# Thread: Prove or disprove nonnegativity of the integrand

1. ## 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 $\Omega$.

bkarpuz

2. ## 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.

3. ## Re: Prove or disprove nonnegativity of the integrand

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

4. ## Re: Prove or disprove nonnegativity of the integrand

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 $t$).