Hilbert Space Direct Sum is Hilbert Space

Theorem
Let $\left({H_i}\right)_{i \in I}$ be a $I$-indexed family of Hilbert spaces over $\Bbb F \in \left\{{\R, \C}\right\}$.

Let $H = \displaystyle \bigoplus_{i \in I} H_i$ be their Hilbert space direct sum.

Then $H$ is a Hilbert space.

$H$ is a Vector Space
From the definition of Hilbert space direct sum, we see that $H$ is a nonempty subset of a vector space (namely, the direct sum of the $H_i$ as vector spaces).

From the Vector Subspace Test it follows that it is to be shown that:


 * $(1): \qquad \forall h_1, h_2 \in H: \displaystyle \sum \left\{{ \left\Vert{ \left({h_1 + h_2}\right) \left({i}\right) }\right\Vert^2_{H_i}: i \in I }\right\} < \infty$
 * $(2): \qquad \forall \lambda \in \Bbb F, h \in H: \displaystyle \sum \left\{{ \left\Vert{ \left({\lambda h}\right) \left({i}\right) }\right\Vert^2_{H_i}: i \in I }\right\} < \infty$

Considering $(1)$, have the following:

For $(2)$, observe that:

Thus, by the Vector Subspace Test, $H$ is a vector space.

$\left\langle{\cdot, \cdot}\right\rangle$ is an Inner Product
It suffices to check well-definedness of $\left\langle{\cdot, \cdot}\right\rangle$, and subsequently the five properties of an inner product.

Well-definedness
It is necessary to verify that for $g, h \in H$, in fact $\left\langle{g, h}\right\rangle \in \Bbb F$.

That is, it is required to show that $\left\langle{g, h}\right\rangle = \displaystyle \sum \left\{{ \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}$ converges in $\Bbb F$.

Absolutely Convergent Generalized Sum Converges applies to the Banach space $\Bbb F$ and the $I$-indexed subset $\left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}$ of $\Bbb F$.

Hence it will suffice to show that $\displaystyle \sum \left\{{ \left\vert{ \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i} }\right\vert: i \in I}\right\}$ converges in $\R$.

For brevity, denote already $\left\Vert{h}\right\Vert^2$ for the expression $\displaystyle \sum \left\{{\left\Vert{h \left({i}\right)}\right\Vert_{H_i}^2: i \in I}\right\}$.

Define $g' \in H$ by $g' \left({i}\right) = \begin{cases} g \left({i}\right) & \text{if } \left\Vert{g \left({i}\right)}\right\Vert_{H_i} \ge \left\Vert{h \left({i}\right)}\right\Vert_{H_i} \\ \mathbf{0}_{H_i} & \text{otherwise} \end{cases}$.

Note that $\left\Vert{g'}\right\Vert^2 \le \left\Vert{g}\right\Vert^2$ by Generalized Sum Preserves Inequality.

Similarly, let $h' \in H$ be defined by $h' \left({i}\right) = \begin{cases} h \left({i}\right) & \text{if } \left\Vert{h \left({i}\right)}\right\Vert_{H_i} \ge \left\Vert{g \left({i}\right)}\right\Vert_{H_i} \\ \mathbf{0}_{H_i} & \text{otherwise} \end{cases}$.

By Generalized Sum Preserves Inequality again, have $\left\Vert{h'}\right\Vert^2 \le \left\Vert{h}\right\Vert^2$.

More significantly, by construction of $g', h'$:


 * $(3): \qquad \left\Vert{g \left({i}\right)}\right\Vert_{H_i}, \left\Vert{h \left({i}\right)}\right\Vert_{H_i} \le \left\Vert{\left({g' + h'}\right) \left({i}\right)}\right\Vert_{H_i}$

As $H$ is a vector space, $g' + h' \in H$, and we can establish:

Hence, for all $g, h \in H$, $\left\langle{g, h}\right\rangle \in \Bbb F$ by the comment on Generalized Sum Preserves Inequality.

Conclusion
$\left\langle{\cdot, \cdot}\right\rangle$ is checked to be a mapping from $H \times H$ to $\Bbb F$, satisfying the five conditions for an inner product.

That is, $\left\langle{\cdot, \cdot}\right\rangle$ is an inner product on $H$.

$H$ is complete
A Hilbert space is a complete inner product space.

Thus, it remains to verify that $H$ is complete.

Suppose $\left({h_n}\right)_{n\in\N}$ is a Cauchy sequence in $H$.

Let $N \in \N$ such that $n, m \ge N \implies \left\Vert{h_n - h_m}\right\Vert < \epsilon$.

That is, $\displaystyle \sum \left\{{ \left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2: i \in I}\right\} < \epsilon^2$.

From Generalized Sum is Monotone obtain that, for all $i \in I$:


 * $\left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2 < \epsilon^2$

It follows that $\left({h_n \left({i}\right) }\right)_{n\in\N}$ is a Cauchy sequence in $H_i$.

$H_i$ is a Hilbert space, hence complete.

Hence there is some $h_i \in H_i$ such that $\displaystyle \lim_{n\to\infty} h_n \left({i}\right) = h_i$.

Now let $h$ be defined by $h \left({i}\right) = h_i$; it is the only candidate for $\displaystyle \lim_{n\to\infty} h_n = h$.

It remains to be shown that indeed $\displaystyle \lim_{n\to\infty} h_n = h$, and then that $h \in H$.

So, for any $\epsilon > 0$, an $N \in \N$ is to be found such that for all $n \ge N$:


 * $(4): \qquad \displaystyle \sum \left\{{ \left\Vert{ \left({h_n - h}\right) \left({i}\right) }\right\Vert_{H_i}^2: i \in I}\right\} < \epsilon^2$

To this end, let $N \in \N$ be such that:


 * $(5): \qquad n, m \ge N \implies \left\Vert{h_n - h_m}\right\Vert^2 < \frac {\epsilon^2} 2$

Such an $N$ exists as $\left({h_n}\right)_{n\in\N}$ is a Cauchy sequence.

Now observe that, for any finite $G \subseteq I$ and $n \ge N$:

The last inequality follows from $(5)$, as $m \ge N$ eventually when $m \to \infty$.

From Bounded Generalized Sum Converges, it now follows that:


 * $\displaystyle \sum \left\{ {\left\Vert{ \left({h_n - h}\right) \left({i}\right) }\right\Vert_{H_i}^2 : i \in I}\right\} \le \frac {\epsilon^2} 2 < \epsilon^2$

This precisely establishes the inequality desired in $(4)$ for $n \ge N$.

It follows that $\displaystyle \lim_{n \to \infty} h_n = h$.

To show that $h \in H$, it is to be shown that $\left\Vert{h}\right\Vert^2 < \infty$.

This is done as follows:

The latter sum converges by Square-Summable Indexed Sets Closed Under Addition, yielding convergence of $\left\Vert{h}\right\Vert^2$.

Therefore, $h \in H$.

That is, every Cauchy sequence in $H$ converges to a limit in $H$, hence $H$ is complete.

By definition, $H$ is a Hilbert space.