Consider Rn with the Lebesgue measure. Denote by Ff(ξ)=∫Rne−2iπx⋅ξf(x)dx the Fourier transform of f∈L1(Rn) and by f^ the Fourier-Plancherel transform of f∈L2(Rn). Let χR(ξ):=(1−R∣ξ∣)χ∣ξ∣⩽R for R>0 and define for s∈R+
Hs(Rn):={f∈L2(Rn)∣(1+∣⋅∣2)s/2f^(⋅)∈L2(Rn)}
(i) Prove that Hs(Rn) is a vector subspace of L2(Rn), and is a Hilbert space for the inner product ⟨f,g⟩:=∫Rn(1+∣ξ∣2)sf^(ξ)g^(ξ)dξ, where zˉ denotes the complex conjugate of z∈C.
(ii) Construct a function f∈Hs(R),s∈(0,1/2), that is not almost everywhere equal to a continuous function.
(iii) For f∈L1(Rn), prove that FR:x↦∫RnFf(ξ)χR(ξ)e2iπx⋅ξdξ is a well-defined function and that FR∈L1(Rn) converges to f in L1(Rn) as R→+∞.
[Hint: Prove that FR=KR∗f where KR is an approximation of the unit as R→+∞.]
(iv) Deduce that if f∈L1(Rn) and (1+∣⋅∣2)s/2Ff(⋅)∈L2(Rn) then f∈Hs(Rn).
[Hint: Prove that: (1) there is a sequence Rk→+∞ such that KRk∗f converges to f almost everywhere; (2) KR∗f is uniformly bounded in L2(Rn) as R→+∞.]