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

    Thanks for your interest.
    bkarpuz
    Last edited by bkarpuz; Dec 18th 2011 at 08:41 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 $\displaystyle \Omega$ is a wedge below $\displaystyle y=x$ in the first quadrant.
    The function $\displaystyle Y(t,s):=\int_{s}^{t}X(t,u)\mathrm{d}u$ integrates the function $\displaystyle X$ along the line $\displaystyle (t,s)\rightarrow (t,t)$.
    It seems to me that the integrand $\displaystyle X$ could take negative values as long as it overshoots fast enough close to the region $\displaystyle y=x$ since you always have to integrate that part.
    I think you can "build" such a function slice by slice ($\displaystyle t$ by $\displaystyle t$). You could take some function that starts at zero, assume negative values until $\displaystyle y=x/2$ where it goes back to zero. Then it assumes large positive values from $\displaystyle y=x/2$ to $\displaystyle y=x$. Find some way to make that $\displaystyle t$-dependent, and you should have a counter-example. You could take for example a $\displaystyle \sin$ which period (and amplitude) varies with $\displaystyle t$ for the lower part and a sharp $\displaystyle \tanh$ 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 $\displaystyle \Omega$ is a wedge below $\displaystyle y=x$ in the first quadrant.
    The function $\displaystyle Y(t,s):=\int_{s}^{t}X(t,u)\mathrm{d}u$ integrates the function $\displaystyle X$ along the line $\displaystyle (t,s)\rightarrow (t,t)$.
    It seems to me that the integrand $\displaystyle X$ could take negative values as long as it overshoots fast enough close to the region $\displaystyle y=x$ since you always have to integrate that part.
    I think you can "build" such a function slice by slice ($\displaystyle t$ by $\displaystyle t$). You could take some function that starts at zero, assume negative values until $\displaystyle y=x/2$ where it goes back to zero. Then it assumes large positive values from $\displaystyle y=x/2$ to $\displaystyle y=x$. Find some way to make that $\displaystyle t$-dependent, and you should have a counter-example. You could take for example a $\displaystyle \sin$ which period (and amplitude) varies with $\displaystyle t$ for the lower part and a sharp $\displaystyle \tanh$ for the upper part.
    Thank you for the comments vincisonfire, a counter example is $\displaystyle X(t,s)=-\sin \big(2 \pi s/t \big)$ for $\displaystyle (t,s)\in\Omega$.
    What if we have the property that $\displaystyle X$ takes $\displaystyle 1$ along the bisector $\displaystyle t=s$?
    By the way, the result is true if $\displaystyle X(t,s)=X(s)$ for all $\displaystyle (t,s)\in\Omega$.
    Last edited by bkarpuz; Dec 18th 2011 at 09:10 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 $\displaystyle X$ takes $\displaystyle 1$ along the bisector $\displaystyle t=s$?
    By the way, the result is true if $\displaystyle X(t,s)=X(s)$ for all $\displaystyle (t,s)\in\Omega$.
    I would think the result is true if $\displaystyle X(t,u) = X(u)$. 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 $\displaystyle t$ since the domain of the line integral becomes infinitely small.
    I think the construction still holds if you impose $\displaystyle X(t,t) = 1$.
    It seems the $\displaystyle t$-dependence is crucial here. In fact, if you are given the $\displaystyle t$-dependence of $\displaystyle X$, you can make everything positive around the origin, and then do the dive-and-overshoot business once you have enough space (at larger $\displaystyle t$).
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