and $\ell_{p'}^n$ isometrically, but for $p=4$ that gives a lower bound of only $n^{1/4}$.</p>