Using the definitions of convolution and of Fourier transform show that, for \(f,g\in L^1(\mathbb R)\), \[ \widehat {(f*g)}(t)=\hat f(t)\hat g(t) \] for \(t\in \mathbb R\).
Question
Math 6210 • Real Analysis
Fall 1996 • Question 5
Question
Fall 1996 • Question 5
Using the definitions of convolution and of Fourier transform show that, for \(f,g\in L^1(\mathbb R)\), \[ \widehat {(f*g)}(t)=\hat f(t)\hat g(t) \] for \(t\in \mathbb R\).