Consider a Banach space $E$ and a discrete set $X$. For an operator $T$ on $\ell^2(X)$ I can consider and induced operator $T'$ on the Bochner-Lebesgue space $\ell^2(X;E)$ of $E$-valued square-summable functions. This operator is defined by the same matrix representation, but simply acting on a larger Banach space. If we view $\ell^2(X;E)$ as a sort of tensor product of $\ell^2(X)$ and $E$ then $T'$ can be viewed as ``$T\otimes I$''.


My question is: are there Banach spaces $E$ other than the Hilbert space for which there is constant $C=C(E)$ and a norm estimate 
$$\Vert T'\Vert\_{B(\ell^2(X;E))}\le C\,\Vert T\Vert\_{B(\ell^2(X))}$$
for all $T\in B(\ell^2(X))$? Help will be greatly appreciated!



Norm of "tensoring" with the identity
No. This is Proposition 4.1 in Kwapien, "Isomorphic characterizations of inner product spaces by orthogonal series with vector valued coefficients." Studia Math. 44 (1972), 583–595. [MathSciNet](http://www.ams.org/mathscinet-getitem?mr=341039) See [Studia Math Archive](http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-smv44i1p50bwm?q=bwmeta1.element.bwnjournal-number-sm-1972-44-6;5&qt=CHILDREN-STATELESS)


To be precise, if $\mathcal F:L^2(\mathbb R) \rightarrow L^2(\mathbb R)$ is the Fourier transform, then $\mathcal F: L^2(\mathbb R) \otimes E \rightarrow L^2(\mathbb R, E)$ is bounded if only if $E$ is isomorphic to a Hilbert space. As $L^2(\mathbb R)$ is isomorphic to $\ell^2$ via a choice of basis, this gives an example operator $T$ (all be it hard to write down as an operator on $\ell^2$).


There is more in the book by Defant and Floret on Tensor Norms. Or see their survey paper "Continuity of Tensor Product Operators" (which, I'm afraid, I think is hard to get hold of).


