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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.

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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 $g(x,\xi) = \frac{1}{(2\pi)^{n/2}} f(x) \varphi(\xi) e^{-\xi \cdot x}$. – t.b. Jul 20 '11 at 2:09
@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? – Jack Jul 20 '11 at 2:13
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. – t.b. Jul 20 '11 at 2:15
@Theo: Fair enough. That's good enough to be an answer. – Jack Jul 20 '11 at 2:16
up vote 6 down vote accepted

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.

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