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Math Help - Prove or disprove nonnegativity of the integrand

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation Prove or disprove nonnegativity of the integrand

    Dear MHF members,

    I have the following problem.
    Problem. Let Y(t,s):=\int_{s}^{t}X(t,u)\mathrm{d}u for (t,s)\in\Omega:=\{(t,s):\ t\geq s\geq 0\},
    where X\in\mathrm{C}(\Omega,\mathbb{R}) with X(t,t)\equiv1 for all t\geq0.
    Prove or disprove that Y>0 in \Omega^{\circ} (inside of \Omega) implies X\geq0 in \Omega.

    Thanks for your interest.
    bkarpuz
    Last edited by bkarpuz; December 18th 2011 at 09:41 AM. Reason: Updated the problem due to the comments by vincisonfire
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    Senior Member vincisonfire's Avatar
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    Re: Prove or disprove nonnegativity of the integrand

    Hi, I just gonna give some thoughts.
    The region \Omega is a wedge below y=x in the first quadrant.
    The function Y(t,s):=\int_{s}^{t}X(t,u)\mathrm{d}u integrates the function X along the line (t,s)\rightarrow (t,t).
    It seems to me that the integrand X could take negative values as long as it overshoots fast enough close to the region y=x since you always have to integrate that part.
    I think you can "build" such a function slice by slice ( t by t). You could take some function that starts at zero, assume negative values until y=x/2 where it goes back to zero. Then it assumes large positive values from y=x/2 to y=x. Find some way to make that t-dependent, and you should have a counter-example. You could take for example a \sin which period (and amplitude) varies with t for the lower part and a sharp \tanh for the upper part.
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    Senior Member bkarpuz's Avatar
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    Re: Prove or disprove nonnegativity of the integrand

    Quote Originally Posted by vincisonfire View Post
    Hi, I just gonna give some thoughts.
    The region \Omega is a wedge below y=x in the first quadrant.
    The function Y(t,s):=\int_{s}^{t}X(t,u)\mathrm{d}u integrates the function X along the line (t,s)\rightarrow (t,t).
    It seems to me that the integrand X could take negative values as long as it overshoots fast enough close to the region y=x since you always have to integrate that part.
    I think you can "build" such a function slice by slice ( t by t). You could take some function that starts at zero, assume negative values until y=x/2 where it goes back to zero. Then it assumes large positive values from y=x/2 to y=x. Find some way to make that t-dependent, and you should have a counter-example. You could take for example a \sin which period (and amplitude) varies with t for the lower part and a sharp \tanh for the upper part.
    Thank you for the comments vincisonfire, a counter example is X(t,s)=-\sin \big(2 \pi s/t \big) for (t,s)\in\Omega.
    What if we have the property that X takes 1 along the bisector t=s?
    By the way, the result is true if X(t,s)=X(s) for all (t,s)\in\Omega.
    Last edited by bkarpuz; December 18th 2011 at 10:10 AM.
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    Senior Member vincisonfire's Avatar
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    Re: Prove or disprove nonnegativity of the integrand

    Quote Originally Posted by bkarpuz View Post
    What if we have the property that X takes 1 along the bisector t=s?
    By the way, the result is true if X(t,s)=X(s) for all (t,s)\in\Omega.
    I would think the result is true if X(t,u) = X(u). With the above construction, you see that we will always get into trouble close to the origin, i.e. we cannot find a function that will dive and overshoot fast enough for all t since the domain of the line integral becomes infinitely small.
    I think the construction still holds if you impose X(t,t) = 1.
    It seems the t-dependence is crucial here. In fact, if you are given the t-dependence of X, you can make everything positive around the origin, and then do the dive-and-overshoot business once you have enough space (at larger t).
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