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Regarding Schwartz inequality and integration bounds

  1. Feb 2, 2016 #1
    Based on Schwartz inequality, I am trying to figure out why there
    should/can be the "s" variable which is the lower bound of the
    integration in the RHS of the following inequality:
    ## \left \|\int_{-s}^{0} A(t+r)Z(t+r) dr \right \|^{2} \leq s\int_{-s}^{0}\left \| A(t+r)Z(t+r) \right \|^{2} dr, ##
    instead of:
    ## \left \|\int_{-s}^{0} A(t+r)Z(t+r) dr \right \|^{2} \leq \int_{-s}^{0}\left \| A(t+r)Z(t+r) \right \|^{2} dr. ##
     
    Last edited: Feb 2, 2016
  2. jcsd
  3. Feb 2, 2016 #2

    Krylov

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  4. Feb 2, 2016 #3
  5. Feb 2, 2016 #4

    Samy_A

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    Apply the Cauchy-Schwarz inequality to the functions##f,\ g## defined by ##f(r)=A(t+r)Z(t+r)## and ##g(r)=1##.
     
  6. Feb 2, 2016 #5
    Sure. My confusion was because of the example in my mind that s can be a very small value which then the upper bound will be also very small and as a result, physically conservatism but again mathematically correct. Thank you very much.
     
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