Let (e_k) be an orthonormal sequence in a Hilbert space H, and let
M=span{e_k}. Show that for any x \in H we have x \in \overline{M}
\iff x can be represented by \sum_{k=1}^{\infty} \alpha_k e_k, with
coefficients \alpha_k = <x,e_k>.
Let (e_k) be an orthonormal sequence in a Hilbert space H, and let
M=span{e_k}. Show that for any x \in H we have x \in \overline{M}
\iff x can be represented by \sum_{k=1}^{\infty} \alpha_k e_k, with
coefficients \alpha_k = <x,e_k>.
It should be clear that it suffices to prove the following:
Let an orthonormal set in a Hilbert space , then is a Hilbert base ( ) we have
Since is a Hilbert base, we have and so, we divide this in two cases:
1) : We have then where (where is the basefield or ) then . Thus , and as such for every we have and so
2) : We choose a such that . Then . Now we take and using the triangle inequality two times and the Cauchy-Schwarz inequality afterwards we obtain:
Thus we have shown that
Since , we are finished.
Since for all we have then , we have that and so, , and so .
Man, that was hard to type, anyway hope it helps.
Thank you for clarifying that for me Jose!
No I'll manage.
My sincerest gratitude for the solution you provided typed out is such elegance.
I've got plenty to learn about inner product spaces and hilbert spaces and your feedback is helping a great deal!
I was checking this the other day, and realized that there's a mistake in basically, I'm fixing and trying to show that for sufficiently large the series approaches what it has to, and if you notice I concluded with which is clearly false, since I'm letting go to infinity, leaving fixed, not the other way around.
Anyway, I'll try it again later, but in the meantime sorry for the wrong proof. Although the other implication IS right (it IS kind of trivial though ).
Okay, I think I got it now.
Let be a Hilbert Space, and be an othonormal set. Let , then if . And so for we have that is an orthogonal set and so:
by definition
because is an othonormal set.
becuase the norm is always positive.
But this is for an arbitrary and so (in particular it converges).
Suppose that then there exist a sequence such that and it's obvious that (actually, all but finitely many terms are zero for each ).
Remember that in a normed space , we have that is complete iff every absolutely convergent series is convergent. Now since is complete we have that converges. Now it suffices to prove that this series converges to .
And so but since the limit is unique . If is a Hilbert basis the every is in and we're finished.
Okay, I think I got it right this time, but be wary anyway, who knows...