What Causes the Error in Calculating the L2 Norm of Complex Functions?

[divB]

Hi,

I want to show:

<br /> \|f-jg\|^2 = \|f\|^2 - 2 \Im\{&lt;f,g&gt;\} + \|g\|^2<br />

However, as far as I understand, for complex functions &lt;f,g&gt; = \int f g^* dt, right? Therefore:

<br /> \|f-jg\|^2 = &lt;f-jg, f-jg&gt; = \int (f-jg)(f-jg)^* dt = \int (f-jg)(f+jg) dt = \int f^2 + jfg - jfg + g^2 dt = \|f\|^2 + \|g\|^2<br />

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]

<br /> \int (f-jg)(f-jg)^* dt = \int (f-jg)(f+jg) dt<br />

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 = &lt;v,v&gt;, 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:

<br /> \|f-jg\|^2 = &lt;f-jg,f-jg&gt; = &lt;f,f-jg&gt;-&lt;jg,f-jg&gt; \\<br /> = &lt;f,f&gt; - &lt;f,jg&gt; - (&lt;jg,f&gt;-&lt;jg,jg&gt;) \\<br /> = &lt;f,f&gt; - &lt;f,jg&gt; - &lt;jg,f&gt; + &lt;jg,jg&gt; \\<br /> = &lt;f,f&gt; - j&lt;f,g&gt; - j&lt;g,f&gt; - &lt;g,g&gt;<br /> = \|f\|^2 - 2j&lt;f,g&gt; - \|g\|^2<br />

But 2j&lt;f,g&gt; is not 2\Im&lt;f,g&gt;...

 

[CompuChip]

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

 

[Fredrik]

&lt;f,f&gt; - &lt;f,jg&gt; - &lt;jg,f&gt; + &lt;jg,jg&gt; \\<br /> =&lt;f,f&gt; - j&lt;f,g&gt; - j&lt;g,f&gt; - &lt;g,g&gt;\\<br /> = \|f\|^2 - 2j&lt;f,g&gt; - \|g\|^2<br />

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