# Computing inner products on tensor product of Hilbert spaces

Let $$\mathcal{H}_1$$ and $$\mathcal{H}_2$$ be Hilbert spaces with inner products $$\langle \cdot, \cdot \rangle_1$$ and $$\langle \cdot, \cdot \rangle_2$$, respectively. As described in many references (and Wikipedia as well), one can define a corresponding pre-Hilbert tensor product space by taking the (algebraic) tensor product $$\mathcal{H}_1 \times \mathcal{H}_2$$ equipped with the inner product $$\langle x_1 \otimes x_2, y_1 \otimes y_2 \rangle \triangleq \langle x_1, y_1 \rangle_1 \langle x_2, y_2 \rangle_2 \quad \forall x_1,y_1 \in \mathcal{H}_1, \; x_2,y_2 \in \mathcal{H}_1,$$ which extends bilinearly to finite linear combinations. The completion of this construction then yields a Hilbert space, which is a topological tensor product denoted by $$\mathcal{H}_1 \hat{\otimes} \mathcal{H}_2$$.

Now, my question regards the computation of the inner product for a given pair of tensors from $$\mathcal{H}_1 \hat{\otimes} \mathcal{H}_2$$. Let two arbitrary tensors in that space be $$x = \sum_{n=1}^\infty u_n \otimes v_n, \quad \text{and} \quad y = \sum_{n=1}^\infty w_n \otimes z_n.$$ Now, the inner product was defined so as to extend bilinearly for finite linear combinations. Hence, how can one compute the inner product $$\langle x, y \rangle$$? Does it make sense to write $$\langle x, y \rangle = \langle \sum_{n=1}^\infty u_n \otimes v_n, \sum_{m=1}^\infty w_m \otimes z_m \rangle = \sum_{n=1}^\infty \sum_{m=1}^\infty \langle u_n \otimes v_n, w_m \otimes z_m \rangle,$$ as if bilinearity holds also for infinite linear combinations?

• Yes, of course, these things are constructed so that this kind of linear combinations are continuous operations. All necessary continuous estimates are supplied by Cauchy-Schwarz. Jun 5, 2019 at 21:19
• Thank you @GiuseppeNegro for your comment. But could you please elaborate what do you mean by "all necessary continuous estimates"? Jun 8, 2019 at 17:38
• Anyway, let me add that there is nothing deep here, do not fear these things. Jun 9, 2019 at 21:35
• Okay, thanks in advance @GiuseppeNegro. Jun 13, 2019 at 9:42

$$\newcommand{\Hcal}{\mathcal{H}}$$Your question boils down to the following.
If $$(h_1), (h_2)$$ are sequences in $$\Hcal_1, \Hcal_2$$ respectively, is it true that $$\tag{1} h_1(n)\to h_1 \ \text{ and }\ h_2(n)\to h_2\ \quad \Longrightarrow\quad h_1(n)\otimes h_2(n)\to h_1\otimes h_2?$$
Indeed, once (1) is proven, you can just apply it to the sequences $$h_1(n):=\sum_{k=1}^n u_k, \qquad h_2(n):=\sum_{k=1}^n v_k,$$ which you defined in the main text.
The proof of (1) is an immediate consequence of the obvious inequality $$\lVert h_1\otimes h_2\rVert\le \lVert h_1\rVert\lVert h_2\rVert$$ (actually, this is even an identity), and of the bilinearity of $$\otimes$$.