ksoileau
Posts:
85
From:
Houston, TX
Registered:
3/9/08


Re: Estimate the integral of the product of two functions
Posted:
Aug 16, 2012 7:44 PM


On Thursday, August 16, 2012 2:34:29 PM UTC5, nagornyi wrote: > The functions g(x) and w(x) are defined on [a,b], the first one is bounded, the second one is positive bounded: > > g(x)<=M > > 0<=w(x)<=m. > > > > There are two additional conditions: > > 1. int[a,b] w(x) dx = 1 > > 2. int[a,b] g(x) dx = 0 > > > > Question:  int[a,b] g(x) w(x) dx  <= ? > > ========================================= > > > > Using only the first condition, we can write: > >  int[a,b] g(x) w(x) dx  <= g(x)* int[a,b] w(x) dx  <= M. > > Introduction of the second condition should (seems like) improve the estimate. But how?
We consider the case in which m>=1, a=0 and b=1. Fix t in the interval [1/2/M,11/2/M]. Take g(x)=1/(2t) over [0,t) and 1/(2(1t)) over [t,1]. Then g(x)<=M and int[a,b] g(x) dx = 0. Next, take w(x)=m(m1)/t over [0,t) and m over [t,1]. Then int[a,b] w(x) dx = 1. Finally  int[a,b] g(x) w(x) dx  = int[0,t] 1/(2t)(m(m1)/t) dx +int[t,1] 1/(2(1t))m dx  = t/(2t)(m(m1)/t) +(1t)(1/(2(1t))m)  =(1m)/2/t=(m1)/2/t<=(m1)/2*2M=(m1)*M so the upper bound on the integral is at least (m1)*M.

