- #1
muzialis
- 166
- 1
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
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