MHB Closed infinite orthonormal system

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An infinite orthonormal system in a Euclidean space is closed if the norm of any vector can be expressed as the sum of the squares of its projections onto the basis vectors. The discussion explores the proof of this property using optimal approximations from finite-dimensional subspaces. It confirms that the equality between the sums of squares of the projections holds true, as both expressions represent the same quantity. The clarification provided emphasizes that for real numbers, the square of a number equals the square of its absolute value. The conversation concludes with the participant expressing understanding of the concept.
mathmari
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Hey! :o

I am lokking at the proof of the following sentence.

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

We suppose a subspace of $H$, that is produced by the basis $H_m=span \{e_1, e_2,..., e_m \} \subset H$ and we suppose $y_m \in H_m$ the optimal approximation of $x \in H$ from $H_m$.
So $(x-y_m, u)=0, \forall u \in H_m$
For $u=y_m:$
$(x-y_m, y_m)=0$
$x-y_m \perp y_m$
From the Pythagorean Theorem:
$||(x-y_m)+y_m||^2=||x-y_m||^2+||y_m||^2$
$||x||^2= ||x-y_m||^2+||y_m||^2$

$y_m$ can be written as $\sum_{i=1}^m{(x,e_i)e_i}$

So
$||x-\sum_{i=1}^m{(x,e_i)e_i}||^2+||\sum_{i=1}^m{(x,e_i)e_i}||^2=||x||^2$

Taking the limit $m \rightarrow \infty$ we have:

(From the definition of closed orthonormal system we have that $\lim_{n \rightarrow \infty}{||x-\sum_{i=1}^m{(x,e_i)e_i}||}=0, \forall x \in H$)

$||x||^2=\sum_{i=1}^{\infty}{|(x,e_i)|^2}$I tried to get the result by doing the calculations:

$||x||^2=|| \sum_{i=1}^{\infty}{(x,e_i)e_i}||^2=( \sum_{i=1}^{ \infty }{(x,e_i)e_i}, \sum_{i=1}^{\infty}{(x,e_i)e_i})=\sum_{i=1}^{ \infty }{((x,e_i)e_i, (x,e_i)e_i))}=\sum_{i=1}^{\infty}{(x,e_i)(e_i, e_i)(x,e_i)}= \sum_{i=1}^{\infty}{(x,e_i)(x,e_i)}= \sum_{i=1}^{\infty}{(x,e_i)^2}$

But is this the same as $\sum_{i=1}^{\infty}{|(x,e_i)|^2}$?
 
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Hey! :)

If I understand correctly, you're asking whether the following equality holds?
$$\sum_{i=1}^{\infty}{(x,e_i)^2} = \sum_{i=1}^{\infty}{|(x,e_i)|^2}$$

If so, you may want to consider that for any real number $a$, we have that:
$$a^2 = |a|^2$$
This is true for both positive and negative $a$.
 
I like Serena said:
Hey! :)

If I understand correctly, you're asking whether the following equality holds?
$$\sum_{i=1}^{\infty}{(x,e_i)^2} = \sum_{i=1}^{\infty}{|(x,e_i)|^2}$$

If so, you may want to consider that for any real number $a$, we have that:
$$a^2 = |a|^2$$
This is true for both positive and negative $a$.

Ok! I got it! Thank you! (Smile)
 
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