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Jose27 · #2 May 31st, 2009, 01:18 PM

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.

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