1. ## Fourier Coefficients

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>.

2. It should be clear that it suffices to prove the following:

Let $B=\{e_k: k \in \mathbb{N}\}$ an orthonormal set in a Hilbert space $H$, then $B$ is a Hilbert base ( $H=\overline{lin(B)}$) $\Longleftrightarrow$ $\forall u \in H$ we have $u = \sum_{i=1}^ \infty \ (u \circ e_i)e_i$
$\Longrightarrow)$

Since $B$ is a Hilbert base, we have $H=\overline{lin(B)}$ and so, we divide this in two cases:

1) $u \in lin(B)$: We have then $u = \sum_{i=1}^ m \ a_i(e_i)$ where $a_i \in \mathbb{F}$ (where $\mathbb{F}$ is the basefield $\mathbb{R}$ or $\mathbb{C}$) then $u \circ e_k = (\sum_{i=1}^m \ a_i(e_i)) \circ e_k = \sum_{i=1}^m \ a_i(e_i \circ e_k) = a_k$. Thus $u \circ e_k = a_k$, and as such for every $n>m$ we have $(u \circ e_n) = 0$ and so $u = \sum_{i=1}^ \infty \ (u \circ e_i)e_i.$
2) $u \in \overline{lin(B)}$: We choose a $v \in lin(B)$ such that $\|
{u-v}\|
<\epsilon$
. Then $v = \sum_{i=1}^ l \ (v \circ e_k)e_k$. Now we take $m>l$ and using the triangle inequality two times and the Cauchy-Schwarz inequality afterwards we obtain:

$\|{\sum_{i=1}^ m \ (u \circ e_i)e_i} -u \| \leq \|{\sum_{i=1}^ m \ ( u \circ e_i)e_i}-{\sum_{i=1}^l \ ( v \circ e_i)e_i}\| + \| {v-u}\|$ $\leq \|{\sum_{i=1}^ m \ ( u-v \circ e_i)e_i}\| +\epsilon \leq {\sum_{i=1}^ m \ {\vert(u-v \circ e_i)\vert}} + \epsilon \leq \sum_{i=1}^ m \ {\|{u-v}\|} + \epsilon \leq m\epsilon + \epsilon \rightarrow 0.
$

Thus we have shown that $u = \sum_{i=1}^ \infty \ ( u \circ e_i) e_i.$

Since $H=\overline{lin(B)}$, we are finished.

$\Longleftarrow)
$

Since for all $m \in \mathbb{N}$ we have $u_m = \sum_{i=1}^ m \ ( u \circ e_i)e_i$ then $u_m \in lin(B)$, we have that $u_m \rightarrow u$ and so, $u \in \overline{lin(B)}$, and so $H=\overline{lin(B)}$.

Man, that was hard to type, anyway hope it helps.

3. Just a quick question what is $\overline{lin(B)}$ ? is it same as span ? i.e. all possible linear combinations of e_n ?

4. Originally Posted by frater_cp
Just a quick question what is \overline{lin(B)} ? is it same as span ? i.e. all possible linear combinations of e_n ?
No, it means the topological closure (obviously with the topology induced from the norm) of the linear span of B.

By the way check it out, I fixed the first post and now everything is legible...Yay.

5. Youre the best Jose ! My brain is a bit tired has been a long day, but I'm gonna work through it slowly in the morning when I'm fresh. Thanks so much for the assistance :-)

6. can you just help me with something small... I am not exactly sure what the
operator $\circ$ means.

Never seen it before is it an inner product or something else?

7. Yes, it's the inner product, I was going to use the normal brackets, but my instructions got all mixed up, and decided to change the brackets for this notation. Hope it doesn't confuse you much.

8. 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!

9. Originally Posted by Jose27
It should be clear that it suffices to prove the following:

Let $B=\{e_k: k \in \mathbb{N}\}$ an orthonormal set in a Hilbert space $H$, then $B$ is a Hilbert base ( $H=\overline{lin(B)}$) $\Longleftrightarrow$ $\forall u \in H$ we have $u = \sum_{i=1}^ \infty \ (u \circ e_i)e_i$
$\Longrightarrow)$

Since $B$ is a Hilbert base, we have $H=\overline{lin(B)}$ and so, we divide this in two cases:

1) $u \in lin(B)$: We have then $u = \sum_{i=1}^ m \ a_i(e_i)$ where $a_i \in \mathbb{F}$ (where $\mathbb{F}$ is the basefield $\mathbb{R}$ or $\mathbb{C}$) then $u \circ e_k = (\sum_{i=1}^m \ a_i(e_i)) \circ e_k = \sum_{i=1}^m \ a_i(e_i \circ e_k) = a_k$. Thus $u \circ e_k = a_k$, and as such for every $n>m$ we have $(u \circ e_n) = 0$ and so $u = \sum_{i=1}^ \infty \ (u \circ e_i)e_i.$
2) $u \in \overline{lin(B)}$: We choose a $v \in lin(B)$ such that $\|
{u-v}\|
<\epsilon$
. Then $v = \sum_{i=1}^ l \ (v \circ e_k)e_k$. Now we take $m>l$ and using the triangle inequality two times and the Cauchy-Schwarz inequality afterwards we obtain:

