7
$\begingroup$

I learned the following from Hunter's Applied Analysis. Denote the Schwartz space $${\mathcal S}({\mathbb R}^n):=\{\varphi\in C^{\infty}({\mathbb R}^n):\sup_{x\in{\mathbb R}^n}|x^{\alpha}\partial^{\beta}\varphi(x)|<\infty\quad\text{for every}\quad\alpha,\beta\in{\mathbb Z}_+^n\}.$$ If $\varphi\in{\mathcal S}({\mathbb R}^n)$, then the Fourier transform $\hat{\varphi}:{\mathbb R}^n\to{\mathbb C}$ is the function defined by $$\hat{\varphi}(\xi):=\frac{1}{(2\pi)^{n/2}}\int_{{\mathbb R}^n}\varphi(x)e^{-i\xi\cdot x}dx,\quad \xi\in{\mathbb R}^n.$$

Suppose that $f,\varphi\in{\mathcal S}$. Then we have $$\begin{align}\int\hat{f}(\xi)\varphi(\xi)d\xi&=\int\frac{1}{(2\pi)^{n/2}}\bigg(\int f(x)e^{-i\xi\cdot x}dx\bigg)\varphi(\xi)d\xi\\ &=\int f(x)\frac{1}{(2\pi)^{n/2}}\bigg(\int\varphi(\xi)e^{-i\xi\cdot x }d\xi\bigg)dx\\ &=\int f(x)\hat{\varphi}(x)dx. \end{align}$$ This is the motivation of the definition of the Fourier transform of tempered distributions.

Here is my question:

How can I get the second equality by Fubini's theorem?

The form I know about the theorem is $$\int_{A}\bigg(\int_B f(x,y)dy\bigg)dx =\int_{B}\bigg(\int_A f(x,y)dx\bigg)dy =\int_{A\times B}f(x,y)d(x,y)$$ But I am wondering what is $f(x,y)$ in the case above.

$\endgroup$
3
  • 2
    $\begingroup$ The trick is to move everything inside the integrals and what is there will be the thing to which you'll have to apply Fubini. So you'll apply it to $g(x,\xi) = \frac{1}{(2\pi)^{n/2}} f(x) \varphi(\xi) e^{-\xi \cdot x}$. $\endgroup$ Commented Jul 20, 2011 at 2:09
  • 1
    $\begingroup$ @Theo: +1.Thanks for your answer! I focus too much on the "difference" to notice the trick. Could you write your comment as an answer? $\endgroup$ Commented Jul 20, 2011 at 2:13
  • $\begingroup$ Okay, will do. But I don't have much more to say. By the way, there is a $dx$ missing in the equation you ask about. $\endgroup$ Commented Jul 20, 2011 at 2:15

1 Answer 1

Reset to default
6
$\begingroup$

The function to which you'll have to apply Fubini is $$g(x,\xi) = \frac{1}{(2\pi)^{n/2}} f(x) \varphi(\xi) e^{-\xi \cdot x}.$$ I think you can check for yourself that this is a function in $\mathcal{S}(\mathbb{R}^{n}\times\mathbb{R}^{n}) \subset L^{1}(\mathbb{R}^n \times \mathbb{R}^n)$, so it is a function to which we can apply Fubini.

The calculation itself is straightforward: start with the first equation, move everything inside the integrals, switch the order of integration and pull the things out, so as to arrive at the second equation, the one you're asking about.

$\endgroup$

You must log in to answer this question.