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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?2011-07-29
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    Rajesh D, what is the definition of "space spanned by a set"?2011-07-29
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    @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.2011-07-29
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    @Jonas Meyer : the vector space composed of all possible linear combinations of the given set of vectors.2011-07-29
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    @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 $L^2(\mathbb R)$.)2011-07-29
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    @Jonas Meyer : In that case why name it as $\mathcal{L}^2(\mathbb{R})$, the more appropriate name would be the 'span of the Fourier basis' ? I am even more puzzled.2011-07-29
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    @Rajesh D: I don't understand. $L^2(\mathbb R)$ is [$L^2(\mathbb R)$](http://en.wikipedia.org/wiki/Lp_space#Lp_spaces_2); what is wrong with this name?2011-07-29
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    By definition, the span of a basis is the entire space, *period*. *However*, a *Hilbert basis* need not be a basis of the vector space in the linear sense (those are sometimes called "Hamel bases"). The linear span of a Hilbert basis is guaranteed to be a *dense subspace* of the vector space; in the Hilbert base case, we usually consider only *closed* subspaces, and talk about the 'span' to mean the closure of the linear span. I suspect that you are looking at a Hilbert basis of $L^2(\mathbb{R})$, rather than at a Hamel basis; your original statement holds for Hamel bases.2011-07-29
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    @Jonas Meyer : I've read in some books on linear algebra and wavelets, that the Fourier basis is a complete orthonormal basis of $L^2(\mathbb{R})$ and they treat the $L^2(\mathbb{R})$ space with this basis as some kind of a holy grail. Instead why don't they consider the space spanned by the Fourier basis (set of sinusoids of all frequencies) ?2011-07-29
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    @Rajesh: Don't confuse [Hilbert basis](http://en.wikipedia.org/wiki/Orthonormal_basis) with [Hamel basis](http://en.wikipedia.org/wiki/Hamel_basis#Related_notions).2011-07-29
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    @Rajesh: If Qiaochu's answer hasn't answered your question, could you please give a precise reference to a treatment of $L^2(\mathbb R)$ that you find confusing? Again, sinusoids are not even in $L^2(\mathbb R)$, so they do not form a basis for $L^2(\mathbb R)$.2011-07-29
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    @Jonas : I had read the book, http://www.amazon.com/Introduction-Wavelets-Through-Undergraduate-Mathematics/dp/0387986391 some time back and had only remembered the final equation of Fourier transform for functions of $L^2(\mathbb{R})$ and was tempted to think that it is an orthonormal expansion as in Fourier series, but now i realize that the integral of inverse Fourier transform is actually a principal value. There isn't anything wrong in book.2011-07-29
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    the integral of Fourier transform is actually a principal value.2011-07-29
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    We 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

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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}$2011-07-29
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    @Rajesh: okay. None of those functions lie in $L^2(\mathbb{R})$, so they can't span it in any reasonable sense.2011-07-29
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    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. http://eom.springer.de/r/r082340.htm2011-07-29
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    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?2011-07-29
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    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.2011-07-29
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    @Qiaochu Yuan : What do we call the representation of functions in $\mathcal{L}^2(\mathbb{R})$ with Fourier Transform ? Now it is clear that it is not an orthonormal expansion in the conventional sense.2011-08-12
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    @Rajesh: to be honest, I don't know. I vaguely remember hearing that this is a "direct integral," or something like that.2011-08-12
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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 !!!