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L2 norm of complex functions?

by divB
Tags: complex, functions, norm
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divB
#1
Jan15-13, 12:19 AM
P: 81
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.
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CompuChip
#2
Jan15-13, 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 ||f|| = ||Re f|| + ||Im f||, which makes sense, right?
Fredrik
#3
Jan15-13, 08:40 AM
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Quote Quote by divB View Post
[tex]
\int (f-jg)(f-jg)^* dt = \int (f-jg)(f+jg) dt
[/tex]
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
#4
Jan15-13, 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]
\|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]...
CompuChip
#5
Jan15-13, 03:04 PM
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What is <cf, g> and what is <f, cg> if c is a complex number?
Fredrik
#6
Jan15-13, 06:31 PM
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Quote Quote by divB View Post
[tex]<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]
These steps are both wrong. What are the properties of an inner product on a complex vector space?


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