# Basis for a product Hilbert space.

Let $H_1, H_2, \ldots, H_n$ be a countable family of Hilbert spaces. Let H be the set of tuples $x = (x_1, \ldots, x_n,\ldots)\in \prod_n H_n$ with the property that $$\|x \| ^2 =\sum_n \| x_n \| _{H_n}^2 <\infty.$$ Then H is also a Hilbert space.

Prove that H is non-separable and determine an orthonormal basis in this space.

If I take a sequence $$(x^{(k)})$$ such that: $$x^{(1)} = (1,0,0,\ldots), x^{(2)} = (0,1,0,\ldots)$$ then this sequence has no converging subsequnce, hence not separable (?). I have no good idea how to find a basis.

• If $H$ is a separable Hilbert space, then the sequence which consists of elements of a Hilbert basis has no (strongly) converging subsequences, so your argument doesn't work. Commented Nov 22, 2012 at 16:44
• Is that so? I thought that since we had a norm we get a metric? Commented Nov 22, 2012 at 22:28

I think you are getting confused with the notation here. For each of your $x_n$, its coordinates are vectors in $H_n$; so "$1$" makes no sense there.

Note that we cannot expect the construction to always give a non-separable Hilbert space. Because we can take $H_n=\mathbb C$ for all $n$, and then $\prod_nH_n=\ell^2(\mathbb N)$, which is separable.

Note also that you never defined what the inner product in the direct sum is, but your condition on the norms suggests that it is the canonical one, $$\langle x,y\rangle = \sum_n\langle x_n,y_n\rangle.$$

To make up a basis of $\prod_nH_n$, the natural way is to use bases from each of the $H_n$. So, for each $n$, fix an orthonormal basis $B_n=\{e_{k,n}\}_{k\in K_n}$ of $H_n$.

Let $$B=\{\,x\in\prod_nH_n:\ \exists m\text{ with }x_m\in B_m\text{ and }x_r=0\text{ if }r\ne m\}$$ It is clear that $B$ is an orthonormal set. Now suppose that $x\in B^\perp$. Then, for any $m$ and any $k\in K_m$, $$0=\langle x,e_{k,m}\rangle=\langle x_m,e_{k,m}\rangle.$$ As $k\in K_m$ was arbitrary, $x_m=0$; as $m$ was arbitrary, $x=0$. So $B$ is total, and it is thus a basis.

Now, if all the $H_n$ are separable, then $K_n$ is countable for all $n$. As the family $H_1,H_2,\ldots$ is countable, we have $$B=\bigcup_{m\in\mathbb N}\{x:\ x_m\in B_m\text{ and zero elsewhere }\}.$$ As each $B_m$ is countable, $B$ is a countable union of countable sets, and is itself counatble. So $H$ is separable.

If any of the spaces $H_1,H_2,\ldots$ is non-separable, or if the family $\{H_n\}$ is uncountable, then $\prod H_n$ will be non-separable.