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Math Help - Nonnegativity of y implies the same for x

  1. #1
    Senior Member bkarpuz's Avatar
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    Red face [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.
    This is a contradiction.
    But my method fails to deliver the case \mu=0.
    I need help at this point.
    Your help is highly appreciated!
    Last edited by bkarpuz; November 4th 2009 at 03:05 AM.
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  2. #2
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by bkarpuz View Post
    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.
    This is a contradiction.
    But my method fails to deliver the case \mu=0.
    I need help at this point.
    Your help is highly appreciated!
    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!
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  3. #3
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    Opalg's Avatar
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    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 \\<br />
&= \int_{0}^{t}\left(\int_v^tK(t,u)K(u,v)\,du\right)x  (v)\,dv <br />
= \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 \\<br />
&= \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.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Red face

    Quote Originally Posted by Opalg View Post
    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.
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  5. #5
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by bkarpuz View Post
    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.
    This is a contradiction.
    But my method fails to deliver the case \mu=0.
    I need help at this point.
    Your help is highly appreciated!
    I got it now!

    Proof. We proceed 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 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}
    Last edited by bkarpuz; November 7th 2009 at 08:48 PM.
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  6. #6
    Senior Member bkarpuz's Avatar
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    Talking

    Quote Originally Posted by bkarpuz View Post
    I got it now!

    Proof. We proceed 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\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).
    Last edited by bkarpuz; November 4th 2009 at 06:55 AM.
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