T∈S(given)
To prove-for all ψ∈S,F(T∗ψ)=(2π)n/2.F(ψ).F(T)
F(T(t))=fourier transform of T(t)
(T∗ψ)(t)=∫−∞∞T(x)ψ(t−x)dx
F(T(t))=(2π)−1/2(∫−∞∞T(t)exp(−iwt)dt
F((T∗ψ)(t))=(2π)−1/2(∫−∞∞(T∗ψ)(t)exp(−iwt)dt
F((T∗ψ)(t))=(2π)−1/2(∫−∞∞(∫−∞∞T(x)ψ(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∗ψ)(t))=(∫−∞∞T(x)(2π)−1/2(∫−∞∞ψ(t−x)exp(−iwt)dt)dx
(2π)−1/2(∫−∞∞ψ(t−x)exp(−iwt)dt)=F(ψ(t−x))
F((T∗ψ)(t))=(∫−∞∞T(x)F(ψ(t−x))dx ----(i)
By properties of Fourier Transform,
F(ψ(t−x))=exp(−iwt)F(ψ(x))
Using this in equation (i)
F((T∗ψ)(t))=∫−∞∞T(x)exp(−iwt)F(ψ(t)dx
F((T∗ψ)(t))=(2π)1/2F(ψ(t).(2π)−1/2.∫−∞∞T(x)exp(−iwt)dx
F((T∗ψ)(t))=(2π)1/2F(ψ(t).F(T(t))
So,
F(T∗ψ)=(2π)n/2.F(ψ).F(T) where n=1
(Proved)
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