Before I ask my actual question about direct sums and tensor products of Hilbert spaces, let's first talk about direct sums and tensor products of vector spaces.
We might define direct sums of vector spaces by the corresponding universal mapping property, but for later, it's more useful to work with a constructive definition. For a family $\{V_i\}_{i \in I}$ of vector spaces, we may define the direct sum $\bigoplus_{i \in I} V_i$ to be the set of all sequences $(v_i)_{i \in I}$ for which $v_i \in V_i$ for all $i \in I$ and $v_i = 0$ for all but finitely many $i \in I$, endowed with the component-wise vector addition $(v_i)_{i \in I} + (w_i)_{i \in I} = (v_i + w_i)_{i \in I}$ and the scalar multiplication $\alpha (v_i)_{i \in I} = (\alpha v_i)_{i \in I}$.
I won't repeat the definition of the tensor product of vector spaces here.
For the direct sum and the tensor product of vector spaces, it holds that the they are distributive in the sense that there is a canonical isomorphism
\begin{align*}
W \otimes \bigoplus_{i \in I} V_i \simeq \bigoplus_{i \in I} (W \otimes V_i) \,,
\end{align*}
where $w \otimes (v_i)_{i \in I} \mapsto (w \otimes v_i)_{i \in I}$, as explained, for example, in Theorem 5.4 in this document about tensor products.
For Hilbert spaces, the direct sum explained above does not lead necessarily lead to Hilbert spaces since the (vector space) direct sum of infinitely many Hilbert spaces is not complete. Thus, for Hilbert spaces, one usually defines the direct sum as follows. For a family $\{H_i\}_{i \in I}$ of Hilbert spaces, we may define the direct sum $\bigoplus_{i \in I} H_i$ to be the set of all sequences $(v_i)_{i \in I}$ for which $v_i \in H_i$ for all $i \in I$, $v_i = 0$ for all but countably many $i \in I$ and $\sum_{i \in I} \Vert v_i \Vert^2 < \infty$, endowed with the component-wise vector addition $(v_i)_{i \in I} + (w_i)_{i \in I} = (v_i + w_i)_{i \in I}$ and the scalar multiplication $\alpha (v_i)_{i \in I} = (\alpha v_i)_{i \in I}$. The inner product on this space is given by $\langle (v_i)_{i \in I}, (w_i)_{i \in I} \rangle = \sum_{i \in I} \langle v_i, w_i \rangle$.
For Hilbert spaces, the (vector space) tensor product does not necessarily lead to Hilbert spaces since the tensor product of infinite-dimensional Hilbert spaces is not complete. However, one may define the tensor product of two Hilbert spaces $H$ and $K$ to be the metric completion of the vector space tensor product with respect to the inner product $\langle v_1 \otimes w_1, v_2 \otimes w_2 \rangle = \langle v_1, v_2 \rangle \langle w_1, w_2 \rangle$ on $H \otimes K$. (See the Wikipedia article on tensor products of Hilbert spaces.)
My question: For this "Hilbert space direct sum" and "Hilbert space tensor product", is there also a canonical isomorphism
\begin{align*}
K \otimes \bigoplus_{i \in I} H_i \simeq \bigoplus_{i \in I} (K \otimes H_i)?
\end{align*}
I know that the tensor product for Hilbert spaces fails to have the universal property (at least for continuous linear functions). Hence, if this would be needed in the proof of the distributivity, then distributivity fails.
Best Answer
Let's denote the tensor product of Hilbert spaces by $\widehat{\otimes}$, the tensor product of vector spaces by $\otimes$.
We have a natural bilinear map $K \times \bigoplus_i H_i \to \bigoplus_i (K \widehat{\otimes} H_i)$ which sends $(a,(h_i))$ to $(a \otimes h_i)$. It extends to a linear map $K \otimes \bigoplus_i H_i \to \bigoplus_i (K \widehat{\otimes} H_i)$. This map is isometric because $$\langle (a \otimes h_i),(a' \otimes h'_i) \rangle = \sum_i \langle a,a' \rangle \cdot \langle h_i,h'_i \rangle = \langle a,a' \rangle \langle (h_i),(h'_i) \rangle.$$ Hence it extends to an isometric linear map on the completion $K \widehat{\otimes} \bigoplus_i H_i \to \bigoplus_i (K \widehat{\otimes} H_i)$. One checks that its image is dense. The image is also closed since it is complete. Thus it is an isometric isomorphism.