The problem I am working on is:
"Let H 'be a separable Hilbert space and M a countable dense subset of H. Show that H contains a total orthonormal sequence which can be obtained from M by the Gram-Schmidt process."
The Gram Schmidt process preserves span, so if I can extract from M a linearly independent sequence whose span is also dense in H, then I am done. It seems intuitive that such a sequence should exist since M is dense, but I'm not sure how to prove that result.
Best Answer
List out the elements of $M$ as $x_1,x_2,x_3,\dots$, and WLOG assume none of them are the zero vector. Start with $L_1=\{x_1\}$, then add (i.e union) to $L_1$ the first term in the sequence not already contained in $\text{span}(L_1)$. Call this new set $L_2$. Then, you once again find the first element not already in span of $L_2$. Call this new set $L_3$. Keep going and define $L:=\bigcup_{n=1}^{\infty}L_n$. By construction, $M\subset \text{span}(L)$ (the set of all finite linear combinations), so taking closures implies $\overline{\text{span}(L)}=H$. We also ensured linear independence of each $L_n$, thus $L$ is also linearly independent. I leave it to you to write this out precisely as a recursive proof. Now apply Gram-Schmidt to $L$ to get an orthonormal basis.