{{Proofread}} {{wtd|I think some of the partial results are up on their own pages now, and thus can be replaced by a reference (or this should become the case). Furthermore, this is *so* long, transclusions and/or foldable templates are a must}} === $H$ 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: :$(1): \qquad \forall h_1, h_2 \in H: \displaystyle \sum \left\{{ \left\Vert{ \left({h_1 + h_2}\right) \left({i}\right) }\right\Vert^2_{H_i}: i \in I }\right\} < \infty$ :$(2): \qquad \forall \lambda \in \Bbb F, h \in H: \displaystyle \sum \left\{{ \left\Vert{ \left({\lambda h}\right) \left({i}\right) }\right\Vert^2_{H_i}: i \in I }\right\} < \infty$ Considering $(1)$, have the following: {{begin-eqn}} {{eqn|l = \left\Vert{ \left({h_1 + h_2}\right) \left({i}\right) }\right\Vert^2_{H_i}      |r = \left\Vert{ h_1 \left({i}\right) + h_2 \left({i}\right) }\right\Vert^2_{H_i} }} {{eqn|o = \le      |r = \left({ \left\Vert{h_1 \left({i}\right)}\right\Vert_{H_i} + \left\Vert{h_2 \left({i}\right)}\right\Vert_{H_i} }\right)^2      |c = Triangle inequality for $\left\Vert{\cdot}\right\Vert_{H_i}$ }} {{eqn|ll= \implies      |l = \sum \left\{ { \left\Vert{ \left({h_1 + h_2}\right) \left({i}\right) }\right\Vert^2_{H_i}: i \in I }\right\}      |o = \le      |r = \sum \left\{ { \left({ \left\Vert{h_1 \left({i}\right)}\right\Vert_{H_i} + \left\Vert{h_2 \left({i}\right)}\right\Vert_{H_i} }\right)^2: i \in I }\right\}      |c = Generalized Sum Preserves Inequality }} {{eqn|o = <      |r = \infty      |c = $h_1, h_2 \in H$, Square-Summable Indexed Sets Closed Under Addition }} {{end-eqn}} For $(2)$, observe that: {{begin-eqn}} {{eqn|l = \left\Vert{ \left({\lambda h}\right) \left({i}\right) }\right\Vert^2_{H_i}      |r = \left\vert{\lambda}\right\vert^2 \left\Vert{h \left({i}\right)}\right\Vert^2_{H_i}      |c = $\left\Vert{\cdot}\right\Vert_{H_i}$ is a norm }} {{eqn|ll= \implies      |l = \sum \left\{ { \left\Vert{ \left({\lambda h}\right) \left({i}\right) }\right\Vert^2_{H_i}: i \in I }\right\}      |o = \le      |r = \left\vert{\lambda}\right\vert^2 \sum \left\{ { \left\Vert{h \left({i}\right)}\right\Vert^2_{H_i}: i \in I }\right\}      |c = Generalized Sum is Linear }} {{eqn|o = <      |r = \infty      |c = As $h \in H$ }} {{end-eqn}} Thus, by the Two-Step Vector Subspace Test, $H$ is a vector space. {{qed|lemma}} === $\left\langle{\cdot, \cdot}\right\rangle$ 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: {{begin-eqn}} {{eqn|l = \left\vert{ \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i} }\right\vert      |o = \le      |r = \left\Vert{g \left({i}\right)}\right\Vert_{H_i} \left\Vert{h \left({i}\right)}\right\Vert_{H_i}      |c = Cauchy-Bunyakovsky-Schwarz Inequality }} {{eqn|o = \le      |r = \left\Vert{\left({g' + h'}\right) \left({i}\right)}\right\Vert_{H_i}^2      |c = Equation $(1)$ }} {{eqn|ll= \implies      |l = \sum \left\{ { \left\vert{ \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i} }\right\vert: i \in I}\right\}      |o = \le      |r = \sum \left\{ { \left\Vert{\left({g' + h'}\right) \left({i}\right)}\right\Vert_{H_i}^2: i \in I}\right\}      |c = Generalized Sum Preserves Inequality }} {{end-eqn}} Hence, for all $g, h \in H$, $\left\langle{g, h}\right\rangle \in \Bbb F$ by the comment on Generalized Sum Preserves Inequality. {{qed|lemma}} ==== Property 1: $\left\langle{g, h}\right\rangle = \overline{\left\langle{h, g}\right\rangle}$ ==== {{begin-eqn}} {{eqn|l = \left\langle{g, h}\right\rangle      |r = \sum \left\{ { \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}      |c = Definition of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{eqn|r = \sum \left\{ { \overline{\left\langle{ h \left({i}\right), g \left({i}\right) }\right\rangle_{H_i} }: i \in I}\right\}      |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an inner product }} {{eqn|r = \overline{ \sum \left\{ { \left\langle{ h \left({i}\right), g \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\} }      |c = Convergence of Generalized Sum of Complex Numbers: Corollary }} {{eqn|r = \overline{ \left\langle{h, g}\right\rangle }      |c = Definition of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{end-eqn}} {{qed|lemma}} ==== Property 2: $\left\langle{\lambda g, h}\right\rangle = \lambda \left\langle{g, h}\right\rangle$ ==== {{begin-eqn}} {{eqn|l = \left\langle{\lambda g, h}\right\rangle      |r = \sum \left\{ { \left\langle{\left({\lambda g}\right) \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}      |c = Definition