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Inequality: Integration of Norm

  1. Jul 31, 2013 #1
    I am struggling with this question. I need a different perspective. Any recommendation is appreciated.
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  3. Aug 1, 2013 #2

    pasmith

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    Homework Helper

    What exactly is the question? I see
    [tex]
    \|\dot f(t)\|^2 \leq \int_{t-\tau}^t \|\dot f(\theta)\|^2\,\mathrm{d}\theta
    [/tex]
    with [itex]\tau \neq 0[/itex].

    The inequality does not hold for all [itex]\tau > 0[/itex] unless [itex]\|\dot f(t)\| = 0[/itex], since the right hand side can be made arbitrarily small by taking [itex]\tau > 0[/itex] sufficiently small.

    The inequality does not hold for any [itex]\tau < 0[/itex] unless [itex]\|\dot f(\theta)\|[/itex] vanishes identically on [itex](t,t+|\tau|)[/itex] and [itex]\|\dot f(t)\| = 0[/itex], since otherwise the right hand side is non-positive ([itex]\int_{t+|\tau|}^t = -\int_{t}^{t + |\tau|}[/itex]) and the left hand side is non-negative.
     
  4. Aug 6, 2013 #3
    Follow up question

    Thanks pasmith,

    Yeah, by using definition of Riemann integration rule I already proved that the inequality is wrong. I tried to find a domain in which the inequality holds; however, there is no such domain.

    How about we multiply only left-hand side with τ (tau)? Will this inequality be hold in some domain? What do you think?
     
  5. Aug 6, 2013 #4
    what if 0<tau<1 ?
     
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