• $E$ be a $\mathbb R$-Banach space
  • $H$ be a $\mathbb R$-Hilbert space
  • $E\:\hat\otimes_\pi\:H$ denote the completion of the tensor product of $E$ and $H$ with respect to the projective norm

By Riesz' representation theorem, $H$ is isometrically isomorphic to $H'$. Are we able to conclude that $E\:\hat\otimes_\pi\:H$ is isometrically isomorphic to $E\:\hat\otimes_\pi\:H'$?

Let $u\in E\:\hat\otimes_\pi\:H$. There are $(x_n)_{n\in\mathbb N}\subseteq E$ and $(h_n)_{n\in\mathbb N}\subseteq H$ with $$\sum_{n=1}^\infty\left\|x_n\right\|_E\left\|h_n\right\|_H<\infty\tag1$$ and $$\left\|u-\sum_{n=1}^Nx_n\otimes h_n\right\|_{E\:\hat\otimes_\pi\:H}\xrightarrow{N\to\infty}0\tag2.$$ By Riesz' representation theorem, there is a unique $(\varphi_n)_{n\in\mathbb N}\subseteq H'$ with $$\varphi_n=\langle\;\cdot\;,h_n\rangle_H\tag3$$ and $$\left\|\varphi_n\right\|_{H'}=\left\|h_n\right\|_H\tag4$$ for all $n\in\mathbb N$. Clearly, $\sum_{n=1}^\infty\left\|x_n\right\|_E\left\|\varphi_n\right\|_{H'}<\infty$ and hence there is a unique $v\in E\:\hat\otimes_\pi\:H'$ with $$\left\|v-\sum_{n=1}^Nx_n\otimes\varphi_n\right\|_{E\:\hat\otimes_\pi\:H'}\xrightarrow{N\to\infty}0\tag5.$$ So, the candidate for the desired isometric isomorphism is $$\iota:E\:\hat\otimes_\pi\:H\to E\:\hat\otimes_\pi\:H'\;,\;\;\;u\mapsto v\tag6.$$

At the moment, it's not clear to me that $\iota$ is even well-defined, i.e. that $v$ doesn't depend on the choice of $((x_n,h_n))_{n\in\mathbb N}$. How can we show that and how can we show that $\iota$ is an isometric isomorphism?

Remark: Maybe the question can be generalized as follows: Let $F,\tilde F$ be $\mathbb R$-Banach spaces such that $F$ is isometrically isomorphic to $\tilde F$. Is $E\:\hat\otimes_\pi\:F$ isometrically isomorphic to $E\:\hat\otimes_\pi\:\tilde F$?

  • 4
    $\begingroup$ The answer to your final question in the remark is yes, just by functoriality (or even just a direct check using the definitions). Separability is irrelevant. Basically, the projective tensor product of Banach spaces respects isomorphisms of Banach spaces (otherwise we wouldn't bother with it) $\endgroup$
    – Yemon Choi
    May 2, 2018 at 21:22
  • $\begingroup$ @YemonChoi If the answer to the question in the remark is yes, then the answer of the original question is yes too. But as I wrote in the question, I fail to see why $\iota$ is even well-defined (and the approach I've chosen uses the definitions) So, how do we see that? (BTW, I didn't intend to include the separability assumption on $E,F$ and removed it now.) $\endgroup$
    – 0xbadf00d
    May 2, 2018 at 23:15
  • $\begingroup$ The key is to think of the usual (uncompleted) tensor product as having the right kind of universal property, so that any linear map of vector spaces $F \to G$ will always give rise to a well-defined linear map $E\otimes F \to E\otimes G$. Then you just check that this map extends continuously to the completed projective tensor products. (Actually, you can go straight to the completed tensor products if you are OK with their characterisations as suitable universal objects for the category of Banach spaces, but I'm guessing that's not your preferred way of thinking about them) $\endgroup$
    – Yemon Choi
    May 3, 2018 at 0:51
  • 2
    $\begingroup$ Isn't it just $id_E\otimes S: E\hat{\otimes}_\pi F\to E\hat{\otimes}_\pi G$ is an isometry if $S:F\to G$ is an isometry? $\endgroup$ May 3, 2018 at 7:12
  • 1
    $\begingroup$ @JochenWengenroth Actually, the projective tensor product need not respect isometries: if $S:F\rightarrow G$ is an isometry, all we can say in general is that $id_E\otimes S$ is contractive. However, if $S$ is an isometric isomorphism, then $S^{-1}$ exists and is an isometry, so $id_E\otimes S^{-1}$ is also contractive, and is easily seen to be the inverse to $id_E\otimes S$. $\endgroup$ May 3, 2018 at 10:49


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