# Thread: Nonnegativity of y implies the same for x

1. ## [Solved] Nonnegativity of y implies the same for x

Dear friends,

Let me introduce you the problem killing me.

Assume that $x,y\in\mathrm{C}\big([0,\infty),(-\infty,\infty)\big)$ and consider the following equation
$x(t)=\int_{0}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$ for $t\geq0$, where $p\in\mathrm{C}\big([0,\infty),[0,\infty)\big)$.
Prove: $y\geq0$ implies $x\geq0$.
Note: We do not know whether y is differentiable or not.

The first method comes into my mind is proving by contradiction.
Suppose the contrary that $y\geq0$ but $x<0$ for some values in $[0,\infty)$.
Then we may pick $s\in[0,\infty)$ satisfying $x\geq0$ on $[0,s]$ and $x<0$ on $(s,s+\varepsilon)$ for some $\varepsilon>0$.
For simplicity, set $\mu:=\int_{0}^{s}\mathrm{e}^{-\textstyle\int_{\eta}^{s+\varepsilon}p(\zeta)\math rm{d}\zeta}x(\eta)\mathrm{d}\eta$ and $x_{\min}:=\min\nolimits_{\eta\in[s,s+\varepsilon]}\{x(\eta)\}$.
Then, for all $t\in(s,s+\varepsilon)$, we have
$0>x(t)=\int_{0}^{s}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta$ $+\int_{s}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$
............. $\geq\mu+x_{\min}\int_{s}^{t}\mathrm{e}^{-\textstyle\int_{s}^{t}p(\zeta)\mathrm{d}\zeta}\mat hrm{d}\eta$
............. $\geq\mu+x_{\min}\int_{s}^{t}\mathrm{d}\eta$ $=\mu+(t-s)x_{\min}$.
If $\mu>0$, then letting $\delta>0$ with $\delta\leq\min\{-\mu/x_{\min},\varepsilon/2\}$, we get $s+\delta\in(s,s+\varepsilon)$ and thus $0>x(s+\delta)\geq\mu+\delta x_{\min}\geq0$.
But my method fails to deliver the case $\mu=0$.
I need help at this point.

2. Originally Posted by bkarpuz
Dear friends,

Let me introduce you the problem killing me.

Assume that $x,y\in\mathrm{C}\big([0,\infty),(-\infty,\infty)\big)$ and consider the following equation
$x(t)=\int_{0}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$ for $t\geq0$, where $p\in\mathrm{C}\big([0,\infty),[0,\infty)\big)$.
Prove: $y\geq0$ implies $x\geq0$.
Note: We do not know whether y is differentiable or not.

