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I'm studying tensor products of Hilbert spaces following the construction given in Folland's A Course in Abstract Harmonic Analysis. Let $\mathcal{H}_1$ and $\mathcal{H_2}$ two Hilbert spaces. Construct the tensor product $\mathcal{H}_1 \otimes \mathcal{H}_2$ as the set of all bounded antilinear operators from $\mathcal{H_2}$ to $\mathcal{H_1}$ such that the norm $\Vert A \Vert_\otimes := \sqrt{\sum_\beta \Vert Av_\beta \Vert^2}$ is finite, where $\lbrace v_\beta \rbrace$ is any orthonormal basis of $\mathcal{H}_2$.

I have some trouble showing that this space is complete. The author observes that $\forall A \in \mathcal{H}_1 \otimes \mathcal{H}_2$ we have $\Vert A \Vert \leq \Vert A \Vert_\otimes$ (where the first norm is the operatorial one), and so any sequence $\lbrace A_n \rbrace_{n\in\mathbb{N}}$ which is Cauchy with respect to the "tensor norm" converges to an operator $A$ in the operatorial norm topology. Then it is stated that $A$ is also the limit of $\lbrace A_n \rbrace_{n\in\mathbb{N}}$ in the "tensor topology", but I can't prove it: I hoped I could evaluate $\Vert A - A_n \Vert_\otimes^2 = \sum_\beta\Vert Av_\beta - A_nv_\beta\Vert^2$ and find an upper bound but I have no idea how to go further.

I also tried another approach, constructing an inner product space isomorphism between this space and the completion of the tensor product of vector spaces:

$\Phi:\overline{\mathcal{H}_1 \otimes_{\mathbb{C}Mod} \mathcal{H}_2} \rightarrow \mathcal{H}_1 \otimes \mathcal{H}_2$

$\Phi : \qquad u \otimes v\;\;\quad \mapsto \langle v,\cdot\rangle u$

but I can't show that this map is surjective.

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First you should check that $\|A\|_\otimes<\infty$. For this purpose note that $\sup_n\|A_n\|_\otimes<\infty$. Thus $$ \sum_{\beta\in B}\|A v_\beta\|^2=\sup_{F\subset B\text{ finite}}\sum_{\beta\in F}\|A v_\beta\|^2=\sup_{F\subset B\text{ finite}}\lim_{n\to\infty}\sum_{\beta\in F}\|A_n v_\beta\|^2\leq \sup_n \|A_n\|_\otimes^2. $$ Similarly you can prove that $\|A-A_n\|_\otimes\to 0$: $$ \sum_{\beta\in B}\|(A-A_n)v_\beta\|^2=\sup_{F\subset B\text{ finite}}\lim_{m\to\infty}\sum_{\beta\in F}\|(A_m-A_n)v_\beta\|^2\leq\sup_{m}\|A_m-A_n\|_\otimes^2. $$ The right side goes to zero as $n\to\infty$ since $(A_n)$ is a Cauchy sequence w.r.t. $\|\cdot\|_\otimes$.

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  • $\begingroup$ Thank you again for your answer. It seems that seeing the sum as the supremum of all finite sums did the trick. $\endgroup$ – M. Rinetti Feb 6 at 21:51
  • $\begingroup$ Instead of taking the supremum over all finite sums you could just as well take the limit of the parial sums, that does not make a difference for the proof. It's just that this definition also works for uncountable bases. $\endgroup$ – MaoWao Feb 7 at 12:05
  • $\begingroup$ I can see your point, but in this case I don't know anything about the cardinality of the basis, do I? $\endgroup$ – M. Rinetti Feb 9 at 11:51

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