Hilbert Space Direct Sum is Hilbert Space

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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


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$H$ 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:

\(\ds \norm {\map {\paren {h_1 + h_2} } i}^2_{H_i}\) \(=\) \(\ds \norm {\map {h_1} i + \map {h_2} i}^2_{H_i}\)
\(\ds \) \(\le\) \(\ds \paren {\norm {\map {h_1} i}_{H_i} + \norm {\map {h_2} i}_{H_i} }^2\) Triangle inequality for $\norm {\, \cdot \,}_{H_i}$
\(\ds \leadsto \ \ \) \(\ds \sum \set {\norm {\map {\paren {h_1 + h_2} } i}^2_{H_i}: i \in I}\) \(\le\) \(\ds \sum \set {\paren {\norm {\map {h_1} i}_{H_i} + \norm {\map {h_2} i}_{H_i} }^2: i \in I}\) Generalized Sum Preserves Inequality
\(\ds \) \(<\) \(\ds \infty\) $h_1, h_2 \in H$, Square-Summable Indexed Sets Closed Under Addition


For $(2)$, observe that:

\(\ds \norm {\map {\paren {\lambda h} } i}^2_{H_i}\) \(=\) \(\ds \size \lambda^2 \norm {\map h i}^2_{H_i}\) $\norm {\, \cdot \,}_{H_i}$ is a norm
\(\ds \leadsto \ \ \) \(\ds \sum \set {\norm {\map {\paren {\lambda h} } i}^2_{H_i}: i \in I}\) \(\le\) \(\ds \size \lambda^2 \sum set {\norm {\map h i}^2_{H_i}: i \in I}\) Generalized Sum is Linear
\(\ds \) \(<\) \(\ds \infty\) As $h \in H$

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

$\Box$


$\innerprod \cdot \cdot$ 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:

\(\ds \size {\innerprod {\map g i} {\map h i}_{H_i} }\) \(\le\) \(\ds \norm {\map g i}_{H_i} \norm {\map h i}_{H_i}\) Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces
\(\ds \) \(\le\) \(\ds \norm {\map {\paren {g' + h'} } i}_{H_i}^2\) Equation $(3)$
\(\ds \leadsto \ \ \) \(\ds \sum \set {\size {\innerprod {\map g i} {\map h i}_{H_i} }: i \in I}\) \(\le\) \(\ds \sum \set {\norm {\map {\paren {g' + h'} } i}_{H_i}^2: i \in I}\) Generalized Sum Preserves Inequality

Hence, for all $g, h \in H$, $\innerprod g h \in \Bbb F$ by the comment on Generalized Sum Preserves Inequality.

$\Box$


Property 1: $\innerprod g h = \overline {\innerprod h g}$

\(\ds \innerprod g h\) \(=\) \(\ds \sum \set {\innerprod {\map g i} {\map h i}_{H_i}: i \in I}\) Definition of $\innerprod \cdot \cdot$
\(\ds \) \(=\) \(\ds \sum \set {\overline {\innerprod {\map h i} {\map g i}_{H_i} }: i \in I}\) $\innerprod \cdot \cdot_{H_i}$ is an inner product
\(\ds \) \(=\) \(\ds \overline{ \sum \set {\innerprod {\map h i} {\map g i}_{H_i}: i \in I} }\) Convergence of Generalized Sum of Complex Numbers: Corollary
\(\ds \) \(=\) \(\ds \overline {\innerprod h g}\) Definition of $\innerprod \cdot \cdot$

$\Box$


Property 2: $\innerprod {\lambda g} h = \lambda \innerprod g h$

\(\ds \innerprod {\lambda g} h\) \(=\) \(\ds \sum \set {\innerprod {\map {\paren {\lambda g} } i} {\map h i}_{H_i}: i \in I}\) Definition of $\innerprod \cdot \cdot$
\(\ds \) \(=\) \(\ds \sum \set {\lambda \innerprod {\map g i} {\map h i}_{H_i}: i \in I}\) $\innerprod \cdot \cdot_{H_i}$ is an inner product
\(\ds \) \(=\) \(\ds \lambda \sum \set {\innerprod {\map g i} {\map h i}_{H_i}: i \in I}\) Generalized Sum is Linear
\(\ds \) \(=\) \(\ds \lambda \innerprod h g\) Definition of $\innerprod \cdot \cdot$

