Can we embed $X'\otimes Y$ into the space of bounded, linear operators $X\to Y$?

Question

Let

• $\mathbb F\in\left\{\mathbb C,\mathbb R\right\}$
• $X$ and $Y$ be normed $\mathbb F$-vector spaces
• $X'$ denote the topological dual space of $X$
• $\mathfrak L(X,Y)$ denote the space of bounded, linear operators from $X$ to $Y$
• $\mathfrak B(X'\times Y,\mathbb F)$ be the space of bilinear forms on $X'\times Y$
• $X\otimes Y$ denote the tensor product of $X$ and $Y$

Can we show that $X'\otimes Y$ can be embedded into $\mathfrak L(X,Y)$, i.e. that there is a

1. injective,
2. continuous and
3. open

mapping $\iota:X'\otimes Y\to\iota(X'\otimes Y)$?

Clearly, we would need to choose a norm on $$X'\otimes Y:=\operatorname{span}\left\{\varphi\otimes y:(\varphi,y)\in X'\times Y\right\}\;,$$ where $$(\varphi\otimes y)(A):=A(\varphi,y)\;\;\;\text{for }B\in\mathfrak B(X'\times Y,\mathbb F)\;.$$ I think that the projective norm $$\pi(u):=\inf\left\{\sum_{i=1}^n\left\|\varphi_i\right\|_{X'}\left\|y_i\right\|_Y:u=\sum_{i=1}^n\varphi_i\otimes y_i\right\}$$ will do it.

My idea is to define $$(\iota u)(x):=\sum_{i=1}^n\varphi_i(x)y_i\;\;\;\text{for }x\in X\tag 1$$ for $u\in X'\otimes Y$ with $u=\sum_{i=1}^n\varphi_i\otimes y_i$.

This $\iota$ is obviously linear. Maybe we can show that it is bounded too (i.e. a bounded, linear operator). This would yield (2.). How can we show this and how can we show (1.) and (3.)?

Answer 1

To see that the linear map $\iota$ defined by (1) is injective, first notice that for any representation $u=\sum^n_{j=1}\phi_j\otimes y_j$ of $u\in X'\otimes Y$ one may assume that the $y_j$'s are linearly independent by simply redefining the $\phi_j$'s. Once this is done, we have that $\iota(u)=0$ implies that $\phi_j=0$ for each $j=1,\ldots,n$ and therefore $u=0$, as desired. This allows us to identify $X'\otimes Y$ as a vector space with the subspace of finite-rank elements of $\mathfrak{L}(X,Y)$.

Moreover, $\iota$ is certainly continuous, for if $T=\sum^n_{j=1}\phi_j(\cdot)y_j$ is a finite-rank linear operator from $X$ to $Y$ (so that $\phi_j\in X'$, $y_j\in Y$, $j=1,\ldots,n$), then $$\|T\|\doteq\sup_{\|x\|=1}\|Tx\|\leq\sum^n_{j=1}\sup_{\|x\|=1}\|\phi_j(x)y_j\|=\sum^n_{j=1}\|\phi_j\|\cdot\|y_j\|\ .$$

Finally, $\iota(X'\otimes Y)$ is certainly a $F_\sigma$ subset of $\mathfrak{L}(X,Y)$ but it cannot be open in general unless $Y$ is finite dimensional, for its (operator) norm closure is contained in the (operator norm closed) subspace of compact linear operators from $X$ to $Y$. After all, the image of the unit ball of $X$ under a finite-rank linear operator is a bounded subset of a finite dimensional normed vector space and thus relatively compact. Therefore, the map $\iota$ cannot be open if $Y$ is infinite dimensional.

Remark: The completion of $X'\otimes Y$ with respect to the projective norm $\pi$ can be identified through $\iota$ with the subspace of nuclear (i.e. trace-class) linear operators from $X$ to $Y$, which are of course compact due to the above inequality.

Remark 2: As a nomenclature aside, the projective norm $\pi$ in finite dimensions is not the Frobenius norm (more usually called "Hilbert-Schmidt norm" in infinite dimensional Hilbert spaces), but the Ky Fan norm (more usually called "nuclear norm" in infinite dimensions). They are equivalent when $X,Y$ (and therefore $X'\otimes Y$) are finite dimensional, of course, but no longer so in the infinite-dimensional case, for instance in the completion of the subspace of finite-rank linear operators in the strongest of both norms. In the case $X=Y$, the nuclear norm is dominated by the trace norm, and both coincide if and only if $X$ is topologically isomorphic to a Hilbert space (see comments below for details), which implies that in this case the nuclear norm is strictly stronger than the Hilbert-Schmidt norm in infinite dimensions.