# L2 norm of complex functions?

by divB
Tags: complex, functions, norm
 P: 69 Hi, I want to show: $$\|f-jg\|^2 = \|f\|^2 - 2 \Im\{\} + \|g\|^2$$ However, as far as I understand, for complex functions $= \int f g^* dt$, right? Therefore: $$\|f-jg\|^2 = = \int (f-jg)(f-jg)^* dt = \int (f-jg)(f+jg) dt = \int f^2 + jfg - jfg + g^2 dt = \|f\|^2 + \|g\|^2$$ Where is my wrong assumption? Thanks.
 HW Helper Sci Advisor P: 4,281 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?
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P: 8,837
 Quote by divB $$\int (f-jg)(f-jg)^* dt = \int (f-jg)(f+jg) dt$$
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?

P: 69

## L2 norm of complex functions?

Hi, thank you. Ok, no integrals, but only use <,>

I am again confused :(

$$\|v\|^2 = <v,v>$$, 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:

$$\|f-jg\|^2 = <f-jg,f-jg> = <f,f-jg>-<jg,f-jg> \\ = <f,f> - <f,jg> - (<jg,f>-<jg,jg>) \\ = <f,f> - <f,jg> - <jg,f> + <jg,jg> \\ = <f,f> - j<f,g> - j<g,f> - <g,g> = \|f\|^2 - 2j<f,g> - \|g\|^2$$

But $2j<f,g>$ is not $2\Im<f,g>$...
 HW Helper Sci Advisor P: 4,281 What is and what is if c is a complex number?
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 Quote by divB $$- - + \\ = - j - j - \\ = \|f\|^2 - 2j - \|g\|^2$$