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I'm reading about direct sums of Hilbert spaces currently, and I've read that you have to supply certain restrictions on the vectors that can be an infinite direct sum of Hilbert spaces. So specifically:

Let $I$ be an arbitrary index set of cardinality at least $\aleph_0$, and let $\{\mathbb{H}_i\}_{i\in I}$ be a family of Hilbert spaces over a field $\mathbb{F}$. Then we write $$ \mathbb{H}\equiv\bigoplus_{i\in I}\mathbb{H}_i = \left\{(h_i)_{i\in I}\ |\ \forall i\ h_i\in\mathbb{H}_i\ \land\ \sum_{i\in I}\|h_i\|^2<\infty\right\} $$ and define $$ \left\langle(h_i),\ (g_i)\right\rangle_{\mathbb{H}} = \sum_{i\in I}\left\langle h_i,\ g_i\right\rangle_{\mathbb{H}_i} $$ so that $$ \|(h_i)\|_{\mathbb{H}}^2 = \sum_{i\in I}\|h_i\|_{\mathbb{H}_i}^2 $$

I'm not quite sure what to make of the extra condition that $\sum_{i\in I}\|h_i\|^2<\infty$, it just seems like an unnecessary condition. I completely understand that the condition implies that the inner product converges, but like, doesn't that mean that the sequence $H_n = \left(\frac1nh_1,\ \frac1nh_2,\ \cdots\right)$, where $h_i\in\mathbb{H}_i$ and $\|h_i\| = 1$, doesn't converge to $(0,\ 0,\ \cdots)$? Intuitively, it totally should, at least to me.

I'm having trouble formulating my difficulty understanding this, because the more I try to write the more I realize that my question basically amounts to "why not just let the inner product sometimes be infinite"? What does that break? Why can't we define convergence, or the inner product, in some other fashion? Is it just because this definition of the inner product follows naturally from the inner product of a finite direct sum of Hilbert spaces?

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    In your example, is $H_n$ meant to have infinitely many nonzero terms, or does it stop at some function of $n$? – 2017-01-22
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    What does having infinite inner product break? The definition of a Hilbert space, for one. Even that if a normed space. These are pretty basic desirable properties that we'd expect of a sum of Hilbert spaces. – 2017-01-22
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    @ErickWong in response to your first question, no it doesn't have infinitely many zero terms; it's $i$ term is $\frac1n h_i$, for all $i\in I$, so there's actually no zero term. – 2017-01-22
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    In response to your second comment: that's what's confusing me. I get it, or at least I should get it: if the inner product is infinite we loose a lot of desirable properties. I'm just getting caught up in the question of "why not?". Why not just let it be infinite sometimes? Will that, for example, cause certain sequences that are technically "Cauchy" to diverge because the inner product is infinite? Or something? I'm unsure, I kind of just want to talk to someone who understands it more intuitively. Furthermore, why are we forcing ourselves to define the inner product this way? – 2017-01-22

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The general concept of a direct sum of Widgets is that:

  • The direct some is also a Widget.
  • The direct sum contains an isomorphic copy of each Widget used in its construction.

As applied to the direct sum of Hilbert spaces, this calls for a construction of $\mathbb{H}$ such that

  • $\mathbb{H}$ has an inner product
  • When restricted to the subspace $V_i=\{(h_j) : h_j=0\ \forall j\ne i\}$, which is a copy of $\mathbb{H}_i$ inside of $\mathbb{H}$, this inner product agrees with the inner product on $\mathbb{H}_i$

The above properties determine what the inner product on $\mathbb{H}$ should be.

Sure, one can write down an alternative definition of a sum, which allows for arbitrary sequences. And define topology in some other way, for example using the original norms as seminorms on the product (as in this MathOverflow question). The result is a locally convex topological vector space (LCS), an important and interesting thing to study. But do recognize that the Hilbert space structure went out of the window along the way: in effect we have ignored it, treating each space simply as LCS and taking the product accordingly. This is not a process one would call a direct sum of Hilbert spaces; it's a direct sum of LCS.