Let $(H_\alpha)_{\alpha\in A}$ be an uncountable family of Hilbert spaces. (the countable case is discussed here: Basis for a product Hilbert space.)

Let H be the set of tuples $x = (x)_{\alpha\in A} \in \prod_{\alpha\in A} H_\alpha$ with the property that $$\|x \| ^2 =\sum_{\alpha\in A} \| x_\alpha \| _{H_\alpha}^2 <\infty.$$ Show that this is an hilbert space. Prove that H is non-separable and determine an orthonormal basis in this space. The triangle inequality for the countable case is discussed here: Countable family of Hilbert spaces is complete. The positivity and Homogenity seems obvious.

How do we find the basis and is the space always non-separable because its uncountable?

  • $\begingroup$ What exactly do you mean by $H_{1},...,H_{n}$ being an uncountable family? At the moment it looks like a finite family of $n\in\mathbb{N}$ many members. Also if $\{H_{i}\}_{i\in I}$ is such that $I$ is uncountable, then how do you define a tuple $(x_{1},x_{2},...)\in\Pi \,H_{i}$, as it is indexed by $\mathbb{N}$? $\endgroup$
    – T. Eskin
    Commented Nov 23, 2012 at 10:07
  • $\begingroup$ Note that in particular, at most countably many terms $x_{\alpha}$ are non-zero. $\endgroup$ Commented Nov 23, 2012 at 10:12
  • 2
    $\begingroup$ Sorry for that, I edited it now. $\endgroup$
    – Johan
    Commented Nov 23, 2012 at 10:12
  • $\begingroup$ Patrick: Yes so I understand it to, What happens to the separability then? And how does that affect the basis? $\endgroup$
    – Johan
    Commented Nov 23, 2012 at 10:14

1 Answer 1


You can define an inner product on this space by letting $$ ( (x_{\alpha} )_{\alpha \in A}), (y_{\alpha})_{\alpha \in A}) \overset{def}= \sum_{\alpha \in A} (x_{\alpha},y_{\alpha})_{H_{\alpha}}. $$ By assumption, there are only at most countably many terms in this sum. Note that if you see those $A$-tuples as subsets of the Hilbert space where you only take the product over those Hilbert spaces where you're currently working with index $\alpha$ such that $x_{\alpha},y_{\alpha}\neq 0$, the Cauchy-Schwarz inequality holds, so that this sum is well defined. You can see that it defines an inner product and that everything works fine.

Of course, the issue with separability is the uncountable amount of Hilbert spaces you're taking the product with. A countable subset is clearly not dense if the product is uncountable, since for every countable subset $\{ (x_{n_{\alpha}})_{\alpha \in A} \}$of $H$, you can find an uncountable set $B \subseteq A$ of indices over which every element of your subset will satisfy $x_{n_{\beta}} = 0$ for all $\beta \in B$, which shows $\{x_n\}$ is not dense.

The reason for the existence of $B$ is simple : since every element $x_n$ will vanish on the complement of a countable set of indices $\alpha \in A$, if this set where $x_n$ does not vanish is called $B_n$, then $B_n$ is countable, hence so is $\bigcup_n B_n$, i.e. the set of indices where at least one $x_n$ has a non-zero component. Therefore, outside this countable set, all the $x_n$'s vanish.

As a basis, if $\mathcal E_{\alpha}$ is a basis of $H_{\alpha}$, define $\tilde e_{\alpha_0}$ to be the tuple that is zero in every $\alpha$-component but is $e_{\alpha_0}$ in the $\alpha_0$-component. Using the above-defined inner product you see that you can project on those and get your basis.

Hope that helps,

  • $\begingroup$ very good! thanks! $\endgroup$
    – Johan
    Commented Nov 23, 2012 at 13:40

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