sinc is not L1

The main results used in the proof will be that fL1(A)|f|L1(A) and the dominated convergence theorem.

Let f(x)=|sinc(x)| and suppose it’s Lebesgue integrableMathworldPlanetmath in +.

Consider the intervals Ik=[kπ,(k+1)π] and Uk=i=0kIk=[0,(k+1)π].

and the succession of functions fn(x)=f(x)χUn(x), where χUn is the characteristic functionMathworldPlanetmathPlanetmathPlanetmath of the set Un.

Each fn is a continuous functionMathworldPlanetmathPlanetmath of compact support and will thus be integrable in +. Furthermore fn(x)f(x) (pointwise)

in each Ik, f(x)|sin(x)|(k+1)π.



Suppose f is integrable in +. Then by the dominated convergence theorem limn+fn=+f.

But limn+fnlimnk=0n2(k+1)π=+ and we get the contradictionMathworldPlanetmathPlanetmath +f+.

So f cannot be integrable in +. This implies that f cannot be integrable in and since a function is integrable in a set iff its absolute valueMathworldPlanetmathPlanetmath is


Title sinc is not L1
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