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maze
maze is offline
#7
Jun4-09, 07:07 AM
P: 655
To ease explanation, consider the complex exponentials e-ikx instead of sines and cosines, and consider the complex exponential version of the fourier transform.

If you give it a bit of thought, you will realize that the fourier transform of e-ikx is not actually a function, but rather a delta "function" (really, the delta distribution). Why? e-ikx is a perfect wave of a single frequency, so it's Fourier transform has all of the weight concentrated at a single point k, and no weight at any other frequencies.

Therefore the natural space to think about Fourier transforms of things like e-ikx is a space of distributions. The space commonly used is the dual space to the Schwarz space, and then the Fourier transform is F:S'->S' rather than F:L2->L2.

The complex exponentials might form a Schauder basis for S', though I highly doubt it. They certainly don't form a Hamel basis. This is actually an interesting question.