1
$\begingroup$

I am currently studying the Fourier transform in Euclidean Space $\mathbb{R}^n$, using Knapp's book "Basic Real Analysis".

Upon proving the property that the Fourier transform turns Convolution in $L^1$ into multiplication (that is, if $f,g$ are in $L^1$ then $\mathcal{F}(f*g) = \mathcal{F}(f)\mathcal{F}(g)$ we use Fubini's theorem.

Here the author states that the following interchange is valid:

$\begin{equation} \int \int |f(x-t)g(t) e^{-2\pi i x \cdot y}| \, dt \,dx = \int \int |f(x-t)g(t) e^{-2\pi i x \cdot y}| \, dx \,dt \end{equation}$

I have trouble justifying this on my own - since I am integrating over $\mathbb{R}^n$, I think I cannot use Fubini's theorem immediately because it is a statement about $\sigma$ - finite measure spaces, is that correct ?

EDIT: As of the comment below, this is not a problem because Euclidean space is $\sigma$ - finite with respect to Lebesgue measure

So then the question becomes how do I apply Fubini's theorem, not knowing whether the product of the two functions is absolutely integrable ?

In case the left hand side is finite I can indeed use Fubini's theorem. Now if I suppose that the left hand side is infinite - I must show that the right hand side is also infinite and this is where I struggle with the argument ..

(I am sorry for being so confused, somehow I cannot get my head around Fubini's theorem and Lebesgue - integrable functions, I always mix assumptions, properties etc up. Thanks for your patience!)

  • 0
    @harlekin: Right.2012-05-20

0 Answers 0