Closed infinite orthonormal system

[mathmari]

Hey!

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

 

[I like Serena]

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 $a$.

 

[mathmari]

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)