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L2 norm of complex functions? 
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#1
Jan1513, 12:19 AM

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Hi,
I want to show: [tex] \fjg\^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] \fjg\^2 = <fjg, fjg> = \int (fjg)(fjg)^* dt = \int (fjg)(f+jg) dt = \int f^2 + jfg  jfg + g^2 dt = \f\^2 + \g\^2 [/tex] Where is my wrong assumption? Thanks. 


#2
Jan1513, 03:57 AM

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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 f² = Re f² + Im f², which makes sense, right? 


#3
Jan1513, 08:40 AM

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#4
Jan1513, 03:01 PM

P: 81

L2 norm of complex functions?
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] \fjg\^2 = <fjg,fjg> = <f,fjg><jg,fjg> \\ = <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]... 


#5
Jan1513, 03:04 PM

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What is <cf, g> and what is <f, cg> if c is a complex number?



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