Questions: Fourier Analysis for PDEs and Sobolev Embedding
4 questions to test your understanding
Score: 0 / 4
Question 1 Multiple Choice
The Sobolev space H^s(ℝⁿ) is characterized via the Fourier transform as:
Au ∈ H^s iff ∫(1+|ξ|²)^s |û(ξ)|² dξ < ∞
Bu ∈ H^s iff |û(ξ)| ≤ C|ξ|^{-s}
Cu ∈ H^s iff û has compact support
Du ∈ H^s iff û ∈ L^s
The H^s norm is ||u||²_{H^s} = ∫(1+|ξ|²)^s|û(ξ)|²dξ. The weight (1+|ξ|²)^s penalizes high frequencies when s > 0 (requiring smoothness) and allows high frequencies when s < 0 (permitting distributions). This is equivalent to the weak derivative definition when s is a non-negative integer.
Question 2 True / False
Fractional Sobolev spaces H^s for non-integer s are naturally defined using the Fourier transform.
TTrue
FFalse
Answer: True
While integer-order Sobolev spaces can be defined using weak derivatives, the Fourier characterization immediately extends to any real s, including negative values. H^{-s} is the dual of H^s₀, and H^{1/2} appears naturally as the trace space for H¹.
Question 3 Short Answer
What does the Sobolev embedding theorem say in terms of Fourier decay?
Think about your answer, then reveal below.
Model answer: If s > n/2, then the Fourier integral ∫(1+|ξ|²)^{-s}dξ converges, so H^s(ℝⁿ) embeds into continuous functions
When s > n/2, functions in H^s have Fourier transforms in L¹ (by Cauchy-Schwarz: ∫|û|dξ ≤ ||u||_{H^s}(∫(1+|ξ|²)^{-s}dξ)^{1/2} < ∞), so u = ∫ûe^{iξ·x}dξ is continuous and bounded. This is the critical condition for pointwise regularity.
Question 4 True / False
A Fourier multiplier operator T_m defined by (T_m u)^∧(ξ) = m(ξ)û(ξ) is bounded on L² for any bounded m.
TTrue
FFalse
Answer: True
By Parseval's theorem, ||T_m u||_{L²} = ||m û||_{L²} ≤ ||m||_{L^∞}||û||_{L²} = ||m||_{L^∞}||u||_{L²}. L² boundedness of Fourier multipliers is automatic. Boundedness on L^p for p ≠ 2 is much deeper and is the subject of the Hormander-Mikhlin multiplier theorem.