Complex conjugation equality problem

[eljose]

let be the equality where "*" is complex conjugation...z=a+ib but z*=a-ib with a and b real, then:

$$\int_{0}^{c}dxf(x)g(x)=\int_{0}^{c}dxf(x)[g(x)]*$$

with c a real number then my question is if the equality above implies necessarily that:

$$g(x)=[g(x)]*$$ so g is real where:

-f(x)>0 or 0 on the interval (0,c) and f is a real-valued function.
-c is an arbitrary and positive real number.

 

[benorin]

Let $$g(x)=u(x)+iv(x)$$ so that $$[g(x)]*=u(x)-iv(x)$$. The equality then becomes

$$\int_{0}^{c}f(x)(u(x)+iv(x)) dx=\int_{0}^{c}f(x)(u(x)-iv(x)) dx \\ \Rightarrow \int_{0}^{c}f(x)u(x) dx+i\int_{0}^{c}f(x)v(x) dx = \int_{0}^{c}f(x)u(x) dx-i\int_{0}^{c}f(x)v(x) dx \Rightarrow i\int_{0}^{c}f(x)v(x) dx = -i\int_{0}^{c}f(x)v(x) dx \Rightarrow 2i\int_{0}^{c}f(x)v(x) dx=0$$

so we require that $$\int_{0}^{c}f(x)v(x) dx=0$$ for said equality to hold, in terms of g(x) this is $$\int_{0}^{c}f(x)\Im [g(x)] dx=0$$

which could hold for any number of different functions f and g.

 

[matt grime]

As stated it is exceptionally trivial to find counter examples and benorin was being far to rigorous in proving something here. g=0 for all but some finite number of points where it is i will do. And you wonder why people don't trust your non-rigorous 'proofs'?