Convolution Theorem for Fourier Transform with Distributions

[mhill]

Homework Statement

using the Fourier Transform convolution theorem should be true that

[tex] i^{m+n}D^{m}\delta (u)D^{n}\delta (u)= A \mathcal F _{u}(\int_{-\infty}^{\infty}dt (t-x)^{m}t^{n} ) [/tex]

Homework Equations

- Fourier transform convolution theorem (would be valid for distributions ? )

The Attempt at a Solution

i have thought that although the integrals are divergent , the Convolution theorem should hold no matter if we are dealing with distributions (in fact if one of the functions is a distribution but the other is not the convolution theorem holds for example the case f(x)=1 it Fourier transform is a dirac delta but the convolution integral is well defined.

 

[mighty2000]

Thank you for your post. The Fourier transform convolution theorem is indeed valid for distributions. In fact, the convolution theorem is often used to define the Fourier transform of distributions. The integral in the convolution theorem may be divergent, but as long as the functions are integrable, the convolution theorem will hold. In the case of distributions, the integral may be interpreted as a generalized integral, and the convolution theorem will still hold.

In your example, the function f(x) = 1 is a well-behaved function, so the convolution theorem holds. The Fourier transform of f(x) is a dirac delta, and the convolution integral is well-defined. In the case of distributions, the convolution theorem will hold even if one of the functions is a distribution. So, in your equation, the convolution theorem will hold as long as the functions are integrable. However, the Fourier transform of a distribution may not always be a well-defined function, so it is important to check the integrability of the functions before using the convolution theorem.

I hope this helps answer your question. If you have any further doubts or concerns, please do not hesitate to ask. Keep up the good work in your scientific endeavors!



Scientist