# The tensor product of two positive operators is a positive operator

Let $$E$$, $$F$$ be two complex Hilbert spaces and $$\mathcal{L}(E)$$ (resp. $$\mathcal{L}(F)$$) be the algebra of all bounded linear operators on $$E$$ (resp. $$F$$).

The algebraic tensor product of $$E$$ and $$F$$ is given by $$E \otimes F:=\left\{\xi=\sum_{i=1}^dv_i\otimes w_i:\;d\in \mathbb{N}^*,\;\;v_i\in E,\;\;w_i\in F \right\}.$$

In $$E \otimes F$$, we define $$\langle \xi,\eta\rangle=\sum_{i=1}^n\sum_{j=1}^m \langle x_i,z_j\rangle_1\langle y_i ,t_j\rangle_2,$$ for $$\xi=\displaystyle\sum_{i=1}^nx_i\otimes y_i\in E \otimes F$$ and $$\eta=\displaystyle\sum_{j=1}^mz_j\otimes w_j\in E \otimes F$$.

The above sesquilinear form is an inner product in $$E \otimes F$$.

It is well known that $$(E \otimes F,\langle\cdot,\cdot\rangle)$$ is not a complete space. Let $$E \widehat{\otimes} F$$ be the completion of $$E \otimes F$$ under the inner product $$\langle\cdot,\cdot\rangle$$.

If $$T\in \mathcal{L}(E)$$ and $$S\in \mathcal{L}(F)$$, then the tensor product of $$T$$ and $$S$$ is denoted $$T\otimes S$$ and defined as $$\big(T\otimes S\big)\bigg(\sum_{k=1}^d x_k\otimes y_k\bigg)=\sum_{k=1}^dTx_k \otimes Sy_k,\;\;\forall\,\sum_{k=1}^d x_k\otimes y_k\in E \otimes F,$$ which lies in $$\mathcal{L}(E \otimes F)$$. The extension of $$T\otimes S$$ over the Hilbert space $$E \widehat{\otimes} F$$, denoted by $$T \widehat{\otimes} S$$, is the tensor product of $$T$$ and $$S$$ on the tensor product space, which lies in $$\mathcal{L}(E\widehat{\otimes}F)$$.

An operator $$A\in\mathcal{L}(E)$$ is said to be positive if $$\langle Ax\mid x\rangle \geq 0$$ for any $$x\in E$$.

If $$T$$ and $$S$$ are positive operators, then clearly $$T\otimes S$$ is positive on $$E \otimes F$$. How to prove that $$T \widehat{\otimes} S$$ is positive on $$E\widehat{\otimes}F$$?

$$E \otimes F$$ is included (isometrically) in $$E \hat \otimes F$$ in an obvious way and we have that $$T \hat \otimes S = T \otimes S$$ on $$E \otimes F$$.
So for $$x \in E \otimes F$$, we have that $$\langle T \hat \otimes S (x), x \rangle \geq 0$$ since $$T \otimes S$$ is positive. For general $$x \in E \hat \otimes F$$, there is a sequence $$x_n$$ in $$E \otimes F$$ such that $$x_n \to x$$ in $$E \hat \otimes F$$. Then $$0 \leq \langle T \hat \otimes S(x_n), x_n \rangle \to \langle T \hat \otimes S(x), x \rangle$$ since $$T \hat \otimes S$$ is continuous and $$\langle \cdot, \cdot \rangle$$ is jointly continuous.
• Please I don't understand why $E \otimes F$ is dense in $E \hat \otimes F$? And the density in under what topology? Thanks a lot for your help. – Student Oct 28 '18 at 8:04
• This is part of the definition of the completion of a metric space. See the definition here. This means $E \otimes F$ is dense in $E \hat \otimes F$ for the topology induced by the inner product on $E \hat \otimes F$ (which induces the original topology on $E \otimes F$ as a subspace). – Rhys Steele Oct 28 '18 at 8:13