We know that $c_0\check\otimes c_0=c_0(c_0)$ where $\check\otimes$ is the injective tensor product. is the following still true? $$c_0\check\otimes l^\infty = l^\infty(c_0).$$
Thank you for your help
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By exercise 4.8 in Tensor Norms and Operator Ideals. A. Defant, K. Floret $\ell_\infty \check\otimes E$ is isometrically isomorphic to a subspace of $\ell_\infty(E)$ consisting of relatively compact sequences. For the case of $c_0$ we have a criterion of relative compactness. Applying it we see that the sequence $(x_n)_{n\in\mathbb{N}}$ with $x_n(k)=\min(1,2^{n-k})\in c_0$ is not relatively compact in $c_0$. Therefore it doesn't belong to a natural copy of $\ell_\infty \check\otimes c_0$ in $\ell_\infty(c_0)$.
Therefore the most obvious map that one may expect to be an isomorphism is not surjective, and a fortiori an isomorphism. May be there exist some weird isomorphism, but I doubt it is true.
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But the sequence you gave doesn't belong to the space $l^\infty(c_0)$, so we didn't answer the question yet, do we have an example of an element in $l^\infty(c_0)$ which doesn't belong to $l^\infty \check\otimes c_0$? – user100478 Feb 20 '15 at 16:47
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Fixed, now it is in $\ell_\infty(c_0)$ – Norbert Feb 20 '15 at 19:45