Prove or disprove nonnegativity of the integrand

**bkarpuz** · December 18th 2011, 01:34 AM

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

**vincisonfire** · December 18th 2011, 07:34 AM

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.

**bkarpuz** · December 18th 2011, 08:44 AM

Originally Posted by vincisonfire

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 $(t,s)\in\Omega$ .

**vincisonfire** · December 18th 2011, 09:03 AM

Originally Posted by bkarpuz

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 $t$ ).

Thread: Prove or disprove nonnegativity of the integrand

LinkBack

Thread Tools

Display

Prove or disprove nonnegativity of the integrand

Re: Prove or disprove nonnegativity of the integrand

Re: Prove or disprove nonnegativity of the integrand

Re: Prove or disprove nonnegativity of the integrand

Similar Math Help Forum Discussions

prove or disprove

Prove or disprove that (ac,b) =1

How do I prove/disprove this?

Prove or Disprove. Please help.

Prove or Disprove

Search Tags