Fourier Trasform of Delta functions

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SUMMARY

The Fourier transform of constant functions results in the Dirac delta distribution, as established through the principle of self-adjointness. This relationship is demonstrated using duality pairing, where equals <δ, g>. Specifically, when g is the constant unit function, the equality <1, F(g)> = g(0) holds true, confirming the connection between the Fourier transform and the Dirac delta function. The discussion references the Wikipedia page on the Dirac delta function for further clarification.

PREREQUISITES
  • Understanding of Fourier transforms
  • Familiarity with Dirac delta distribution
  • Knowledge of self-adjoint operators
  • Basic concepts of duality pairing in functional analysis
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  • Study the properties of self-adjoint operators in functional analysis
  • Explore the implications of duality pairing in Fourier analysis
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Mathematicians, physicists, and engineers seeking a deeper understanding of Fourier transforms and their applications, particularly in relation to the Dirac delta function.

muzialis
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Hi All,

I am trying to understand more rigorously why the Fourier transform of a constant functions equals the Dirac delta distribution.

I found somewhere this is justified by imposing the self-adjointness, so that under a duality pairing <..,..> and indicating with F(f) the transform of a function f, it is required that

<F(δ), g> = < δ , g >

If g equals the constant unitary function my source, http://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_transform, quotes,

<1, F (g)> = g (0)= < δ , g >

I understand the second equality, but not sure about the first...

Many thanks for your help
 
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