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L2 norm of complex functions?

  1. Jan 15, 2013 #1
    Hi,

    I want to show:

    [tex]
    \|f-jg\|^2 = \|f\|^2 - 2 \Im\{<f,g>\} + \|g\|^2
    [/tex]

    However, as far as I understand, for complex functions [itex]<f,g> = \int f g^* dt[/itex], right? Therefore:

    [tex]
    \|f-jg\|^2 = <f-jg, f-jg> = \int (f-jg)(f-jg)^* dt = \int (f-jg)(f+jg) dt = \int f^2 + jfg - jfg + g^2 dt = \|f\|^2 + \|g\|^2
    [/tex]

    Where is my wrong assumption?
    Thanks.
     
  2. jcsd
  3. Jan 15, 2013 #2

    CompuChip

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    Are f and g complex functions? Then you should write f = Re(f) + j Im(f), g = Re(g) + j Im(g).

    Or are they just the real parts of a single function? Because then you've just shown that ||f||² = ||Re f||² + ||Im f||², which makes sense, right?
     
  4. Jan 15, 2013 #3

    Fredrik

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    This one should be $$\int (f-jg)(f-jg)^* dt = \int (f-jg)(f^*+jg^*) dt.$$ But why use the definition of <,> at all? I assume that you have already proved that it's an inner product. So why not just use that?
     
  5. Jan 15, 2013 #4
    Hi, thank you. Ok, no integrals, but only use <,>

    I am again confused :(

    [tex]\|v\|^2 = <v,v>[/tex], as far as I understand also for complex functions. But then, with using only the inner product, I have no chance to obtain an imaginary part only:

    [tex]
    \|f-jg\|^2 = <f-jg,f-jg> = <f,f-jg>-<jg,f-jg> \\
    = <f,f> - <f,jg> - (<jg,f>-<jg,jg>) \\
    = <f,f> - <f,jg> - <jg,f> + <jg,jg> \\
    = <f,f> - j<f,g> - j<g,f> - <g,g>
    = \|f\|^2 - 2j<f,g> - \|g\|^2
    [/tex]

    But [itex]2j<f,g>[/itex] is not [itex]2\Im<f,g>[/itex]...
     
  6. Jan 15, 2013 #5

    CompuChip

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    What is <cf, g> and what is <f, cg> if c is a complex number?
     
  7. Jan 15, 2013 #6

    Fredrik

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    These steps are both wrong. What are the properties of an inner product on a complex vector space?
     
    Last edited: Jan 15, 2013
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