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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 $E\widehat{\otimes}F$?

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Let me give some more details for the approach I suggested in the comments of your other question here.

$E \otimes F$ 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.

This isn't really even a statement about tensor products. It is a special case of the fact that if a bounded operator is positive on a dense subset then it must be positive everywhere.

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  • $\begingroup$ 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. $\endgroup$ – Student Oct 28 '18 at 8:04
  • $\begingroup$ 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). $\endgroup$ – Rhys Steele Oct 28 '18 at 8:13
  • $\begingroup$ Now it is very clear. Thanks a lot. $\endgroup$ – Student Oct 28 '18 at 8:16

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