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Innner product for which derivative operator is bounded

  1. Nov 13, 2009 #1
    Is the derivative operator d/dx bounded with respect to the norm <f,g> defined by

    integral from 0 to 1 of f g* +f'g*'

    where * denotes conjugation. Thank you. (Not homework)
     
  2. jcsd
  3. Nov 13, 2009 #2
    Note that [tex]\frac{d}{dx} : C^1([0,1]) \rightarrow C^0([0,1])[/tex] you defined the inner product (not the norm) on [tex]C^1[/tex], but what is that of [tex]C^0[/tex]? I'm going to assume you give it the inner product: [tex]<f,g> = \int_{0} ^{1} f\overline{g}[/tex]. Then the norms induced by these are [tex]\Vert f \Vert _{C^1} = \left( \int_{0}^{1} \vert f \vert ^2 + \int_{0}^{1} \vert f' \vert ^2 \right) ^{1/2}[/tex] and [tex]\Vert g \Vert _{C^0} = \left( \int_{0}^{1} \vert g \vert ^2 \right) ^{1/2}[/tex]

    Then [tex]\Vert f' \Vert _{C^0} ^2 = \int_{0}^{1} \vert f' \vert ^2 \leq \int_{0}^{1} \vert f \vert ^2 + \int_{0}^{1} \vert f' \vert ^2 = \Vert f \Vert _{C^1} ^2[/tex] so it's bounded with respect to those norms.
     
  4. Nov 14, 2009 #3
    Jose,
    Thank you very much.
    Regards
     
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