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, I want to prove $T\otimes S$ is positive on $E \otimes F$.

Let $X=\sum_{i=1}^nx_i\otimes y_i\in E\otimes F$. Then \begin{align*} \langle (T\otimes S)X,X\rangle & =\sum_{i=1}^n\sum_{j=1}^n \langle Tx_i\mid x_j\rangle_1\langle Sy_i\mid y_j\rangle_2. \end{align*} My goal is to prove that $$\langle (T\otimes S)X,X\rangle \geq 0,$$ for any $X \in E\otimes F$.


I'm not sure that it can be done directly as you want. But if you know that positive operators admit positive square roots, you know that $T=T_0^*T_0$, $S=S_0^*S_0$. Then $$ T\otimes S=T_0^*T_0\otimes S_0^*S_0=(T_0\otimes S_0)^*(T_0\otimes S_0)\geq0. $$

  • $\begingroup$ Thanks a lot for your answer. However, I think you have shown that $T \widehat{\otimes} S$ is a positive operator in $E\widehat{\otimes}F$ because $(T_0\otimes S_0)^*$ does not make sense in $E\otimes F$ because it is not a Hilbert space. $\endgroup$
    – Student
    Jan 31 '19 at 18:57
  • $\begingroup$ Yes, but the restriction to $E\otimes F$ is $T\otimes S$. $\endgroup$ Jan 31 '19 at 19:00
  • $\begingroup$ Yes thank you professor. $\endgroup$
    – Student
    Jan 31 '19 at 19:04

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