# L2 norm of complex functions?

by divB
Tags: complex, functions, norm
 P: 81 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.
 Sci Advisor HW Helper P: 4,300 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?
Emeritus
PF Gold
P: 9,226
 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: 81 L2 norm of complex functions? Hi, thank you. Ok, no integrals, but only use <,> I am again confused :( $$\|v\|^2 =$$, 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: $$\|f-jg\|^2 = = - \\ = - - (-) \\ = - - + \\ = - j - j - = \|f\|^2 - 2j - \|g\|^2$$ But $2j$ is not $2\Im$...
 Sci Advisor HW Helper P: 4,300 What is and what is if c is a complex number?
Emeritus
PF Gold
P: 9,226
 Quote by divB $$- - + \\ = - j - j - \\ = \|f\|^2 - 2j - \|g\|^2$$
These steps are both wrong. What are the properties of an inner product on a complex vector space?

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