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 $S,N\in \mathcal{L}(F)$ be such that

  • $\operatorname{Im} (T)\subseteq \overline{\operatorname{Im} (M)}$.

  • $\operatorname{Im} (S)\subseteq\overline{\operatorname{Im} (N)}$.

I want to prove that $$\operatorname{Im}(T \widehat{\otimes} S)\subseteq \overline{\operatorname{Im}(M \widehat{\otimes} N)}.$$

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


1 Answer 1


I think this is easy. It follows from the simple fact that if $E_0$ is a dense linear subspace of $E$ then $T(E_0)$ is dense in $T(E)$. Thus Im$(T)\otimes$ Im$(S) =$ Im$(T \otimes S) = T\hat{\otimes} S(E\otimes F)$ is dense in Im$(T\hat{\otimes} S)$, and it is easy to see that Im$(T)\otimes$ Im$(S)$ is contained in the closure of Im$(M\otimes N)$.


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