I am trying to prove $\widehat{f*g}=\hat f*\hat g$ for $f,g$ in the Schwartz Space, i.e $\sup(|x|^nf^{(k)})<\infty$ for all $n,k$.
$$\widehat{f*g}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-t)g(t)e^{-2\pi ix\xi}dtdx.$$ I am trying to apply the interchange of integrals.
So we have:
\begin{align}\widehat{f*g} &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-t)e^{-2\pi ix\xi}dx g(t)dt \\&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi}dx e^{-2\pi it\xi}g(t)dt \\&=\hat{f}\hat{g} \end{align}
However, to apply the interchange of integral, I have to make sure $|f(x-t)g(t)e^{-2\pi ix\xi}|=|f(x-t)g(t)|$ is of moderate decrease, i.e $|f(x-t)g(t)|\le\frac{A}{(1+x^2)(1+t^2)}$. I know it is not true that $\exists A>0$ such that $|f(x-t)|\le\frac{A}{1+x^2}$ for all $x,t$. Since here $A$ is independent of $t$, and for each $A$ we can always shift $f$ by $t$ to make some point of $f(x-t)$ strictly larger than $\frac{A}{1+x^2}$. So we cannot prove $f(x-t),g(t)$ separately. We have to estimate them together to get the decay. Any help will be appreciated.
You can show that bound, but you do not need to in order to show that the product is integrable.
First the easy way. $|f(x-t)g(t)|\le \|f\|_{L^\infty} \|g\|_{L^\infty} \lesssim 1$, and similarly $ |(x-t)^2 f(x-t)g(t)| \lesssim 1.$ Together, $$ |(1+(x-t)^2)f(x-t)g(t)| \lesssim 1$$ and similarly $$ |(1+(x-t)^2)(1+t^2)f(x-t)g(t)| \lesssim 1$$
Now note $$ \int_{\mathbb R^2 } \frac1{(1+(x-t)^2)(1+t^2)}dxdt = \int_{\mathbb R^2 } \frac1{(1+y^2)(1+t^2)}dydt < \infty$$ So $f(x-t)g(t)\in L^1(\mathbb R^2)$ and Fubini's theorem applies.
If you really want to prove the bound: Note that if $|x-t|\ge |x|/2$, then of course $4(1+(x-t)^2) \ge 1+x^2$. So for such $x,t$, $$|(1+x^2)(1+t^2)f(x-t)g(t)| \le 4|(1+(x-t)^2)(1+t^2)f(x-t)g(t)|\lesssim 1.$$ But if instead $|x-t|<|x|/2$ then $$|x|\le |x-t|+|t| < \frac{|x|}2 + |t|$$ so that $$|x| \le 2|t|.$$ This implies $$ |(1+x^2)(1+t^2)f(x-t)g(t)| \le 4 |(1+t^2)^2f(x-t)g(t)| \lesssim 1,$$ showing the moderate decrease.