Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
3  
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: en.wikipedia.org/wiki/Orthonormal_basis ,"* 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

1 Answer 1

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.