I used to think that in any Vector space the space spanned by a set of orthogonal basis vectors contains the basis vectors themselves. But when I consider the vector space $\mathcal{L}^2(\mathbb{R})$ and the Fourier basis which spans this vector space, the same is not true ! I'd like to get clarified on possible mistake in the above argument.
Does the vector space spanned by a set of orthogonal basis contains the basis vectors themselves always?
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0We can also say that $L^2(\mathbb R)$ is the Hilbert direct _integral_, not sum, of the one-dimensional spaces $\mathbb C\cdot e^{i\xi x}$. To my perception, the issue is not about Hamel versus Hilbert basis, but about _continuous_ rather than _discrete_ decomposition, and, although the exponentials are not in $L^2$, they are what we want. This is not _elementary_ Hilbert space theory. – 2011-07-29
2 Answers
If by "the Fourier basis" you mean the functions $e^{2 \pi i n x}, n \in \mathbb{Z}$, then these functions do not lie in $L^2(\mathbb{R})$ as they are not square-integrable over $\mathbb{R}$, so in particular they can't span that space in any reasonable sense. The functions $e^{2 \pi i n x}$ do span $L^2(S^1)$ (in the Hilbert space sense).
Perhaps you are getting the Fourier transform for periodic functions mixed up with the Fourier transform on $\mathbb{R}$.
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0@Rajesh: to be honest, I don't know. I va$g$uely remember hearin$g$ that this is a "direct inte$g$ral," or something like that. – 2011-08-12
I want you guys to understand few things clearly:
1> The dirchlet conditions are "sufficient" not necessary for F' transform'bility
2> The sinusoid is DEFINITELY in the span of {exp(j2PIft)} this follows from linear algebra! However It does not mean that the fourier TRANSFORM exists ! .. however in case of the periodic signals with finite discontinuities , and INTEGRABILITY(ABSOLUTE) WITHIN A PERIOD, the fourier SERIES does exist !!! ..
3> for a sine wave , Acos(wt) we (in some engineering contexts) agree that the fourier transform is the delta distribution along +/- w . However strictly we can only talk of the fourier series of such a function ! .
4> the fourier transform is the fourier series , approximated as the fundamental frequency being "epsilon" -->0 ; i.e the funda period being inf.
lastly ! do remember that fourier transformability is not necessary for SPANNING !!
This may not really be an answer to the above but is WORTH THINKING ABOUT: The fourier "BASIS" definitely does not span L2(r) , consider the simple rectangular pulse which is definitely square integrable over R. However If you do consider the infinite linear combination (of coefficients of a sinc(f) function whose inverse fourier transform is the pulse) , in the limit as we go on taking infinite (uncountable) sums , The result does NOT converge to the rect pulse in the point convergence sense , but only in the norm square sense. !
In essence , rect(t) is NOT FOURIER TRANSFORMABLE !! if it were, then all the filter theory we study makes no sense !!!