$\|{\sum_{i=1}^ m \ (u \circ e_i)e_i} -u \| \leq \|{\sum_{i=1}^ m \ ( u \circ e_i)e_i}-{\sum_{i=1}^l \ ( v \circ e_i)e_i}\| + \| {v-u}\|$ $\leq \|{\sum_{i=1}^ m \ ( u-v \circ e_i)e_i}\| +\epsilon \leq {\sum_{i=1}^ m \ {\vert(u-v \circ e_i)\vert}} + \epsilon \leq \sum_{i=1}^ m \ {\|{u-v}\|} + \epsilon \leq m\epsilon + \epsilon \rightarrow 0.
$

Thus we have shown that $u = \sum_{i=1}^ \infty \ ( u \circ e_i) e_i.$

Since $H=\overline{lin(B)}$, we are finished.

$\Longleftarrow)
$

Since for all $m \in \mathbb{N}$ we have $u_m = \sum_{i=1}^ m \ ( u \circ e_i)e_i$ then $u_m \in lin(B)$, we have that $u_m \rightarrow u$ and so, $u \in \overline{lin(B)}$, and so $H=\overline{lin(B)}$.

Man, that was hard to type, anyway hope it helps.
I was checking this the other day, and realized that there's a mistake in $\Longrightarrow )$ basically, I'm fixing $\epsilon$ and trying to show that for sufficiently large $m$ the series approaches what it has to, and if you notice I concluded with $m\epsilon + \epsilon \longrightarrow 0$ which is clearly false, since I'm letting $m$ go to infinity, leaving $\epsilon$ 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 ).

10. Hi Jose. Thanks for alerting me to the fact that there is something wrong with the proof.
Do you think you can rectify it?

11. ## unsolved

Originally Posted by Jose27
I was checking this the other day, and realized that there's a mistake in $\Longrightarrow )$ basically, I'm fixing $\epsilon$ and trying to show that for sufficiently large $m$ the series approaches what it has to, and if you notice I concluded with $m\epsilon + \epsilon \longrightarrow 0$ which is clearly false, since I'm letting $m$ go to infinity, leaving $\epsilon$ 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 ).

I am also seeking a proof for this problem.

12. Okay, I think I got it now.

$\Longrightarrow )$

Let $H$ be a Hilbert Space, $u \in H$ and $B= \{ e_k : k \in \mathbb{N} \} \subset H$ be an othonormal set. Let $a_n = u- \sum_{i=0}^n \ { e_i}$, then $= - \sum_{i=0}^ n \ {}$ $=0$ if $k \leq n$. And so for $k=n$ we have that $C= \{ a_n, e_0,...,e_n \}$ is an orthogonal set and so:

$\Vert u \Vert ^2 = \Vert a_n + \sum_{i=0}^n \ {e_i} \Vert ^2$ by definition

$= \Vert a_n \Vert ^2 + \vert \sum_{i=0}^n \ { \vert \vert ^2} \vert$ because $C$ is an othonormal set.

$\geq \sum_{i=0}^n \ { \vert \vert ^2}$ becuase the norm is always positive.

But this is for an arbitrary $n \in \mathbb{N}$ and so $\sum_{i=0}^{\infty} \ { \vert \vert ^2} \leq \Vert u \Vert ^2$ (in particular it converges).

Suppose that $u \in \overline{linB}$ then there exist a sequence $(x_n) \subset linB$ such that $x_n \longrightarrow u$ and it's obvious that $x_n= \sum_{i=0}^{\infty} \ {e_i}$ (actually, all but finitely many terms are zero for each $x_n$).

Remember that in a normed space $T$, we have that $T$ is complete iff every absolutely convergent series is convergent. Now since $H$ is complete we have that $u_1= \sum_{i=0}^{\infty} \ {e_i}$ converges. Now it suffices to prove that this series converges to $u$.

$\Vert x_n - u_1 \Vert ^2 = \Vert \sum_{i=0}^{\infty} \ {e_i} - \sum_{i=0}^{\infty} \ {e_i} \Vert ^2$

$= \Vert \sum_{i=0}^{\infty} \ {e_i} \Vert ^2$

$\leq \sum_{i=0}^{\infty} \ {\vert \vert ^2}$

$\leq \Vert x_n - u \Vert ^2 \longrightarrow 0
$

And so $x_n \longrightarrow u_1$ but since the limit is unique $u_1=u$. If $B$ is a Hilbert basis the every $u$ is in $\overline{linB}$ and we're finished.

Okay, I think I got it right this time, but be wary anyway, who knows...