Infimum over all vector-valued L^2 spaces

2011-06-04T15:00:16Z

Suppose I have a Banach space $E$ (which may be finite dimensional if you wish), a Hilbert space $H$ and a tensor $\tau \in H\otimes E$ in the algebraic tensor product. There are lots of ways to choose a measure $\mu$ and an isometry (not assumed surjective) $\theta:H\rightarrow L^2(\mu)$. Then $(\theta\otimes\iota)\tau \in L^2(\mu)\otimes E \subseteq L^2(\mu;E)$, and so I can compute the norm of $(\theta\otimes\iota)\tau$ in the vector-valued space $L^2(\mu;E)$.

Is there an intrinsic (or simple, etc.) characterisation of the infimum (over all choices of $\theta$ and $\mu$) of this norm?

(The infimum is non-zero, assuming $\tau\not=0$, as it's always larger than the injective tensor norm. But it's not obvious to me that you actually get a norm on $H\otimes E$ from this).

If $E$ is a Hilbert space, then the norm is independent of the choice of $\mu$ and $\theta$; you just get the Hilbert space tensor product norm. But what if, say, $E$ is a finite-dimensional $\ell^\infty$ space?

Answer by Pietro Majer for Infimum over all vector-valued L^2 spaces

2011-06-06T18:41:31Z

Here is a first step. Let $\tau=\sum_{i=1}^n h_i\otimes u_i$ with $h_1,\dots h_n\in H$ and $u_1,\dots u_n\in E$, where w.l.o.g. $h_i$ are orthonormal.

Then, in your infimum, you may fix $\mu$ to be the counting measure on $\mathbb{N}$, so that $L^2(\mu)=\ell^2$ (this follows from a simple argument using the density of simple functions and the Gram-Schmidt orthonormalization process), and the infimum writes

$$\inf\Bigg( \ \sum_{k=0}^\infty \ \Bigg \| \ \sum_{i=1}^n \lambda_{k,i}u_i \Bigg\|_E^2 \ \Bigg)^{1/2}$$

taken over all $\lambda_{k,i}$ with $\sum_{k=0}^\infty \lambda_{k,i} \lambda_{k,j}=\delta_{i,j} $. Then, it is not clear to me how to make a further reduction, even in the case of $n=2$ vectors.

Infimum over all vector-valued L^2 spaces - MathOverflow

Infimum over all vector-valued L^2 spaces

Answer by Pietro Majer for Infimum over all vector-valued L^2 spaces