Due March 9
Let f be integrable, and for each let . Then
Problem 2 (Riemann-Lebesgue Lemma)
For , let
be its Fourier transform. Then as .
- The convolution is well defined for a.e. x.
There does not exist such that, for all ,
Let in measure.
- If , then
- If there exists such that for all n, then in and .