Answer to Question #97529 in Functional Analysis for Rachel

Question

Let TES^1. Prove that for all φES, (T*φ)^ = (2π)^(n/2)(φ-hat)(T-hat).

Answer 1

$T\in S(given)$

To prove-for all $\psi\in S,F(T*\psi)=(2π)^{n/2}.F(\psi).F(T)$

$F(T(t))= fourier$ transform of T(t)

$(T*\psi)(t)=\int_{-\infty} ^{\infty}T(x)\psi(t-x)dx$

$F(T(t))=(2π)^{-1/2}(\int_{-\infty}^{\infty}T(t)exp(-iwt)dt$

$F((T*\psi)(t))=(2π)^{-1/2}(\int_{-\infty}^{\infty} (T*\psi)(t) exp(-iwt)dt$

$F((T*\psi)(t))=(2π)^{-1/2}(\int_{-\infty}^{\infty} (\int_{-\infty} ^{\infty}T(x)\psi(t-x)exp(-iwt)dx)dt$

We can change the order of the equation and separate T(x) as it will be a constant when integrated with respect to t.

$F((T*\psi)(t))=(\int_{-\infty}^{\infty} T(x) (2π)^{-1/2} (\int_{-\infty} ^{\infty}\psi(t-x)exp(-iwt)dt)dx$

$(2π)^{-1/2} (\int_{-\infty} ^{\infty}\psi(t-x)exp(-iwt)dt)=F(\psi(t-x))$

$F((T*\psi)(t))=(\int_{-\infty}^{\infty} T(x) F(\psi(t-x))dx$ ----(i)

By properties of Fourier Transform,

$F(\psi(t-x))=exp(-iwt)F(\psi(x))$

Using this in equation (i)

$F((T*\psi)(t))=\int_{-\infty}^{\infty} T(x) exp(-iwt)F(\psi(t)dx$

$F((T*\psi)(t))=(2π) ^{1/2}F(\psi(t).(2π)^{-1/2 }. \int_{-\infty}^{\infty} T(x) exp(-iwt)dx$

$F((T*\psi)(t))=(2π) ^{1/2}F(\psi(t).F(T(t))$

So,

$F(T*\psi)=(2π)^{n/2}.F(\psi).F(T)$ where $n=1$

(Proved)