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Let $H$ be a complex Hilbert space and $\mathcal{L}(H)$ be the algebra of all bounded linear operators on $E$.

If $A,B\in \mathcal{L}(H)$, It is true that $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$?

I try as follows:

Let $z\in \overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}$, then $z=\displaystyle\sum_{i=1}^dx_i\otimes y_i$, where $x_i\in\overline{\text{Im}(A)},\;y_i\in\overline{\text{Im}(B)}$ and $d\in\mathbb{N}$. So, there exists $(\alpha_i(n))_{n\in\mathbb{N}}\subset \text{Im}(A)$ and $(\beta_i(n))_{n\in\mathbb{N}}\subset \text{Im}(B)$ such that $x_i=\lim_{n\rightarrow\infty}\alpha_i(n)$ and $y_i=\displaystyle\lim_{n\rightarrow\infty}\beta_i(n)$. Moreover, $\alpha_i(n)=A\tilde{\alpha}_i(n),\;\beta_i(n)=B\tilde{\beta}_i(n)$ and $$\alpha_i(n)\otimes\beta_i(n)=(A\otimes B)(\tilde{\alpha}_i(n)\otimes \tilde{\beta}_i(n))\subset \text{Im}(A\otimes B),$$ where $(\tilde{\alpha}_i(n))_{n\in\mathbb{N}}\subset \mathcal{H}$ and $(\tilde{\beta}_i(n))_{n\in\mathbb{N}}\subset \mathcal{H}$. Let $$z_n=\displaystyle\sum_{i=1}^d\alpha_i(n)\otimes \beta_i(n)\in \text{Im}(A\otimes B).$$ Since $\|x\otimes y\|=\|x\|\|y\|$, then the map $(x,y)\longmapsto x\otimes y$ is continuous. So $z_n\longrightarrow z$ as $n\longrightarrow \infty$, this implies that $z\in \overline{\text{Im}(A\otimes B)}$. As a consequence, $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$.

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    $\begingroup$ The proof is okay. You did not specify the on the tensor product $H\otimes H$ in which you take the closure. Most naturally this would be the Hilbert-space or $\ell^2$ tensor product. But the proof works for all reasonable ones. $\endgroup$ – Peter Michor Jan 18 '18 at 7:31
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The proof is okay. You did not specify the norm on the tensor product $H⊗H$ in which you take the closure. I do not know which one you want. Most naturally this would be the Hilbert-space or $\ell^2$ tensor product (which also describes the space of Hilbert-Schmidt operators). But the proof works for all reasonable norms. Another one describes the space of compact operators. Some of these are called Schatten-norms. See the beginnig of here, or also section 7 of here (and references therein) for more information.

Added:

Details of the proof:

The Hilbert-Schmidt norm on $\mathcal H\otimes\mathcal H$ can be defined as: $$ \|z\|_2 = \inf \Big\{ \|(\|x_i\|)\|_{\ell^2}.\sup_{\|y'\|\le1}\|(\langle y_i,y'\rangle)\|_{\ell^2} : z = \sum x_i\otimes y_i \Big\} $$ Here the infimum is taken over all (finite) representations of $z$ in $\mathcal H\otimes\mathcal H$. This can be seen from section 7 of the second reference I gave above (quite complicated there); best are the references Chevet or Saphar therein which I do not have available now. Given $z$ as in your proof above, choose a finite representation $z = \sum_{i=1}^{N(n)} x_i(n)\otimes y_i(n)$ such that $$ \|(\|x_i(n)\|)\|_{\ell^2}.\sup_{\|y'\|\le1}\|(\langle y_i(n),y'\rangle)\|_{\ell^2} \le \|z\|_2 +\frac{\epsilon}n\,. $$ Now choose your $\alpha_i(n,m)$ and $\beta_i(n,m)$ in such a way that $$ \|(\|x_i(n)-\alpha_i(n,m)\|)\|_{\ell^2}<\frac{\epsilon}{nm}\,,\qquad \|(\|y_i(n)-\beta_i(n,m)\|)\|_{\ell^2}<\frac{\epsilon}{nm}\,. $$ Then you can prove convergence in the $\ell^2$-tensor product.

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  • $\begingroup$ It seems that both links point to the same document. $\endgroup$ – Loïc Teyssier Jan 18 '18 at 23:12
  • $\begingroup$ Link corrected. Sorry. So the tensor product is the one describing Hilbert-Schmidt operators. $\endgroup$ – Peter Michor Jan 19 '18 at 9:29

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