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How do I prove that $B(Y,\ell_2^n)^{**}$ is isometrically isomorphic to $B(Y^{**},\ell_2^n)$?

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  • $\begingroup$ Presumably you are asking for a proof that a particular "natural" map between these two spaces is an isometric isomorphism. Could you tell us what this map is, or is the definition of the map part of what you are trying to figure out? $\endgroup$ – Yemon Choi Jul 24 '17 at 5:02
  • $\begingroup$ @ Choi! We have a natural isometric isomorphic identification $(Y\hat{\otimes}\ell_2^n)^{*}=B(Y,\ell_2^n).$ Therefore, we want to show the following identification $(Y\hat{\otimes}\ell_2^n)^{***}=B(Y^{**},\ell_2^n).$ $\endgroup$ – Mathbuff Jul 24 '17 at 6:18
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    $\begingroup$ There is a very nice proof in Bill Johnson's answer to my question mathoverflow.net/q/145412 (the situation seems a bit different as it considers $B(X,Y)$ with $X$ finite dimensional, but this is the same as $B(Y,\ell_2^n) = B(\ell_2^n,Y^*)$). $\endgroup$ – Mikael de la Salle Jul 24 '17 at 8:03
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    $\begingroup$ Translated in this context, the proof becomes: 1. If $\ell_2^n$ is replaced by $\ell_\infty^N$ the answer is easy because $B(Y,\ell_\infty^N) = \ell_\infty^N(Y^*)$. 2. It follows that $B(Y,X)^{**}=B(Y^{**},X)$ for every subspace of $\ell_\infty^N$. 3. By approximation, the equality follows for every finite-dimensional $X$. $\endgroup$ – Mikael de la Salle Jul 24 '17 at 8:05
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    $\begingroup$ Also, you seem to have missed my point, which is that when you ask if two spaces are isomorphic, you should really specify which map is supposed to be an isomorphism. Think of the question: "is the James space $J$ isomorphic to its bidual?" The answer is yes, but $J$ is not reflexive, because the canonical embedding of $J$ into $J^{**}$ is not surjective. $\endgroup$ – Yemon Choi Jul 24 '17 at 17:37

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