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If $\{\frac{f_{n}}{\|f_{n}\|}\}_{n\in I}$ is an orthonormal basis for a separable Hilbert space $H$, and $\{f_{n}\}_{n\in I}$ is a complete and orthogonal set in $H$, is it true that $\{f_{n}\}_{n\in I}$ is a basis for $H$?

If this is not always true, when it would be?

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Will you please define what you mean by "basis for $H$"? (It is not an orthonormal basis unless $\|f_n\|=1$ for all $n$, and it is not a vector space basis(reference 1 and 2) unless $H$ is finite dimensional, but I don't know whether you have something else in mind.) – Jonas Meyer May 28 '12 at 3:23
If you mean Schauder Basis (every $x$ has a unique representation $\sum\alpha_n f_n$), then yes. More generally, if $\{f_n\}_{n=1}^\infty$ is a Schauder Basis of $H$ (as orthonormal bases are), then so is $\{\alpha_n f_n\}_{n=1}^\infty$ for any sequence of non-zero scalars $\{\alpha_n\}$. – David Mitra May 28 '12 at 3:26
Okay, I found this in wikipedia: ,"* Given a Hilbert space H and a set S of mutually orthogonal vectors in H, we can take the smallest closed linear subspace V of H containing S. Then S will be an orthogonal basis of V; which may of course be smaller than H itself, being an incomplete orthogonal set, or be H, when it is a complete orthogonal set.*" So, since my set is complete it will be an orthogonal basis. Any one help me to prove it! – Liza May 28 '12 at 3:35
@Liza: So if you are meaning to ask, "Is the smallest closed linear subspace of $H$ containing $\{f_n\}$ equal to $H$?" then the answer is "Yes." Such a subspace also contains an orthonormal basis $\{f_n/\|f_n\|\}$ for $H$. It is still not 100% clear to me what your question is asking. The quote you gave never precisely defined "orthogonal basis," but my guess is you meant what I wrote above, which in this case is equivalent to whether it is a Schauder basis as mentioned in David Mitra's comment. – Jonas Meyer May 28 '12 at 3:38

It is true that $\{f_n\}$ is a Schauder basis for $H$, which by definition means that every $x\in H$ has a unique representation as $\sum_{n=1}^\infty \alpha_n f_n$. Indeed, it is a direct consequence of the definition that multiplying each vector of a Schauder basis by a nonzero scalar gives another Schauder basis. (Based on a comment by David Mitra.)

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