# Tensor Product of Hilbert Spaces: how to prove completeness

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.

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. $$\|\cdot\|_\otimes$$.