of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{eqn|r = \sum \left\{ { \lambda \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}      |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an inner product }} {{eqn|r = \lambda \sum \left\{ { \left\langle{g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}      |c = Generalized Sum is Linear }} {{eqn|r = \lambda \left\langle{h, g}\right\rangle      |c = Definition of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{end-eqn}} {{qed|lemma}} ==== Property 3: $\left\langle{g_1 + g_2, h}\right\rangle = \left\langle{g_1, h}\right\rangle + \left\langle{g_2, h}\right\rangle$ ==== {{begin-eqn}} {{eqn|l = \left\langle{g_1 + g_2, h}\right\rangle      |r = \sum \left\{ { \left\langle{\left({g_1 + g_2}\right) \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}      |c = Definition of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{eqn|r = \sum \left\{ { \left\langle{g_1 \left({i}\right), h \left({i}\right)}\right\rangle_{H_i} + \left\langle{g_2 \left({i}\right), h \left({i}\right)}\right\rangle_{H_i}: i \in I}\right\}      |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an inner product }} {{eqn|r = \sum \left\{ { \left\langle{ g_1 \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\} + \sum \left\{ { \left\langle{ g_2 \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}      |c = Generalized Sum is Linear }} {{eqn|r = \left\langle{g_1, h}\right\rangle + \left\langle{g_2, h}\right\rangle      |c = Definition of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{end-eqn}} {{qed|lemma}} ==== Property 4: $\left\langle{h, h}\right\rangle \ge 0$ ==== {{begin-eqn}} {{eqn|l = \left\langle{h, h}\right\rangle      |r = \sum \left\{ { \left\langle{ h \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}      |c = Definition of $\left\langle{\cdot, \cdot}\right\rangle$ }} {{eqn|o = \ge      |r = \sum \left\{ { 0: i \in I}\right\}      |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an inner product, Generalized Sum Preserves Inequality }} {{eqn|r = 0 }} {{end-eqn}} {{qed|lemma}} ==== Property 5: $\left\langle{h, h}\right\rangle = 0$ iff $h = \mathbf{0}_H$ ==== {{begin-eqn}} {{eqn|l = \left\langle{h, h}\right\rangle      |r = 0 }} {{eqn|ll= \iff \forall i \in I:      |l = \left\langle{ h \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}      |r = 0      |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an inner product, Generalized Sum is Monotone }} {{eqn|ll= \iff \forall i \in I:      |l = h \left({i}\right)      |r = \mathbf{0}_{H_i}      |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an inner product }} {{eqn|ll= \iff      |l = h      |r = \mathbf{0}_H }} {{end-eqn}} {{qed|lemma}} ==== 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 $H$. {{qed|lemma}} === $H$ 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 $n \ge N$: {{begin-eqn}} {{eqn | ll= \forall i \in I:       | l = \left\Vert{ \left({h_n - h}\right) \left({i}\right) }\right\Vert_{H_i}^2       | r = \lim_{m \to \infty} \left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2       | c = Definition of $h \left({i}\right)$ }} {{eqn | ll= \implies       | l = \sum_{i \in G} \left\Vert{ \left({h_n - h}\right) \left({i}\right) }\right\Vert_{H_i}^2       | r = \sum_{i \in G} \ \lim_{m \to \infty} \left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2 }} {{eqn | r = \lim_{m \to \infty} \ \sum_{i \in G} \left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2       | c = Sum Rule for Sequences }} {{eqn | o = \le       | r = \lim_{m \to \infty} \ \sum \left\{ {\left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2 : i \in I}\right\}       | c = Generalized Sum is Monotone }} {{eqn | r = \lim_{m \to \infty} \left\Vert{h_n - h_m}\right\Vert^2       | c = Definition of $\left\Vert{\cdot}\right\Vert$ }} {{eqn | o = \le       | r = \frac {\epsilon^2} 2       | c = Upper and Lower Bounds of Sequences }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \left\Vert{h}\right\Vert^2       | r = \sum \left\{ {\left\Vert{ h \left({i}\right) }\right\Vert_{H_i}^2 : i \in I}\right\}       | c = Definition of $\left\Vert{\cdot}\right\Vert$ }} {{eqn | o = \le       | r = \sum \left\{ {\left({ \left\Vert{ \left({h - h_n}\right) \left({i}\right) }\right\Vert_{H_i} + \left\Vert{ h_n \left({i}\right) }\right\Vert_{H_i} }\right)^2 : i \in I}\right\}       | c = Triangle Inequality for all $\left\Vert{\cdot}\right\Vert_{H_i}$, Generalized Sum Preserves Inequality }} {{end-eqn}} 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. {{qed}}