Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!)

share|improve this question
1  
The space R^n is sigma-finite with respect to the Lebesgue measure. –  Did May 20 '12 at 12:18
    
Ah .. I confused finite and $\sigma -finite$, of course these notions are different. I edit the post immediately so that it is not wrong, thanks for this comment! –  harlekin May 20 '12 at 12:22
1  
The next step is that one knows the product is integrable since, by the change of variable $s=x-t$, the integral of its absolute value is exactly the integral of |f| times the integral of |g|. –  Did May 20 '12 at 12:52
    
@Didier Ok, so you say that if we make the change of variable $s = x - t$ then $dtdx = dtd(s + t) = dtds + dtdt = dtds + 0= dtds$ and then I have an integral of the form $\int f(s) g(t) dsdt = \int f(s) ds \int g(t) dt$ where the left hand side is finite because $f$ and $g$ are in $L^1$. Is that correct ? –  harlekin May 20 '12 at 13:10
2  
Another version of Fubini's theorem (also called Tonelli's theorem) states that, assuming $\sigma$-finite measures, interchanging integrals is always valid when the integrand is non-negative; if one integral is infinite then so will be the other. –  Nate Eldredge May 20 '12 at 13:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.