Let us consider E a finite-dimensional normed space on $\mathbb{R}$ and a real number $p\geq 1$. Is it true that the projective tensorial product $E \widehat{\otimes}_\pi L^p(\mathbb{R},\mathbb{R})$ is isometric to $L^p(\mathbb{R},E)$.
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2$\begingroup$ Only for $p=1$. $\endgroup$– alphaCommented Feb 12, 2014 at 13:46
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$\begingroup$ Are you sure ? even if $E$ is finite dimensional ? Do you know any reference ? $\endgroup$– djangoCommented Feb 12, 2014 at 19:30
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$\begingroup$ For $p\neq 1$ and $E$ finite dimensional, they are isomorphic but not isometrically so. If they were isometric, then by a simple compactness argument the same would hold for $E$ infinite dimensional. I don't have access to a library at the moment so I can't give you a precise reference now. $\endgroup$– alphaCommented Feb 12, 2014 at 21:50
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2$\begingroup$ Why close this? It's not a bad question for a non-expert to ask! $\endgroup$– Todd TrimbleCommented Feb 13, 2014 at 8:22
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See section The Natural Norm on the $p$-Integrable Functions in Tensor Norms and Operator Ideals by A. Defant and K. Floret,
