• $X,Y$ be $\mathbb R$-Banach spaces
  • $X\otimes_\pi Y$ denote the projective tensor product and $X\:\hat\otimes_\pi\:Y$ the completion

If $u=\sum_{i=1}^nx_i\otimes y_i\in X\otimes Y$, then $$Y'\ni\psi\mapsto\sum_{i=1}^nx_i\psi(y_i)\tag1$$ is a bounded linear operator from $Y'$ to $X$ with norm at most $\left\|u\right\|_{X\:\otimes_\pi\:Y}$ (is the norm even equal to $\left\|u\right\|_{X\:\otimes_\pi\:Y}$?). So, the mapping $X\otimes_\pi Y\to\mathfrak L(Y',X)$ is a bounded linear operator and hence admits a unique extension to $X\:\hat\otimes_\pi\:Y$.

If $u=\sum_{n=1}^\infty x_n\otimes y_n\in X\:\hat\otimes_\pi\:Y$, the associated operator is $$Y'\ni\psi\mapsto\sum_{n=1}^\infty x_n\psi(y_n)\tag2.$$

It's easy to see that $X\otimes_\pi Y\to\mathfrak L(Y',X)$ is injective, but is $X\:\hat\otimes_\pi\:Y\to\mathfrak L(Y',X)$ injective too?

I think that we need to assume that $Y$ has the approximation property. Indeed, $Y$ has the approximation property if and only if for all $u=\sum_{n=1}^\infty x_n\otimes y_n\in X\:\hat\otimes_\pi\:Y$, $$\sum_{n=1}^\infty x_n\psi(y_n)=0\;\;\;\text{for all }\psi\in Y'\tag3$$ implies $u=0$. Anything I'm missing?

  • $\begingroup$ It seems you need to look at different reformulations of approximation property. $\endgroup$ – Norbert Jan 23 at 15:04

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