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Inner product

  1. Nov 14, 2005 #1

    quasar987

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    Hi guys, there's a problem to which me and my pals just can't seem to get an answer that is congruent the answer on the back of the book. The homwork is due in two days and I just want to make sure the book is truely wrong. The question's simple enough: given f,g continuous on the circle ([-pi,pi)), is

    [tex](f,g)=\int_0^{\pi} fg^*dt[/tex]

    an inner product on the space of continuous functions defined on the circle?

    We're saying yes because the 3 properties are verified:

    i) (af+bg,h) = a(f,h)+b(g,h), as if evident by the properties of the integral.

    ii) [tex](f,g)^* = \left( \int_0^{\pi} gf^*dt \right)^* = \int_0^{\pi} (gf^*)^*dt = \int_0^{\pi} fg^*dt = (g,f)[/tex]

    iii) [tex](f,f) = \int_0^{\pi} ff^*dt = \int_0^{\pi} |f|^2 dt \geq 0[/tex]

    (since |f| >= 0 and = 0 <==> f=0)


    Hence all 3 properties are verified. Any objection?
     
  2. jcsd
  3. Nov 15, 2005 #2

    benorin

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    Are these complex functions of a real variable? Is * complex conjugation or convolution?
     
  4. Nov 15, 2005 #3

    benorin

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    It seems clear by iii that * is complex conjugation.
     
  5. Nov 15, 2005 #4

    quasar987

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    yeah.
    -----------------
     
  6. Nov 15, 2005 #5

    HallsofIvy

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    Is there a reason why the integral is only from 0 to [itex]\pi[/itex] when the functions are defined on the entire unit circle (in the complex plane)?

    Suppose f(x) were define to be 0 for t between 0 and [itex]\pi[/itex] and sin(t) when t is between [itex]\pi[/itex] and [itex]2\pi[/itex].
    What is [itex]\int ff*dt[/itex] then?
     
    Last edited: Nov 15, 2005
  7. Nov 15, 2005 #6

    quasar987

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    Ok, I see. I just received a call from aforementioned pals who just found the 'ick'.
     
    Last edited: Nov 15, 2005
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