Hilbert Space Direct Sum is Hilbert Space

== Theorem ==
Let $\sequence {H_i}_{i \mathop \in I}$ be a $I$-indexed family of Hilbert spaces over $\Bbb F \in \set {\R, \C}$.
Let $H = \ds \bigoplus_{i \mathop \in I} H_i$ be their Hilbert space direct sum.
Then $H$ is a Hilbert space.
== Proof ==
=== $H$ 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 Two-Step Vector Subspace Test it follows that it is to be shown that:
:$(1): \quad \forall h_1, h_2 \in H: \ds \sum \set {\norm {\map {\paren {h_1 + h_2} } i}^2_{H_i}: i \in I} < \infty$
:$(2): \quad \forall \lambda \in \Bbb F, h \in H: \ds \sum \set {\norm {\map {\paren {\lambda h} } i}^2_{H_i}: i \in I } < \infty$
Considering $(1)$, have the following:
For $(2)$, observe that:
Thus, by the Two-Step Vector Subspace Test, $H$ is a vector space.
=== $\innerprod \cdot \cdot$ is an Inner Product ===
It suffices to check well-definedness of $\innerprod \cdot \cdot$, and subsequently the five properties of an inner product.
==== Well-definedness ====
It is necessary to verify that for $g, h \in H$, in fact $\innerprod g h \in \Bbb F$.
That is, it is required to show that $\innerprod g h = \ds \sum \set {\innerprod {\map g i} {\map h i}_{H_i}: i \in I}$ converges in $\Bbb F$.
Absolutely Convergent Generalized Sum Converges applies to the Banach space $\Bbb F$ and the $I$-indexed subset $\innerprod {\map g i} {\map h i}_{H_i}$ of $\Bbb F$.
Hence it will suffice to show that $\ds \sum \set {\size {\innerprod {\map g i} {\map h i}_{H_i} }: i \in I}$ converges in $\R$.
For brevity, denote already $\norm h^2$ for the expression $\ds \sum \set {\norm {\map h i}_{H_i}^2: i \in I}$.
Define $g' \in H$ by:
:$\map {g'} i = \begin{cases}
 \map g i & \text{if } \norm {\map g i}_{H_i} \ge \norm {\map h i}_{H_i} \\
 \mathbf 0_{H_i} & \text{otherwise}
\end{cases}$
Note that $\norm {g'}^2 \le \norm g^2$ by Generalized Sum Preserves Inequality.
Similarly, let $h' \in H$ be defined by:
:$\map {h'} i = \begin{cases}
 \map h i & \text{if } \norm {\map h i}_{H_i} > \norm {\map g i}_{H_i} \\
 \mathbf 0_{H_i} & \text{otherwise}
\end{cases}$
By Generalized Sum Preserves Inequality again, we have:
:$\norm {h'}^2 \le \norm h^2$
More significantly, by construction of $g', h'$:
:$(3): \quad \norm {\map g i}_{H_i}, \norm {\map h i}_{H_i} \le \norm {\map {\paren {g' + h'} } i}_{H_i}$
As $H$ is a vector space, $g' + h' \in H$, and we can establish:
Hence, for all $g, h \in H$, $\innerprod g h \in \Bbb F$ by the comment on Generalized Sum Preserves Inequality.
==== Property 1: $\innerprod g h = \overline {\innerprod h g}$ ====
==== Property 2: $\innerprod {\lambda g} h = \lambda \innerprod g h$ ====
==== Property 3: $\innerprod {g_1 + g_2} h = \innerprod {g_1} h + \innerprod {g_2} h$ ====
==== Property 4: $\innerprod h h \ge 0$ ====
==== Property 5: $\innerprod h h = 0$ $h = \mathbf 0_H$ ====
==== Conclusion ====
$\innerprod \cdot \cdot$ is checked to be a mapping from $H \times H$ to $\Bbb F$, satisfying the five conditions for an inner product.
That is, $\innerprod \cdot \cdot$ 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 $\sequence {h_n}_{n \mathop \in \N}$ is a Cauchy sequence in $H$.
Let $N \in \N$ such that $n, m \ge N \implies \size {h_n - h_m} < \epsilon$.
That is:
:$\ds \sum \set {\norm {\map {\paren {h_n - h_m} } i}_{H_i}^2: i \in I} < \epsilon^2$.
From Generalized Sum is Monotone obtain that, for all $i \in I$:
:$\norm {\map {\paren {h_n - h_m} } i}_{H_i}^2 < \epsilon^2$
It follows that $\sequence {\map {h_n} i}_{n \mathop \in \N}$ 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 $\ds \lim_{n \mathop \to \infty} \map {h_n} i = h_i$.
Now let $h$ be defined by $\map h i = h_i$; it is the only candidate for $\ds \lim_{n \mathop \to \infty} h_n = h$.
It remains to be shown that indeed $\ds \lim_{n \mathop \to \infty} h_n = h$, and then that $h \in H$.
So, for any $\epsilon > 0$, an $N \in \N$ is to be found such that for all $n \ge N$:
:$(4): \quad \ds \sum \set {\norm {\map {\paren {h_n - h} } i}_{H_i}^2: i \in I} < \epsilon^2$
To this end, let $N \in \N$ be such that:
:$(5): \quad n, m \ge N \implies \size {h_n - h_m}^2 < \frac {\epsilon^2} 2$
Such an $N$ exists as $\sequence {h_n}_{n \mathop \in \N}$ 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:
:$\ds \sum \set {\norm {\map {\paren {h_n - h} } i}_{H_i}^2 : i \in I} \le \frac {\epsilon^2} 2 < \epsilon^2$
This precisely establishes the inequality desired in $(4)$ for $n \ge N$.
It follows that $\ds \lim_{n \mathop \to \infty} h_n = h$.
To show that $h \in H$, it is to be shown that $\norm h^2 < \infty$.
This is done as follows:
The latter sum converges by Square-Summable Indexed Sets Closed Under Addition, yielding convergence of $\norm h^2$.
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.
== Sources ==
* $\S I.6$
Category:Hilbert Spaces