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Suppose that $\left\{ \mathcal{H}_n \, \big| \, n\in\mathbb{N} \right\}$ is a set of orthogonal closed subspaces of a Hilbert space $\mathcal{H}$. We define the infinite direct sum $$\bigoplus_{n=1}^{\infty} \mathcal{H}_n := \left\{ \sum_{n=1}^{\infty} x_n \, \big| \, x_n \in \mathcal{H}_n \, \text{and} \, \sum_{n=1}^{\infty} \|\ x_n \|^2 < \infty \right\} . $$ Prove that $\oplus_{n=1}^{\infty} \mathcal{H}_n$ is a closed linear subspace of $\mathcal{H}$.

I feel like I proved this in a round about manner and am looking for critiques or a 'better' way to prove the result. Here is my proof:

Without loss of generality let $\left\{ e_k \, \big| \, k \in A \right\}$ be an orthonormal set in the Hilbert space $\mathcal{H}$. Then, for each $\mathcal{H}_n$ there is some $A_n$ such that $\left\{ e_k \, \big| \, k\in A_n \right\}$ is a orthonormal set. Additionally, $\displaystyle\bigcup_{n=1}^{\infty} A_n = A$. Let $$\mathcal{G}:=\bigoplus_{n=1}^{\infty} \mathcal{H}_n := \left\{ \sum_{n=1}^{\infty} x_n \, \big| \, x_n \in \mathcal{H}_n \, \text{and} \, \sum_{n=1}^{\infty} \|\ x_n \|^2 < \infty \right\} . $$
First we note that $\mathcal{G}$ is well-defined since $\sum_{n=1}^{\infty} x_n$, for $x_n \in \mathcal{H}_n$ converges if and only if $\sum_{n=1}^{\infty} \|\ x_n \|^2 $ converges, which we have by construction of $\mathcal{G}$. Note that the convergent unordered sums can be added term by term, suppose that $$\sum_{n \in A} x_n := x, \quad \sum_{n \in A} y_n := y$$ Take $\epsilon>0$, there are finite sets $I, J \subset A$, such that $$\|\ \sum_{k \in I} x_k - x \|\ < \frac{\epsilon}{2}, \quad \quad \|\ \sum_{k \in J} y_k - y \|\ < \frac{\epsilon}{2}.$$ It follows that if $I \cup J \subset K$ is a finite subset of $A$, then \begin{align*} \|\ \sum_{k \in K} (x_k + y_k) - (x+y) \|\ &\leq \|\ \sum_{k \in K} x_k - x \|\ + \|\ \sum_{k \in K} y_k - y \|\ \\ &< \frac{\epsilon}{2}+\frac{\epsilon}{2} \\ &=\epsilon. \end{align*} If $x,y \in \mathcal{G}$, with $$x=\sum_{k\in A} x_k e_k , \quad y=\sum_{k \in A} y_k e_k $$ then $$x+y = \sum_{k \in A} (x_k + y_k)e_k.$$ Now since $$|x_k + y_k|^2 \leq (|x_k| + |y_k|)^2 \leq 2(\max\left\{|x_k|,|y_k|\right\})^2 \leq 4(|x_k|^2 + |y_k|^2),$$ it follows that $$\sum_{k \in A} |x_k + y_k|^2 \leq 4\left( \sum_{k \in A} |x_k|^2 + \sum_{k\in A} |y_k|^2 \right) < \infty,$$ so $x+y\in \mathcal{G}$. Similarly, we also have that $\lambda x \in \mathcal{G}$ for all $\lambda \in \mathbb{C}$ and $x\in\mathcal{G}$, so $\mathcal{G}$ is a linear subspace. Now to prove that $\mathcal{G}$ is closed, first note that $\overline{\mathcal{G}} \cap \mathcal{G}^{\perp} = \left\{0\right\}$ because if $x\in\overline{\mathcal{G}}\cap\mathcal{G}^{\perp}$, then $\mathcal{G}=\overline{\mathcal{G}}^{\perp}$ we have $x\perp x$ so $x=0$. Now suppose that $x \in \mathcal{G}$. Let $$y:=\sum_{k\in A} \langle e_k , x \rangle e_k.$$ By Bessel's inequality, $$\sum_{k \in A} | \langle e_k , x \rangle |^2 \leq \|\ x \|^2,$$ so $y\in\mathcal{G}$. Moreover, $\langle e_k , x \rangle = \langle e_k , y \rangle$ for all $k\in A$, which implies that $x-y\in\mathcal{G}^{\perp}$. Since $x,y\in\overline{\mathcal{G}}$, it follows that $x-y\in\overline{\mathcal{G}}\cap\mathcal{G}^{\perp}$, so $x=y$, and $x\in\mathcal{G}$, which implies that $\mathcal{G}$ is closed.

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  • $\begingroup$ Describe your question in the title. $\endgroup$ – DonAntonio Apr 10 '18 at 14:53
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Note that each closed subspace $\mathcal{H}_n$ can be viewed as a Hilbert space equipped with the inner product inherited from the underlying Hilbert space $\mathcal{H}.$ Following John Conway's book on Functional Analysis (see p. 24), we define the orthogonal sum $\bigoplus_{n} \mathcal{H}_n$ to be the totality of all sequences $\Psi = (\Psi(n))_{n =1}^\infty,$ such that $\Psi(n) \in \mathcal{H}_n$ for each $n$ and $\sum_{n=1}^\infty \|\Psi(n)\|^2 < \infty.$ The orthogonal sum $\bigoplus_{n} \mathcal{H}_n$ is equipped with the following natural inner product: $$ \langle (\Psi(n))_{n =1}^\infty,(\Phi(n))_{n =1}^\infty\rangle := \sum_{n=1}^\infty \langle \Psi(n),\Phi(n)\rangle. $$ It is easy to show that $\bigoplus_{n} \mathcal{H}_n$ is a Hilbert space when it is equipped with the above inner product. Since your subspaces $\mathcal{H}_n$'s are assumed to be mutually orthogonal, it is also easy to show that the following mapping is well-defined and inner product preserving: $$ \bigoplus_{n} \mathcal{H}_n \ni (\Psi(n))_{n =1}^\infty \longmapsto \sum_{n=1}^\infty \Psi(n) \in \mathcal{H}. $$ It follows that we can identify the Hilbert space $\bigoplus_{n} \mathcal{H}_n$ with the image of the above mapping which is, of course, an isometry. This image coincides with your definition of infinite direct sum. Since any isometry has a closed range, the claim immediately follows.

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