• $E$ 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?

  • $\begingroup$ 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? $\endgroup$ – Yemon Choi Jan 22 '18 at 23:32
  • 1
    $\begingroup$ Secondly, my immediate instinct is to worry if something goes wrong when E does not have the approximation property $\endgroup$ – Yemon Choi Jan 22 '18 at 23:32
  • $\begingroup$ @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 $\mathfrak B(E'\times E')$. $\endgroup$ – 0xbadf00d Jan 23 '18 at 16:53
  • $\begingroup$ @YemonChoi It's fine for me to assume that $E$ has the approximation property; but he doesn't make this assumption. $\endgroup$ – 0xbadf00d Jan 23 '18 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.

  • $\begingroup$ Possibly something related to the infamous/celebrated Pisier space might work? $\endgroup$ – Yemon Choi Jan 23 '18 at 10:53

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.