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 ChoiCommented Jul 24, 2017 at 5:02
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$\begingroup$ @ 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 $(Y\hat{\otimes}\ell_2^n)^{***}=B(Y^{**},\ell_2^n).$ $\endgroup$– MathbuffCommented Jul 24, 2017 at 6:18
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1$\begingroup$ 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 SalleCommented Jul 24, 2017 at 8:03
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1$\begingroup$ 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 SalleCommented Jul 24, 2017 at 8:05
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1$\begingroup$ 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 ChoiCommented Jul 24, 2017 at 17:37
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