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Fall 2013 - Problem 10

Fourier analysis

  1. Let f∈L2(R)f \in L^2\p{\R} and define h(x)=∫Rf(x−y)f(y) dyh\p{x} = \int_\R f\p{x - y} f\p{y} \,\diff{y} for x∈Rx \in \R. Show that then there exists a function g∈L1(R)g \in L^1\p{\R} such that

    h(ξ)=∫Re−iξxg(x) dxh\p{\xi} = \int_\R e^{-i\xi x} g\p{x} \,\diff{x}

    for ξ∈R\xi \in \R, i.e., hh is the Fourier transform of a function in L1(R)L^1\p{\R}.

  2. Conversely, show that if g∈L1(R)g \in L^1\p{\R}, then there is a function f∈L2(R)f \in L^2\p{\R} such that the Fourier transform of gg is given by x↦h(x)=∫Rf(x−y)f(y) dyx \mapsto h\p{x} = \int_\R f\p{x - y} f\p{y} \,\diff{y}.
Solution.

  1. Set g=F−1(f)2g = \mathcal{F}^{-1}\p{f}^2, i.e., the square of the inverse Fourier transform of ff. Since f∈L2f \in L^2, it follows that F−1(f)∈L2\mathcal{F}^{-1}\p{f} \in L^2 as well, so g∈L1g \in L^1. Thus,

    F(g)=F(F−1(f)2)=F(F−1(f∗f))=f∗f,\mathcal{F}\p{g} = \mathcal{F}\p{\mathcal{F}^{-1}\p{f}^2} = \mathcal{F}\p{\mathcal{F}^{-1}\p{f * f}} = f * f,

    which was what we wanted to show.

  2. Set f=F(g)f = \mathcal{F}\p{\sqrt{g}}. Since g∈L1g \in L^1, it follows that F(g)∈L2\mathcal{F}\p{\sqrt{g}} \in L^2, so the Fourier transform is well-defined. Thus,

    f∗f=F(g)∗F(g)=F(g).f * f = \mathcal{F}\p{\sqrt{g}} * \mathcal{F}\p{\sqrt{g}} = \mathcal{F}\p{g}.