Consider a complex Banach space $X$ with a real subspace isometric to $\ell^1\_{\mathbb R}$. What is the best constant $c$ such that $X$ contains a complex subspace $c$-isometric to $\ell^1\_{\mathbb C}$?


I guess this is a very classical question, but I could not find an answer.


In this question, $\ell^1\_{\mathbb R}$ (respectively $\ell^1\_{\mathbb C}$) is the real (resp. complex) Banach space of summable sequences of real (resp. complex) numbers. And two Banach spaces $Y,Z$ are called $c$-isometric if there is a linear isomorphism $T:Y\to Z$ such that $\|T\| \|T^{-1}\| \leq c$.


The motivation is that I want to understand the complex Banach spaces of trivial (Rademacher) type. A famous theorem of Maurey and Pisier, usually stated for real Banach spaces, shows that $\ell^1\_{\mathbb R}$ is finitely representable in any (real) space of trivial type $\mathbb R$. My question is essentially equivalent to "what happens when the field of scalars is $\mathbb C$?".



Banach-Mazur distance to complex $\ell^1$ of a space containing real $\ell^1$
Schechtman and I discussed your question this morning and have these comments.


You can get $1+\epsilon$. The usual argument for improving the constant works in the complex case as well as the real case; i.e., if a complex Banach space contains a subspace complex isomorphic to $\ell\_1$, then for all $\epsilon >0$ it contains a subspace $1+\epsilon$-complex isomorphic to $\ell\_1$. 


To see that a complex Banach space that contains a real subspace isomorphic to $\ell\_1$ also contains a complex subspace that is complex isomorphic to $\ell\_1$, apply Dor's extension to the complex case of Rosenthal's $\ell\_1$ theorem. The full theorem says that a (real or complex) Banach space contains an isomorphic copy of $\ell\_1$ iff the space contains a bounded sequence that has no weakly Cauchy subsequence.


For other values of $p$ the situation is different. By the Odell-Schlumprecht distortion theorem, for every $M$ there is a an equivalent renorming of $\ell\_p$, $1<p<\infty$, s.t. no subspace of the resulting space is $M$-isomorphic to $\ell\_p$. 


Something we don't see is the following (not that we spent much time thinking about it). Suppose a complex Banach space contains a real subspace isomorphic to real $\ell\_p$, $1<p<\infty$. Does it contain a complex subspace isomorphic to complex $\ell\_p$? However, the local version looks OK; i.e., if real $\ell\_p$ is $C$-finitely representable in a complex space $X$ (and hence $1+\epsilon$-finitely representable by Krivine's theorem) then complex $\ell\_p$ is something like $4+\epsilon$-finitely representable in $X$. For that you use spreading model theory to get a suppression $1$ unconditional basis in a spreading model s.t. the basis is isometrically $\ell\_p$ for real coefficients. It looks like with a bit of work you can change the $4+\epsilon$ to $1+\epsilon$, but we did not check it to the end. It looks like you can make the norm invariant for multiplication of expansions in every coordinate by $\pm 1$ (different $\pm$ in the various coordinates) by passing to a block basis, and multiplication of coordinates by complex numbers having modulus one should also work by passing to a block basis.


