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Fall 2018 - Problem 6

Fourier analysis

Let f∈L2(R)f \in L^2\p{\R} and assume the Fourier transform satisfies ∣f^(ξ)∣>0\abs{\hat{f}\p{\xi}} > 0 for Lebesgue almost every ξ∈R\xi \in \R. Prove that the set of finite linear combinations of the translates fy(x)=f(x−y)f_y\p{x} = f\p{x - y} is norm dense in L2(R)L^2\p{\R}.
Solution.

Suppose there exists g∈L2g \in L^2 such that

∫Rg(x)f(x−y)‾ dx=⟨g,fy⟩=0\int_\R g\p{x}\conj{f\p{x - y}} \,\diff{x} = \inner{g, f_y} = 0

for all y∈Ry \in \R. Let f~(x)=f(−x)‾\tilde{f}\p{x} = \conj{f\p{-x}}. Then

(f~∗g)(y)=∫Rf~(y−x)g(x) dx=∫Rg(x)f(x−y)‾ dx=0,\p{\tilde{f} * g}\p{y} = \int_\R \tilde{f}\p{y - x}g\p{x} \,\diff{x} = \int_\R g\p{x} \conj{f\p{x - y}} \,\diff{x} = 0,

by assumption. Thus h=f~∗g=0h = \tilde{f} * g = 0, so

h^(ξ)=f~^(ξ)g(ξ)=0,\hat{h}\p{\xi} = \hat{\tilde{f}}\p{\xi} g\p{\xi} = 0,

where

f~^(ξ)=∫Re−ixξf(−x)‾ dx=∫Re−ixξf(x) dx‾=f^(ξ)‾,\hat{\tilde{f}}\p{\xi} = \int_\R e^{-ix\xi} \conj{f\p{-x}} \,\diff{x} = \conj{\int_\R e^{-ix\xi} f\p{x} \,\diff{x}} = \conj{\hat{f}\p{\xi}},

so

f(ξ)^‾g(ξ)=0\conj{\hat{f\p{\xi}}} g\p{\xi} = 0

everywhere. Since f(ξ)≠0f\p{\xi} \neq 0 almost everywhere, it follows that g(ξ)=0g\p{\xi} = 0 almost everywhere. Thus, because the Fourier-Plancherel transform is an isometry on L2L^2, it follows that g=0g = 0 almost everywhere, so the translates are dense.