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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!

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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 See Studia Math Archive

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).

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    $\begingroup$ Matt, the choice of a basis does not induce an isomorphism from $L^2(\mathbb R;E)$ to $\ell^2(E)$, i.e. is not a regular isomorphism between $L^2$ and $\ell^2$. However, your example is correct because $L^2$ embeds "regularly" in an ultraproduct of $\ell^2$. $\endgroup$
    – Mikael de la Salle
    Commented Aug 28, 2017 at 13:35
  • $\begingroup$ @Mikael de la Salle Right, yes! That's a good observation. (So, as an example on $L^2(\mathbb R)$ this is absolutely fine, but I was hasty to claim it was immediately an example on $\ell^2(X)$ for some $X$). $\endgroup$
    – Matthew Daws
    Commented Aug 30, 2017 at 9:07

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