sinc is not L1
The main results used in the proof will be that f∈L1(A)⇔|f|∈L1(A) and the dominated convergence theorem.
Let f(x)=|sinc(x)| and suppose it’s Lebesgue integrable in ℝ+.
Consider the intervals Ik=[kπ,(k+1)π] and Uk=⋃ki=0Ik=[0,(k+1)π].
and the succession of functions fn(x)=f(x)χUn(x), where χUn is the characteristic function of the set Un.
Each fn is a continuous function of compact support and will thus be integrable in ℝ+. Furthermore fn(x)↗f(x) (pointwise)
in each Ik, f(x)≥|sin(x)|(k+1)π.
So
∫ℝ+fn=n∑k=0∫(k+1)πkπ|sin(x)|x𝑑x≥n∑k=0∫(k+1)πkπ|sin(x)|(k+1)π=n∑k=02(k+1)π.
Suppose f is integrable in ℝ+. Then by the dominated convergence theorem lim.
But and we get the contradiction .
So cannot be integrable in .
This implies that cannot be integrable in and since a function is integrable in a set iff its absolute value is
Title | sinc is not |
---|---|
Canonical name | SincIsNotL1 |
Date of creation | 2013-03-22 15:44:32 |
Last modified on | 2013-03-22 15:44:32 |
Owner | cvalente (11260) |
Last modified by | cvalente (11260) |
Numerical id | 14 |
Author | cvalente (11260) |
Entry type | Result |
Classification | msc 26A06 |