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Let $H$ be a separable Hilbert Space. Prove that exists orthonormal complete sequence and give example for one non-orthonormal sequence.

I thought taking orthonormal basis for $H$ denoted by $\{e_1,\dots ,e_n\}$ and then proving it's also complete, i.e $$(e_n,f)=0\Rightarrow f\equiv 0$$ $f$ can be represented as $f=\sum_{k=1}^\infty(e_k,f)e_k$ so I need to prove that $\forall k\in\mathbb{N}, (e_k,f)=0$ which as far as I see is given above, isn't it? Where am I using the separability?

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  • $\begingroup$ Since "complete orthonormal sequence" is basically another word for "orthonormal basis", saying "take an orthonormal basis" seems circular. You are probably supposed to prove directly that such a sequence exists. Regarding separability, every Hilbert space does have an orthonormal basis, but if the space is not separable, the basis will be uncountable. $\endgroup$ – Nate Eldredge Oct 11 '14 at 2:54
  • $\begingroup$ why that's important for the basis to be countable? $\endgroup$ – user65985 Oct 11 '14 at 3:00
  • $\begingroup$ The problem says "complete orthonormal sequence". A sequence has to be countable. $\endgroup$ – Nate Eldredge Oct 11 '14 at 3:01
  • $\begingroup$ I didn't think about it. thanks. $\endgroup$ – user65985 Oct 11 '14 at 3:05
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Suppose $H$ has a countably dense subset $\{ s_{n}\}_{n=1}^{\infty}$ of non-zero vectors. Using induction we may discard each $s_{n}$ which is a linear combination of the previous vectors in the sequence in order to arrive at a new sequence $\{ t_{n}\}_{n=1}^{\infty}$ of vectors for which the $\{ t_{1},t_{2},\cdots,t_{k}\}$ is linearly independent for each fixed $k=1,2,3,\cdots$. Then we may perform Gram-Schmidt on the resulting set to obtain an orthonormal sequence $\{ e_{1},e_{2},e_{3},\cdots\}$ such that the first $n$ elements of $\{ t_{k}\}$ spans the same subspace as the first $n$ elements of $\{ e_{k}\}$.

Automatically, finite linear combinations of the $e_{k}$ form a dense linear subspace of $H$ because this subspace includes the original dense subset $\{ s_{n}\}$, which is one equivalent to the orthonormal subset $\{ e_{k}\}_{k=1}^{\infty}$ being complete. A simple example of a non-orthonormal sequence is $\{ e_{1},e_{1}+e_{2},e_{1}+e_{3},e_{1}+e_{4},\cdots\}$. The inner product of any two of the elements of this sequence is $1$.

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