Is the conjugate of an integral the integral of the conjugate in L2 space?

[gonzo]

In the complex conjugate of an integral equal to the integral of the complex conjugate?

If so, is there an easy way to show this?

Thanks.

 

[matt grime]

In general, the answer is no. Just consider analytic functions integrated round loops and apply Morera's Theorem (if the integral of a function round all closed paths is zero then the function is analytic).

 

[gonzo]

Well, what about when you look at the inner product defined for L2 space. Here the claim is made that:

$$<f,g> = \int f\overline{g}$$$$<f,g> = \overline{<g,f>}$$$$\int f\overline{g}=\overline{\int g\overline{f}}$$

But this is the same as saying in this case that the conjugate of the integral is the integral of the conjugate. How is this supported if this isn't true in general?