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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.