$\Box$


Property 3: $\innerprod {g_1 + g_2} h = \innerprod {g_1} h + \innerprod {g_2} h$

\(\ds \innerprod {g_1 + g_2} h\) \(=\) \(\ds \sum \set {\innerprod {\map {\paren {g_1 + g_2} } i} {\map h i}_{H_i}: i \in I}\) Definition of $\innerprod \cdot \cdot$
\(\ds \) \(=\) \(\ds \sum \set {\innerprod {\map {g_1} i} {\map h i}_{H_i} + \innerprod {\map {g_2} i} {\map h i}_{H_i}: i \in I}\) $\innerprod \cdot \cdot_{H_i}$ is an inner product
\(\ds \) \(=\) \(\ds \sum \set {\innerprod {\map {g_1} i} {\map h i}_{H_i}: i \in I} + \sum \set {\innerprod {\map {g_2} i} {\map h i}_{H_i}: i \in I}\) Generalized Sum is Linear
\(\ds \) \(=\) \(\ds \innerprod {g_1} h + \innerprod {g_1} h\) Definition of $\innerprod \cdot \cdot$

$\Box$


Property 4: $\innerprod h h \ge 0$

\(\ds \innerprod h h\) \(=\) \(\ds \sum \set {\innerprod {\map h i} {\map h i}_{H_i}: i \in I}\) Definition of $\innerprod \cdot \cdot$
\(\ds \) \(\ge\) \(\ds \sum \set {0: i \in I}\) $\innerprod \cdot \cdot_{H_i}$ is an inner product, Generalized Sum Preserves Inequality
\(\ds \) \(=\) \(\ds 0\)

$\Box$


Property 5: $\innerprod h h = 0$ if and only if $h = \mathbf 0_H$

\(\ds \innerprod h h\) \(=\) \(\ds 0\)
\(\ds \leadstoandfrom \ \ \) \(\ds \forall i \in I: \, \) \(\ds \innerprod {\map h i} {\map h i}_{H_i}\) \(=\) \(\ds 0\) $\innerprod \cdot \cdot_{H_i}$ is an inner product, Generalized Sum is Monotone
\(\ds \leadstoandfrom \ \ \) \(\ds \forall i \in I: \, \) \(\ds \map h i\) \(=\) \(\ds \mathbf 0_{H_i}\) $\innerprod \cdot \cdot_{H_i}$ is an inner product
\(\ds \leadstoandfrom \ \ \) \(\ds h\) \(=\) \(\ds \mathbf 0_H\)

$\Box$


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 $H$.

$\Box$


$H$ 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 $n \ge N$:

\(\ds \forall i \in I: \, \) \(\ds \norm {\map {\paren {h_n - h} } i}_{H_i}^2\) \(=\) \(\ds \lim_{m \mathop \to \infty} \norm {\map {\paren {h_n - h_m} } i}_{H_i}^2\) Definition of $\map h i$
\(\ds \leadsto \ \ \) \(\ds \sum_{i \mathop \in G} \norm {\map {\paren {h_n - h} } i}_{H_i}^2\) \(=\) \(\ds \sum_{i \mathop \in G} \lim_{m \mathop \to \infty} \norm {\map {\paren {h_n - h_m} } i}_{H_i}^2\)
\(\ds \) \(=\) \(\ds \lim_{m \mathop \to \infty} \sum_{i \mathop \in G} \norm {\map {\paren {h_n - h_m} } i}_{H_i}^2\) Sum Rule for Sequences
\(\ds \) \(\le\) \(\ds \lim_{m \mathop \to \infty} \sum \set {\norm {\map {\paren {h_n - h_m} } i}_{H_i}^2 : i \in I}\) Generalized Sum is Monotone
\(\ds \) \(=\) \(\ds \lim_{m \mathop \to \infty} \size {h_n - h_m}^2\) Definition of $\norm {\, \cdot \,}$
\(\ds \) \(\le\) \(\ds \frac {\epsilon^2} 2\) Upper and Lower Bounds of Sequences

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:

\(\ds \norm h^2\) \(=\) \(\ds \sum \set {\norm {\map h i}_{H_i}^2 : i \in I}\) Definition of $\norm {\, \cdot \,}$
\(\ds \) \(\le\) \(\ds \sum \set {\paren {\norm {\map {\paren {h - h_n} } i}_{H_i} + \norm {\map {h_n} i}_{H_i} }^2 : i \in I}\) Triangle Inequality for all $\norm {\, \cdot \,}_{H_i}$, Generalized Sum Preserves Inequality

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.

$\blacksquare$


Sources

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