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

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?

• Just to clarify: you are not trying to prove that $E\hat\otimes_\pi E$ is isomorphic to a closed subspace of ${\rm Bil}(E' \times E')$ are you? Jan 22, 2018 at 23:32
• Secondly, my immediate instinct is to worry if something goes wrong when E does not have the approximation property Jan 22, 2018 at 23:32
• @YemonChoi I've found the claim in the book Semimartingales: A Course on Stochastic Processes by Michel Métivier on page 138. As I indicated in the question, I thought he means that $E\:\hat\otimes_\pi\:E$ is embedded into $\mathfrak B(E'\times E')$. Jan 23, 2018 at 16:53
• @YemonChoi It's fine for me to assume that $E$ has the approximation property; but he doesn't make this assumption. Jan 23, 2018 at 16:54

As suspected by Yemon Choi the question is very closely related to the approximation property: As can be seen e.g. in the book Tensor Norms and Operator Ideals of Defant and Floret (page 64 combined with the remark 5.4) a Banach space $E$ has the approximation property if and only if the canonical mapping $E\tilde{\otimes}_\pi F \to (E' \otimes_\pi F')'$ is injective for all Banach spaces $F$ (or only for $F=E'$). I don't know if it is written somewhere but I would be very surprised if the mapping would always be injective for $F=E$.
The question as stated (whether $E\tilde{\otimes}_\pi E$ is isomorphic to some subspace of $(E' \otimes_\pi E')'$ in a possibly non-canonical way) is, of course, very different.
Let $P$ be one of the Banach spaces constructed by Theorem 3.2 in Pisier, "Counterexamples to a conjecture of Grothendieck". That is, $P \widehat\otimes P = P \check\otimes P$. Now, $P\check\otimes P$ embeds isometrically into $B(P',P) \subseteq B(P',P'') = (P' \widehat\otimes P')'$, and this embedding is exactly that described in the OP. So for $P$, we obtain a positive answer.
As Jochen Wengenroth says, if $E$ has the approximation property then $E\widehat\otimes E \rightarrow B(E',E'')$ is injective. It seems to me that for "nice" spaces this is unlikely to be bounded below.