The first method comes into my mind is proving by contradiction.
Suppose the contrary that $y\geq0$ but $x<0$ for some values in $[0,\infty)$.
Then we may pick $s\in[0,\infty)$ satisfying $x\geq0$ on $[0,s]$ and $x<0$ on $(s,s+\varepsilon)$ for some $\varepsilon>0$.
For simplicity, set $\mu:=\int_{0}^{s}\mathrm{e}^{-\textstyle\int_{\eta}^{s+\varepsilon}p(\zeta)\math rm{d}\zeta}x(\eta)\mathrm{d}\eta$ and $x_{\min}:=\min\nolimits_{\eta\in[s,s+\varepsilon]}\{x(\eta)\}$.
Then, for all $t\in(s,s+\varepsilon)$, we have
$0>x(t)=\int_{0}^{s}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta$ $+\int_{s}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$
............. $\geq\mu+x_{\min}\int_{s}^{t}\mathrm{e}^{-\textstyle\int_{s}^{t}p(\zeta)\mathrm{d}\zeta}\mat hrm{d}\eta$
............. $\geq\mu+x_{\min}\int_{s}^{t}\mathrm{d}\eta$ $=\mu+(t-s)x_{\min}$.
If $\mu>0$, then letting $\delta>0$ with $\delta\leq\min\{-\mu/x_{\min},\varepsilon/2\}$, we get $s+\delta\in(s,s+\varepsilon)$ and thus $0>x(s+\delta)\geq\mu+\delta x_{\min}\geq0$.
But my method fails to deliver the case $\mu=0$.
I need help at this point.
When the domain satisfies $[0,\lambda]\subset[0,1)$, then a solution comes from Operator theory as follows.
Set
$k(t,s):=\mathrm{e}^{-\textstyle\int_{s}^{t}p(\zeta)\mathrm{d}\zeta}$ for $(t,s)\in[\lambda,s]\times[0,\lambda]$
and
$(Kx)(t):=\int_{0}^{t}k(t,\eta)x(\eta)\mathrm{d}\et a$ for $t\in[0,\lambda]$.
It is not hard to see that $1\geq k(t,s)\geq0$ for all $(t,s)\in[\lambda,s]\times[0,\lambda]$.
And thus $\|Kx\|\leq\lambda\|x\|<\|x\|$, which implies that $\|K\|<1$ and that $I-K$ has the inverse operator $\sum_{i=0}^{\infty}K^{n}$.
Rewritting the given equation, we have
$(I-K)x=y$, which gives $x=(I-K)^{-1}y=\sum_{i=0}^{\infty}K^{n}y$.
On the other hand, $y\geq0$ implies $Ky\geq0$, and therefore $x\geq0$.
However, this method does not help when the domain is a superset of the interval $[0,1]$.
Help!

3. Define $(Hx)(t):=\int_{0}^{t}K(t,u)x(u)\,du$ for $T\geqslant t\geqslant0$, where $K(t,u):=\textstyle\exp\left(-\int_{u}^{t}p(\zeta)\,d\zeta\right)$ for $T\geqslant t\geqslant u\geqslant0$. Then H is an integral operator with kernel K. For this operator, it is possible to determine explicitly its powers $H^n$, which are also integral operators, in terms of their kernels.

In fact (changing the order of integration in the repeated integral),

\begin{aligned}H^2x(t) &= \int_{0}^{t}K(t,u)Hx(u)\,du \\ &= \int_{0}^{t}\int_0^uK(t,u)K(u,v)x(v)\,dvdu \\
&= \int_{0}^{t}\left(\int_v^tK(t,u)K(u,v)\,du\right)x (v)\,dv
= \int_0^tK_2(t,v)x(v)\,dv,\end{aligned}

where

\begin{aligned}K_2(t,v) &:= \int_v^tK(t,u)K(u,v)\,du \\ &= \int_v^t \exp\left(-\int_{u}^{t}p(\zeta)\,d\zeta\right) \exp\left(-\int_{v}^{u}p(\zeta)\,d\zeta\right)du \\
&= \int_v^tK(t,v)\,du = (t-v)K(t,v).\end{aligned}

Thus $H^2$ is the integral operator with kernel $K_2(t,v) = (t-v)K(t,v)$. In the same way, you can prove by induction that $H^n$ is the integral operator with kernel $K_n(t,u) = \frac{(t-u)^{n-1}}{(n-1)!}K(t,u)$.

The next thing you need to know is that the operator norm of an integral operator is less than or equal to its Hilbert–Schmidt norm, the square of which is the integral of the square of the kernel. In the case of the operator $H^n$, the Hilbert–Schmidt norm is given by

$\|H^n\|_{\textsc{hs}}^2 = \int_0^T\int_0^T|K_n(t,u)|^2dtdu = \int_0^T\int_0^T \frac{(t-u)^{2n-2}}{((n-1)!)^2}|K(t,u)|^2dtdu.$

I assume that the function $p(\zeta)$ is positive, so that $|K(t,u)|\leqslant1$. Then

$\|H^n\|_{\textsc{hs}}^2 \leqslant \int_0^T\int_0^T \frac{(t-u)^{2n-2}}{((n-1)!)^2}\,dtdu = \frac{T^{2n}}{2n(2n-1)((n-1)!)^2} \to0$ as $n\to\infty$.

