Let

- $E$ 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 $E\:\hat\otimes_\pi\:\tilde F$?

projectivetensor 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 $id_E\otimes S$. $\endgroup$ – Matthew Daws May 3 '18 at 10:49