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?

  • $\begingroup$ 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. $\endgroup$ Commented Jun 5, 2019 at 21:19
  • $\begingroup$ Thank you @GiuseppeNegro for your comment. But could you please elaborate what do you mean by "all necessary continuous estimates"? $\endgroup$
    – user809418
    Commented Jun 8, 2019 at 17:38
  • $\begingroup$ Anyway, let me add that there is nothing deep here, do not fear these things. $\endgroup$ Commented Jun 9, 2019 at 21:35
  • $\begingroup$ Okay, thanks in advance @GiuseppeNegro. $\endgroup$
    – user809418
    Commented Jun 13, 2019 at 9:42

1 Answer 1


$\newcommand{\Hcal}{\mathcal{H}}$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 $\otimes$.

  • $\begingroup$ Sorry, I still do not see exactly how this suffices. If h_1(n) and h_2(n) are defined as the partial sums like you did, then h_1(n) \otimes h_2(n) is going to be a sum of n^2 tensor products (with all cross-products), instead of a sum of tensor products u_k \otimes v_k (with same k) as I had written. Anyway, I wonder whether the answer to my question is yes simply because the scalar product defined for finite linear combinations of elementary tensors is continuous? $\endgroup$
    – user809418
    Commented Feb 5, 2020 at 16:52

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