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A function of moderate decrease is a map from $\mathbb{R}$ into $\mathbb{C}$ such that there exists $A \in \mathbb{R}$ such that $\forall x\in \mathbb{R}, \ |f(x)| \lt \frac{A}{1 + |x|^{1+\epsilon}}$. And I want to prove that $$(\hat{f\ast g)}(t)=\hat{f}(t)\hat{g}(t).$$ I know the the proof when the function is in Schwarz Space. But the proof cannot work when working in such space (of moderate decrease), since the condition is much weaker. If I use the function $\ F(x,y) =f(y)g(x-y)e^{-2i \pi xt}$, and use double integral in $ x $ and $ y $. The main problem is the feasibility to change the order of integral, I don't know how to deal with $\frac{A}{1 + |x-y|^{1+\epsilon}}$ , since there is a convolution.

Curran
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  • May i ask you how the proof goes when functions are in the schwartz space? i'm have the same problem – Mathcho Nov 29 '16 at 01:22

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Notice that you can apply Tonelli's theorem to the non-negative integrand $\frac{A}{1 + |x - y|^{1 + \epsilon}}\frac{B}{1 + |y|^{1 + \epsilon}}$. Indeed this gives

\begin{align} \int_{\mathbb{R}}\int_{\mathbb{R}}\frac{A}{1 + |x - y|^{1 + \epsilon}}\frac{B}{1 + |y|^{1 + \epsilon}}\,dy\,dx = &\ \int_{\mathbb{R}}\frac{B}{1 + |y|^{1 + \epsilon}}\int_{\mathbb{R}}\frac{A}{1 + |x - y|^{1 + \epsilon}}\,dx\,dy \\ \overset{z = x - y}{=} &\ \int_{\mathbb{R}}\frac{B}{1 + |y|^{1 + \epsilon}}\int_{\mathbb{R}}\frac{A}{1 + |z|^{1 + \epsilon}}\,dz\,dy < \infty. \end{align}

This proves that the function $F$ defined in the question is integrable, hence you can apply Fubini's theorem to interchange the integrals and procede as in the case of rapidly decreasing functions.

Giovanni
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  • Sorry, I am considering the situation based on Riemann integration, so Fubini's theorem may not work. – Curran Oct 20 '15 at 04:44
  • Well, the Lebesgue integral of a bounded Riemann integrable function is the Riemann integral, so this can also be applied in your case. How do you interchange Riemann integrals usually? Could you point me to a theorem? – Giovanni Oct 20 '15 at 04:52
  • I remember the Lebesgue integral equals to the Riemann integral when the domain is a closed interval, but this may fail when working on the whole plane. Actually, I don't know how to interchange Riemann integrals when the domain is unbounded. – Curran Oct 20 '15 at 05:01
  • Good point, I forgot about the "bounded interval" condition. I don't think we can get the desired result without decoupling the double integral, we really need a Fubini-type theorem here. Here is what I have found so far: Fubini's theorem for Riemann integral – Giovanni Oct 20 '15 at 05:06
  • But that theorem works on rectangle, which is bounded, we need a theorem that can deal with improper integrals. – Curran Oct 20 '15 at 05:14
  • Yeah, I have noticed, but it contains references to two books in which apparently the Fubini's theorem for the Riemann integral is discussed. – Giovanni Oct 20 '15 at 05:20