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

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.)?

• I'm a little confused as to what you are trying to show here? You need to first construct a map from $X'\times Y$ into $\mathfrak L(X,Y)$. The $\iota$ you define above is not well-defined, the image of such a map should be a linear map from $X$ into $Y$. – SamM May 18 '16 at 12:14
• @SamM The elements of the image of $\iota$ are bounded, linear operators from $X$ to $Y$. Note that in $(1)$ each $\varphi_i$ is a bounded, linear operator from $X$ to $\mathbb F$. – 0xbadf00d May 18 '16 at 15:58
• So each of the terms in (1) is a bounded linear functional multiplied by an element of $Y$? Such a multiplication (as it is written) is not well defined. You need to state how such a object should act on elements of $x$, you are trying to define a linear operator after all. – SamM May 18 '16 at 16:20
• @SamM $\varphi_iy_i$ is the function $$X\to Y\;,\;\;\;x\mapsto\varphi_i(x)y_i\;.$$ That's common notation for function objects, but I know that $\varphi_iy_i$ usually means $\varphi_i(y_i)$ (which is undefined here) in the context of operators. – 0xbadf00d May 18 '16 at 16:26
• I see; in that case you are quite right. However, it would be better to define the action of such an object on an element of $X$, thus removing any ambiguity as to a definition. (For instance, stating $\iota(u)(x)=\sum_i \varphi_i(x)y_i$ ($x\in X$) make it clear that you are defining some kind of function, and not defining some action of elements in $X'$ on $Y$. – SamM May 18 '16 at 16:33

## 1 Answer

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.

• how do you see that the closure of $X' \otimes Y$ for $\|u\| = \pi(u)$ becomes by $\iota$ the space of compact operators ? – reuns May 18 '16 at 18:11
• no ok, it is obvious that it is a subset of the space of compact operators, since if $\iota(u)$ is not compact then $u = \sum_{i=1}^\infty \varphi_i \otimes y_i$ where $\pi(\varphi_i \otimes y_i) \not \to 0$ and $\pi(u)$ cannot be finite – reuns May 18 '16 at 18:15
• This follows from the following, standard characterization of compact operators: $T\in\mathfrak{L}(X,Y)$ is compact iff $T$ is the norm limit of a sequence of finite rank operators. – Pedro Lauridsen Ribeiro May 18 '16 at 18:16
• and we can think to $\pi$ (the norm that the OP defined on $X' \otimes Y$) as some sort of Frobenius norm on $X\to Y$ – reuns May 18 '16 at 18:22
• Sorry, what I said in my previous comment is true only if $X$ and $Y$ are Hilbert spaces (Enflo provided a counterexample for Banach spaces). I'll fix that in my answer. In any case, the claim that $\iota$ is not open in general still stands - since the subspace of compact operators is norm closed, the closure of the subspace of finite rank operators is contained in the subspace of compact operators. – Pedro Lauridsen Ribeiro May 18 '16 at 18:22