MHB Why is (x,e_i) a zero sequence?

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An infinite orthonormal system in a Hilbert space is closed if the norm of any vector can be expressed as the sum of the squares of its inner products with the basis elements. This leads to the conclusion that the sequence of inner products, denoted as (x,e_i), is a zero sequence. The convergence of (x,e_i) to zero as i approaches infinity is supported by the nth term test for convergence. The discussion highlights the relationship between the properties of orthonormal systems and the behavior of sequences derived from them. Understanding this connection is crucial for grasping the fundamentals of functional analysis.
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Hey! :o

An infinite orthonormal system $\{e_1, e_2, ... \} \subset H$ is closed in $H$ iff $\forall x \in H$
$$||x||^2=\sum_{i=1}^{n}{|(x,e_i)|^2}$$

From the summability of the right part of the relation above, we conclude to that the sequence $(x,e_i)$ is a zero sequence.

Could you explain me how we conlude to that?
 
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Do you mean why $(x,e_i)\to0$ as $i\to\infty$? This follows from the nth term test.
 
Evgeny.Makarov said:
Do you mean why $(x,e_i)\to0$ as $i\to\infty$? This follows from the nth term test.

Aha! I got it! Thank you! (Smirk)
 
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