Hilbert space sum

Question

Let $\{x_i\}_{i\in I}$ as orthonormal set in Hilbert space,

Now define $x = \sum_{i\in I}a_ix_i$ , can we prove $a_i = (x,x_i)$? If $I$ is uncountable set?

I know if we further impose $\sum_{i\in I}|a_i|^2 <\infty$ then set of $i$ such that $a_i \ne 0$ is almost countable（then taking limit,due to continuous of inner product）,what about the case without this assumption?

score 1 · Accepted Answer · answered Oct 07 '20 at 13:12

1

It depends on your definition of "$x = \sum_{i \in I} a_i \, x_i$". I would say that for any reasonable definition, this series converges if and only if $\sum_{i \in I} |a_i|^2 < \infty$.

answered Oct 07 '20 at 13:12

gerw

31,359

How to show the only if part? – yi li Oct 07 '20 at 13:15
What is your definition of $\sum_{i \in I} a_i , x_i$? – gerw Oct 07 '20 at 13:16
Thank you I find one here https://math.stackexchange.com/a/1413886/360262 – yi li Oct 07 '20 at 13:16
2

For the countable case, it is easy and follows from $| \sum_{i = 1}^n a_i , x_i|^2 = \sum_{i = 1}^n |a_i|^2$ (due to orthonormality). In the uncountable case, it should work along the same lines. – gerw Oct 07 '20 at 13:20
Hi I still don't know how to define uncountable sum . – yi li Oct 07 '20 at 14:11

Hilbert space sum

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