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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 $\ell_p^{n^2}$.

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 $$\sup_{n\in \mathbb{N}} \;d_{{\rm BM}}\big( \mathscr{B}(P_n(E)) ), P_{n^2}(E) \big)=\infty.$$

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I presume you mean $<\infty$ rather than $=\infty$ in your second question. As for the first question, probably someone computed the GL constant for $B(\ell_p^n)$. That might give the order of magnitude for the B-M distance you are seeking. I would look in Tomczak's book but my copy is loaned out. –  Bill Johnson Oct 6 '12 at 16:46
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The following might be too trivial for what you want, but how about considering the 1-complemented isometric copy of $\ell_\infty^n$ that lives as diagonal matrices? there must be known lower bounds for the "constant of openness" of any injective operator from $\ell_\infty^n$ to $\ell_p$ –  Yemon Choi Oct 6 '12 at 22:44
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That only gives order $n^{1/p}$ for $p\ge 2$ and $n^{1/2}$ for $1\le p\le 2$,Yemon (using cotype constants on $n$ vectors). –  Bill Johnson Oct 7 '12 at 0:34
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I don't know the answer to the second question, but a related infinite dimensional question is whether there is a Banach space $X$ with a Schauder basis s.t. $X$ is isomorphic to the compact operators on $X$. The universal basis space of Pelczynski has this strange property. –  Bill Johnson Oct 7 '12 at 19:59
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1 Answer

up vote 3 down vote accepted

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 $n^{1/2}$.
$$ $$ 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 $2<p<\infty$ you get a lower bound from the fact that $B(\ell_p^n)$ contains $\ell_\infty^n$ and $\ell_{p'}^n$ isometrically, but for $p=4$ that gives a lower bound of only $n^{1/4}$.

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