Therefore the operator norm of $H^n$ also goes to 0 as $n\to\infty$. The argument that you had previously should then show that $x = H^{n+1}x + \sum_{i=0}^nH^iy \to \sum_{i=0}^\infty H^iy$, so that x is positive if y is positive.

4. Originally Posted by Opalg
Define $(Hx)(t):=\int_{0}^{t}K(t,u)x(u)\,du$ for $T\geqslant t\geqslant0$, where $K(t,u):=\textstyle\exp\left(-\int_{u}^{t}p(\zeta)\,d\zeta\right)$ for $T\geqslant t\geqslant u\geqslant0$. Then H is an integral operator with kernel K. For this operator, it is possible to determine explicitly its powers $H^n$, which are also integral operators, in terms of their kernels.

[snip]

I assume that the function $p(\zeta)$ is positive, so that $|K(t,u)|\leqslant1$. Then

$\|H^n\|_{\textsc{hs}}^2 \leqslant \int_0^T\int_0^T \frac{(t-u)^{2n-2}}{((n-1)!)^2}\,dtdu = \frac{T^{2n}}{2n(2n-1)((n-1)!)^2} \to0$ as $n\to\infty$.

Therefore the operator norm of $H^n$ also goes to 0 as $n\to\infty$. The argument that you had previously should then show that $x = H^{n+1}x + \sum_{i=0}^nH^iy \to \sum_{i=0}^\infty H^iy$, so that x is positive if y is positive.
Dear Professor Opalg, thanks alot!

I wonder to learn that if the problem becomes much more harder by changing the equation with one of the following ones?
$x(t)=\int_{{\color{red}\sqrt{t}}}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$ for $T\geq t\geq0$
or
$x(t)=\int_{{\color{red}\sqrt{t}}}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{{\color{red}\sqrt{t}}}p(\ze ta)\mathrm{d}\zeta}x(\eta)\mathrm{d}\eta+y(t)$ for $T\geq t\geq0$

By the way, which book should I refer for "the operator norm of an integral operator is less than or equal to its Hilbert–Schmidt norm".
It seems I need to learn about (Volterra) integral operators and their properties (compactness, spectral radius) for coping with the problems I encounter.

Many many thanks for your guidence.

5. Originally Posted by bkarpuz
Dear friends,

Let me introduce you the problem killing me.

Assume that $x,y\in\mathrm{C}\big([0,\infty),(-\infty,\infty)\big)$ and consider the following equation
$x(t)=\int_{0}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$ for $t\geq0$, where $p\in\mathrm{C}\big([0,\infty),[0,\infty)\big)$.
Prove: $y\geq0$ implies $x\geq0$.
Note: We do not know whether y is differentiable or not.

The first method comes into my mind is proving by contradiction.
Suppose the contrary that $y\geq0$ but $x<0$ for some values in $[0,\infty)$.
Then we may pick $s\in[0,\infty)$ satisfying $x\geq0$ on $[0,s]$ and $x<0$ on $(s,s+\varepsilon)$ for some $\varepsilon>0$.
For simplicity, set $\mu:=\int_{0}^{s}\mathrm{e}^{-\textstyle\int_{\eta}^{s+\varepsilon}p(\zeta)\math rm{d}\zeta}x(\eta)\mathrm{d}\eta$ and $x_{\min}:=\min\nolimits_{\eta\in[s,s+\varepsilon]}\{x(\eta)\}$.
Then, for all $t\in(s,s+\varepsilon)$, we have
$0>x(t)=\int_{0}^{s}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta$ $+\int_{s}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$
............. $\geq\mu+x_{\min}\int_{s}^{t}\mathrm{e}^{-\textstyle\int_{s}^{t}p(\zeta)\mathrm{d}\zeta}\mat hrm{d}\eta$
............. $\geq\mu+x_{\min}\int_{s}^{t}\mathrm{d}\eta$ $=\mu+(t-s)x_{\min}$.
If $\mu>0$, then letting $\delta>0$ with $\delta\leq\min\{-\mu/x_{\min},\varepsilon/2\}$, we get $s+\delta\in(s,s+\varepsilon)$ and thus $0>x(s+\delta)\geq\mu+\delta x_{\min}\geq0$.
But my method fails to deliver the case $\mu=0$.
I need help at this point.
I got it now!

