Let

- $E$ be a $\mathbb R$-Banach space
- $E\:\hat\otimes_\pi\:E$ denote the projective tensor product

How can we show that $E\:\hat\otimes_\pi\:E$ is isomorphic to a subspace of $\left(E'\:\hat\otimes_\pi\:E'\right)'$?

Clearly, if $\mathfrak B(E'\times E')$ denotes the space of bounded bilinear forms on $E'\times E'$, then $\mathfrak B(E'\times E')$ is isometrically isomorphic to $\left(E'\:\hat\otimes_\pi\:E'\right)'$.

So, we could conclude, if we would be able to show that $E\:\hat\otimes_\pi\:E$ can be embedded into $\mathfrak B(E'\times E')$.

Actually, I know that $E\otimes E$ (the algebraic tensor product) can be embedded into $\mathcal B(E^\ast\times E^\ast)$ (the set of bilinear forms on the cartesian product of the algebraic dual space $E^\ast$ with itself).

A canonical choice for this embedding would be $$\sum_{i=1}^nx_i\otimes y_i\mapsto\left((\varphi,\psi)\mapsto\sum_{i=1}^n\varphi(x_i)\psi(y_i)\right)\tag1\;.$$ If $\iota$ denotes this embedding, then it's easy to see that $\iota$ is a bounded linear operator from $E\otimes_\pi E$ (the algebraic tensor product $E\otimes E$ equipped with the projective norm) to $\mathfrak B(E'\times E')$ and hence admits a unique extension to a bounded linear operator $\overline\iota$ from $E\:\hat\otimes_\pi\:E$ to $\mathfrak B(E'\times E')$.

If this approach is sensible at all, the only thing I need to conclude is the injectivity of $\overline\iota$. How can we show that?

closedsubspace of ${\rm Bil}(E' \times E')$ are you? $\endgroup$ – Yemon Choi Jan 22 '18 at 23:32Semimartingales: A Course on Stochastic Processesby Michel Métivier on page 138. As I indicated in the question, I thought he means that $E\:\hat\otimes_\pi\:E$ is embedded into $\mathfrak B(E'\times E')$. $\endgroup$ – 0xbadf00d Jan 23 '18 at 16:53