ksoileau
Posts:
83
From:
Houston, TX
Registered:
3/9/08
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Re: Estimate the integral of the product of two functions
Posted:
Aug 16, 2012 7:44 PM
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On Thursday, August 16, 2012 2:34:29 PM UTC-5, 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:
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> |g(x)|<=M
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> 0<=w(x)<=m.
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> There are two additional conditions:
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> 1. int[a,b] w(x) dx = 1
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> 2. int[a,b] g(x) dx = 0
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>
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> Question: | int[a,b] g(x) w(x) dx | <= ?
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> =========================================
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> Using only the first condition, we can write:
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> | int[a,b] g(x) w(x) dx | <= |g(x)|*| int[a,b] w(x) dx | <= M.
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> 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,1-1/2/M].
Take g(x)=1/(2t) over [0,t) and -1/(2(1-t)) over [t,1]. Then |g(x)|<=M and int[a,b] g(x) dx = 0.
Next, take w(x)=m-(m-1)/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-(m-1)/t) dx +int[t,1] -1/(2(1-t))m dx |
=| t/(2t)(m-(m-1)/t) +(1-t)(-1/(2(1-t))m) |
=|(1-m)/2/t|=(m-1)/2/t<=(m-1)/2*2M=(m-1)*M
so the upper bound on the integral is at least (m-1)*M.
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