2
$\begingroup$

in this question, it is explained how one can rigorously define the Fourier transform over $L_1$ functions using a Lebesgue integral. But what about when we transform functions which are not in $L_1$?

Specifically, in many branches of physics and engineering it is common to take Fourier transforms of functions such as $cos(t), \space sin(t),\space f(t)=1$

The above examples all result somehow in Dirac Delta functions, which are actually distributions and not functions proper. So it seems we need to come up with a rather different definition of what a Fourier transform is to include these cases.

How can we do this?

$\endgroup$
  • $\begingroup$ You are asking for description of an entire theory which is easily available in books. What sort of answer can one post for this question. You should begin by studying Fourier transforms so $L^{2}$ functions from, say, Rudin's book. Incidentally, this does not involve Dirac delta. $\endgroup$ – Kavi Rama Murthy Nov 14 '18 at 12:04
1
$\begingroup$

In short a tempered distribution $T$ is uniquely defined by its pairing with gaussians : $$T_{a} = e^{-\pi a^2 x^2} (T \ast \frac1a e^{-\pi x^2/a^2})$$ Note $T_{a}$ is $C^\infty \cap L^1$

(when $a \to 0$, $\frac1a e^{-\pi x^2/a^2} \to \delta$ and $T \ast \frac1a e^{-\pi x^2/a^2} \to T$ in the sense of distributions).

Given a tempered distribution, compute the Fourier transform $\hat{T}_{a} = FT[T_a] \in C^\infty \cap L^1$. and let $a \to 0$, you obtain $\hat{T} = \lim_{a \to 0} \hat{T}_a$ (again limit in the sense of distributions) and $\hat{T}_a = \frac1a e^{-\pi x^2/a^2} \ast (e^{-\pi \xi^2 a^2} \hat{T})$, which, if true for one $a$, is true for every $a$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.