sinc is L2

Our objective will be to prove the integral f2(x)𝑑x exists in the Lebesgue sense when f(x)=sinc(x).

The integrand is an even function and so we can restrict our proof to the set +.

Since f is a continuous functionMathworldPlanetmathPlanetmath, so will f2 be and thus for every a>0, fL2([0,a]).

Thus, if we prove fL2([π,[), the result will be proved.

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

and the succession of functions fn(x)=f2(x)χUn(x), where χUn is the characteristic functionMathworldPlanetmathPlanetmathPlanetmath of the set Un.

Each fn is a continuous function of compact support and will thus be integrable in +. Furthermore fn(x)f2(x) (pointwise) in this set.

In each Ik,0f2(x)sin2(x)(kπ)2, for k>0.


xπfn(x)𝑑x=k=1nkπ(k+1)πsin(x)2x2𝑑xk=1nkπ(k+1)πsin(x)2(kπ)2=k=1n12k2π 11we have used the well known result 0πsin2(x)𝑑x=π2

So: limnxπfn(x)𝑑xlimnk=1n12k2π and since the series on the right side converges22asymptotic behaviour as k-2 and fnf2 we can use the monotone convergence theoremMathworldPlanetmath to state that f2L([π,[).

So we get the result that sincL2()

Title sinc is L2
Canonical name SincIsL2
Date of creation 2013-03-22 15:44:44
Last modified on 2013-03-22 15:44:44
Owner cvalente (11260)
Last modified by cvalente (11260)
Numerical id 9
Author cvalente (11260)
Entry type Result
Classification msc 26A06