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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 $\mathcal{L}(E\widehat{\otimes}F)$.

Let $\operatorname{Im} X$ and $\overline{\operatorname{Im} X}$ denote respectively the range of an operator $X$ and the closure of its range.

Let $T,M\in \mathcal{L}(E)$ and $S,N\in \mathcal{L}(F)$. If $\operatorname{Im} (T)\subseteq \overline{\operatorname{Im} (M)}$ and $\operatorname{Im} S\subseteq\overline{\operatorname{Im} (N)}$. I want to show that $$\operatorname{Im}(T \widehat{\otimes} S)\subseteq \overline{\operatorname{Im}(M \widehat{\otimes} N)}.$$

Note that I show that $$\overline{\operatorname{Im} M}\otimes\overline{\operatorname{Im} N}\subseteq\overline{\operatorname{Im}(M \otimes N)}.$$

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Firstly, it is immediate that $$\operatorname{Im}(T \otimes S) = \operatorname{Im}(T) \otimes \operatorname{Im}(S) \subseteq\overline{\operatorname{Im}(M)}\otimes \overline{\operatorname{Im}(N)} \subseteq \overline{\operatorname{Im}(M\otimes N)}$$ and so $\overline{\operatorname{Im}(T \otimes S)} \subseteq \overline{\operatorname{Im}(M\otimes N)}$

Now, $\operatorname{Im}(T \otimes S) = T \hat \otimes S (E \otimes F)$. But $T \hat \otimes S$ is a continuous map on $E \hat \otimes F$ and so $$T \hat \otimes S(E \hat \otimes F) = T \hat \otimes S( \overline{E \otimes F}) \subseteq \overline{T \hat \otimes S(E \otimes F)}.$$ Therefore $\operatorname{Im}(T \hat \otimes S) \subseteq \overline{\operatorname{Im}(T\otimes S)} \subseteq \overline{\operatorname{Im}(M\otimes N)}$

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  • $\begingroup$ Thank you for your answer but I want to show that $$\operatorname{Im}(T \widehat{\otimes} S)\subseteq \overline{\operatorname{Im}(M \widehat{\otimes} N)}.$$ However you show that $$\operatorname{Im}(T \widehat{\otimes} S)\subseteq \overline{\operatorname{Im}(M \otimes N)}.$$ $\endgroup$ Commented Aug 6, 2018 at 11:01
  • $\begingroup$ @Student What I show is a stronger result. $M \hat \otimes N$ is an extension of $M \otimes N$ so obviously $\operatorname{Im}(M \otimes N) \subseteq \operatorname{Im}(M \hat \otimes N)$... $\endgroup$ Commented Aug 6, 2018 at 11:03
  • $\begingroup$ Could you please explain me why $$T \hat \otimes S(E \hat \otimes F) = T \hat \otimes S( \overline{E \otimes F}) ?$$ Thanks a lot. $\endgroup$ Commented Aug 6, 2018 at 12:42
  • $\begingroup$ @Student $E \hat \otimes F = \overline{E \otimes F}$ by definition of the completion (noting that the closure takes place in $E \hat \otimes F$). $\endgroup$ Commented Aug 6, 2018 at 12:44
  • $\begingroup$ Please my last question is: why $\operatorname{Im}(T \otimes S) = T \hat \otimes S (E \otimes F)$? Because by definition $$\operatorname{Im}(T \otimes S) = T \otimes S (E \otimes F).$$ Thanks a lot for your help. $\endgroup$ Commented Aug 7, 2018 at 10:58

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