L2 norm of complex functions?

[divB]

Hi,

I want to show:

$$\|f-jg\|^2 = \|f\|^2 - 2 \Im\{<f,g>\} + \|g\|^2$$

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

$$\|f-jg\|^2 = <f-jg, f-jg> = \int (f-jg)(f-jg)^* dt = \int (f-jg)(f+jg) dt = \int f^2 + jfg - jfg + g^2 dt = \|f\|^2 + \|g\|^2$$

Where is my wrong assumption?
Thanks.

 

[CompuChip]

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?

 

[Fredrik]

$$\int (f-jg)(f-jg)^* dt = \int (f-jg)(f+jg) dt$$

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?

 

[divB]

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 $2\Im<f,g>$...

 

[CompuChip]

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

 

[Fredrik]

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

These steps are both wrong. What are the properties of an inner product on a complex vector space?