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Suppose I am told that $a_j(t)b_i(x)$ for $i,j=1,2,...$ is a orthonormal basis for a Hilbert space $H$.

I want to write an element $h= \sum_{k=1}^\infty c_kh_k$ where $h_k$ is a basis for $H$ and $c_k$ are coefficients.

How do I write $h_k$ in terms of the $a_jb_i$? I don't want two indices, I want only 1 index. I guess I can reorder $a_jb_i$ in some way but I can't see how.

matt.w
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    Any bijection of $\mathbb{Q}$ with $\mathbb{Q}^2$ will do, but will be slightly messy to do explicitly. You might be better off living with two indices, or cheating and using a multi-index as a notational convenience. – copper.hat Jun 27 '13 at 17:00
  • @copper.hat can I say: we know that $a_ib_j$ is a basis. Let us simply denote $f_i$ for the basis functions instead. So every $h= \sum c_if_i$, where the $f_i \in {a_ib_j :i,j=1,...}$? – matt.w Jun 27 '13 at 17:16
  • Not exactly, just because the index $i$ in the $f_i$ and the set definition is a bit confusing. Either you come up with some bijection $\phi:\mathbb{Q} \to \mathbb{Q}^2$ and write something like $h = \sum_i c_{\phi_1(i),\phi_2(i)} f_{\phi_1(i),\phi_2(i)}$, or let $I={(i,j) | i,j \ge 1 }$ and write $h = \sum_{\alpha \in I} c_\alpha f_\alpha$, if you see what I mean. – copper.hat Jun 27 '13 at 17:22
  • Well, actually, what you wrote could work; let $\Phi = {a_i b_j | i,j=1,... }$ and then write $h = \sum_{f \in \Phi} c_f f$. Note that $c$ is now indexed by the function $f$. – copper.hat Jun 27 '13 at 17:24
  • Oops, I meant $f_k \in {...}$ instead. It seems multiindex is nice way to do it. Thanks. – matt.w Jun 27 '13 at 17:37

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