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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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    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.2011-07-20

1 Answers 1

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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.