6-6

=== Problem 6.6 ===
For f\in L\left(\mathbb{R}\right), define the Fourier transform \widehat{f} of f by\widehat{f}(x)=\dfrac{1}{2\pi}\int^\infty_{-\infty} f(t)e^{-ixt} \ dt \ \ \ \ \ \ \ \left(x\in\mathbb{R}\right).(For a complex-valued function F=F_0+iF_1 whose real and imaginary parts F_0 and F_1 are integrable, we define \int F=\int F_0+i\int F_1.) Show that if f and g belong to L\left(\mathbb{R}\right), then\widehat{f \ast g}(x)=2\pi \widehat{f}(x)\widehat{g}(x).
=== Solution. ===
\begin{align} 
\widehat{f\ast g}(x)
&=\dfrac{1}{2\pi}\int_{\mathbb{R}}\left[\int_{\mathbb{R}} f\left(u-t\right)g(t) \ dt\right]e^{-ixu} \ du\\
&=\dfrac{1}{2\pi}\int_{\mathbb{R}} g(t)e^{-ixt} \left[\int_{\mathbb{R}}f(u-t)e^{-ix(u-t)} \ du \right]\ dt\\
\end{align} Denote the change of variable v=u-t , we have\begin{align}
\widehat{f\ast g}(x)
&=\dfrac{1}{2\pi}\int_{\mathbb{R}} g(t)e^{-ixt} \left[\int_{\mathbb{R}}f(v)e^{-ixv} \ dv \right]\ dt\\
&=\dfrac{1}{2\pi}\int_{\mathbb{R}} g(t)e^{-ixt} \ dt \cdot \int_{\mathbb{R}}f(v)e^{-ixv} \ dv\\
&=2\pi\cdot \left[\dfrac{1}{2\pi}\int_{\mathbb{R}} g(t)e^{-ixt} \ dt\right] \cdot \left[\dfrac{1}{2\pi}\int_{\mathbb{R}}f(v)e^{-ixv} \ dv \right]\\
&=2\pi\widehat{f}(x)\ast \widehat{g}(x),
\end{align} where the second line holds using Fubini's theorem, as desired.\blacksquare