Suppose the contrary that $y\geq0$ but $x<0$ for some values in $[0,\infty)$.
Then, we may pick $s\in[0,\infty)$ satisfying $x\geq0$ on $[0,s]$ and $x<0$ for some values in $(s,s+\varepsilon)$ for all $\varepsilon\in(0,1/3)$.
Noting that the nonnegativity of $p$ implies that the exponential function can not exceed $1$.
Then, for all $t\in(s,s+\varepsilon)$, we have
$x(t)=\int_{0}^{s}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta$ $+\int_{s}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$
....... $\geq\int_{s}^{t}x(\eta)\mathrm{d}\eta$
....... $\geq(t-s)\min_{\eta\in[s,s+\varepsilon]}x(\eta)$. (1)
Clearly, because of the continuity of $x$, we may find $r\in(s,s+\varepsilon)$ such that
$x(r)\leq2\min_{\eta\in[s,s+\varepsilon]}x(\eta)$ and $x(r)<0$. (2)
Then using (1) and (2), we get
$x(r)=(r-s)\min_{\eta\in[s,s+\varepsilon]}x(\eta)$
....... $\geq\frac{2}{3}x(r)$,
which implies $1\leq2/3$ by canceling the negative terms $x(r)$ on both sides.
This contradiction completes the proof. $\rule{0.2cm}{0.2cm}$

6. Originally Posted by bkarpuz
I got it now!

Suppose the contrary that $y\geq0$ but $x<0$ for some values in $[0,\infty)$.
Then, we may pick $s\in[0,\infty)$ satisfying $x\geq0$ on $[0,s]$ and $x<0$ on $(s,s+\varepsilon]$ for some $\varepsilon\in(0,1/2)$.
Noting that the nonnegativity of $p$ implies that the exponential function can not exceed $1$.
Then, for all $t\in[s,s+\varepsilon]$, we have
$x(t)=\int_{0}^{s}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta$ $+\int_{s}^{t}\mathrm{e}^{-\textstyle\int_{\eta}^{t}p(\zeta)\mathrm{d}\zeta}x (\eta)\mathrm{d}\eta+y(t)$
....... $\geq\int_{s}^{t}x(\eta)\mathrm{d}\eta$
....... $\geq(t-s)\min_{\eta\in[s,s+\varepsilon]}x(\eta)$. (1)
Clearly, because of the continuity of $x$, we may find $r\in(s,s+\varepsilon]$ such that
$x(r)=\min_{\eta\in[s,s+\varepsilon]}x(\eta)$. (2)
Then using (1) and (2), we get
$x(r)=(r-s)\min_{\eta\in[s,s+\varepsilon]}x(\eta)$
....... $\geq\frac{1}{2}x(r)$,
which implies $1\leq1/2$ by canceling the negative terms $x(r)$ on both sides.
This contradiction completes the proof. $\rule{0.2cm}{0.2cm}$
I would like to mention that the proof above can be easily extended to the following equation:
$x(t)=\int_{\tau_{+}(t)}^{t}K(t,\eta)x(\eta)\mathrm {d}\eta+y(t)$ for $t\in[0,\infty)$,
where $K:\mathbb{D}:=\{(t,s)\in\mathbb{R}_{0}^{2}:t\geq0\ \text{and}\ t\geq s\geq\tau_{+}(t)\}\}\to\mathbb{R}_{0}^{+}$ is bounded,
$\tau:[0,\infty)\to\mathbb{R}$ satisfies $\tau(t)\leq t$ for all $t\in[0,\infty)$ and $\tau_{+}(t):=\max\{\tau(t),0\}$ for $t\in[0,\infty)$.