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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.

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"The same is not true" — Why? Could you explain? – ShreevatsaR Jul 29 '11 at 3:56
Rajesh D, what is the definition of "space spanned by a set"? – Jonas Meyer Jul 29 '11 at 3:59
@ShreevatsaR : i thought that a sinusoid is not part of the vector space $\mathcal{L}^2(\mathbb{R})$, as it is not square integrable over the entire real line. – Rajesh Dachiraju Jul 29 '11 at 4:00
@Jonas Meyer : the vector space composed of all possible linear combinations of the given set of vectors. – Rajesh Dachiraju Jul 29 '11 at 4:01
@Rajesh D: In that case, consider linear combinations where all but one of the coefficients are $0$. You can also think of the span of a set as the smallest subspace containing the set. (It is true that sinusoids are not in $L^2(\mathbb R)$, and therefore they cannot be basis vectors in $L^2(\mathbb R)$.) – Jonas Meyer Jul 29 '11 at 4:04
up vote 4 down vote accepted

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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@Qiacochu : by "the Fourier basis" i mean the functions $e^{2 \pi i f x}, f \in \mathbb{R}$ – Rajesh Dachiraju Jul 29 '11 at 4:20
@Rajesh: okay. None of those functions lie in $L^2(\mathbb{R})$, so they can't span it in any reasonable sense. – Qiaochu Yuan Jul 29 '11 at 4:23
You might look up "rigged Hilbert space", which provides a rigourous way of dealing with such "basis" functions that are not members of the space. See e.g. – Robert Israel Jul 29 '11 at 5:33
Of course physicists do this all the time. Tell them the functions are not in the space, and they still go on. If the final result can be checked by physical experiment, who cares if they didn't understand some of the mathematical steps to get there? – GEdgar Jul 29 '11 at 13:23
I would be deliberately evasive, and say that functions in $L^2(\mathbb R)$ are superpositions of exponentials. Certainly the spirit is that they "span", but, yes, awkward that they're not in the space. Gelfand et al called them "generalized eigenfunctions" (for $d^2/dx^2$, for example). And, yes, this is a provocative situation, in many regards. It's not so easy to arrange natural, explicit examples of "continuous spectrum", perhaps exactly because the eigenfunctions are not "genuine". Zonal spherical functions on reductive groups have the same feature. $GL(n,\mathbb C)$'s Plancherel. – paul garrett Jul 29 '11 at 15:22

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 !!!

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