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(Analysis Now, Pederson) Let $H_\alpha$ be a collection of Hilbert spaces. Let $\sum_\alpha H_\alpha $ be the algebraic direct sum of the $H_\alpha$, i.e. the space of $x \in \prod_\alpha H_\alpha $ such that $x_\alpha = 0$ except finitely many $\alpha$. Put the inner product $\left< x, y \right> = \sum_\alpha \left< x_\alpha, y_\alpha \right>$ on the algebraic direct sum. Call the completion of the algebraic direct sum $\bigoplus_\alpha H_\alpha$ the orthogonal sum of the $H_\alpha$.

"We may identify each $H_\alpha$ with a closed subspace of the orthogonal sum such that $H_\alpha \perp H_\beta$ for $\alpha \neq \beta$." What is this identification, specifically?

I see that $H_\alpha^\prime := \sum_\beta H_\beta^\prime$ is a subspace of the algebraic direct sum, where $H_\beta^\prime$ equals $H_\alpha$ if $\beta=\alpha$ and is $0$ otherwise, and $H_\alpha^\prime \perp H_\gamma^\prime$. I am thinking then to take the corresponding completion isometry $J: \sum_\beta H_\beta \rightarrow \bigoplus_\beta H_\beta$ and consider the $J(H_\alpha^\prime)$, but I'm not so sure.

Lucas
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Turns out that $\bigoplus_\beta H_\beta$ actually equals the collection of $x \in \prod_\beta H_\beta$ such that $\sum_\beta \left\Vert x_\beta \right\Vert^2 < \infty$, where $\left< , \right>$ given above induces the 2-norm $\left\Vert x \right\Vert_2 = \left( \sum_\beta \left\Vert x_\beta \right\Vert \right)^{1/2}$ on $\sum_\beta H_\beta $. So, with $H_\alpha^\prime$ given above, it is now a subspace of $\bigoplus_\beta H_\beta$. Continuing, it is closed (under the 2-norm) since for $x^n \in H_\alpha^\prime$ such that $x^n \rightarrow y \in \bigoplus_\beta H_\beta$, since $\left\Vert x^n - y \right\Vert_2^2 = \sum_\beta \left\Vert x^n_\beta - y \right\Vert^2$ and $x_\beta^n=0$ for $\beta\neq\alpha$, it is then necessary that $y_\beta=0$ for $\beta\neq\alpha$, and hence $y\in H_\alpha^\prime$.

Lucas
  • 282