Suppose I have two functions continuous functions $f,g \in \mathcal{S}(\Bbb{R})$, the Schwartz space. Now I know that the following multiplicative formula holds. Namely, if $\hat{g}$ denotes the Fourier transform of $g$ and similarly for $f$, we have that

$$\int_{\Bbb{R}} f(x)\hat{g}(x)dx = \int_{\Bbb{R}} \hat{f}(y)g(y) dy.$$

Now I am reading the proof of this fact in Stein and Shakarchi volume 1 and it seems that we do not even need $f,g$ to be in the Schwartz space. Them being of moderate decrease is enough, namely there is a constant $A$ such that

$$|f(x)| \leq \frac{A}{1+x^2}$$

and similarly for $g$. Am I right in saying this?


  • 1
    $\begingroup$ Yes, and we can check it by Fubini's theorem. $\endgroup$ Oct 22 '12 at 8:21
  • $\begingroup$ Sure, $x^{-2}$ is integrable right? The only issue is $0$. So add the $1$ and done. Fubini only requires integrability. $\endgroup$
    – JT_NL
    Oct 22 '12 at 13:21

Let $H(x,y):=f(x)g(y)e^{-ixy}$. As $|H(x,y)|\leq \frac{A^2}{(1+x^2)(1+y^2)}$, $H$ is integrable over $\mathbb R^2$. So, by Fubini's theorem, $$\int_{\Bbb R}f(x)\widehat g(x)dx=\int_{\Bbb R^2}f(x)g(y)e^{-ixy}dydx=\int_{\Bbb R^2}f(x)e^{-ixy}g(y)dxdy=\int_{\Bbb R}\widehat f(y) g(y)dy.$$


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