Is the fallowing expression an inner product of $H^\infty\left(\mathbb{C}^+\right)$?
$$\left<f,g\right>=\int_{-\infty}^{\infty}f\left(i\omega\right)\overline{g\left(i\omega\right)}d\omega$$
This satisfies the demands:
- $\left<f+h,g\right>=\left<f,g\right>+\left<h,g\right>$
- $\left<{\lambda}f,g\right>=\lambda\left<f,g\right>$
- $\overline{\left<f,g\right>}=\left<g,f\right>$
The problem is with the demand:
- $\left<f,f\right>=0{\Leftrightarrow}f\equiv0$
Since all the functions in $H^\infty\left(\mathbb{C}^+\right)$ are analytical:
$$\int_{-\infty}^{\infty}f\left(i\omega\right)\overline{f\left(i\omega\right)}d\omega=\int_{-\infty}^{\infty}\left|f\left(i\omega\right)\right|^2d\omega=0{\Leftrightarrow}f\left(s\right)=0{\forall}s=i\omega$$
From the characteristics of $H^\infty\left(\mathbb{C}^+\right)$ I know that $sup\left(f\left(s\right)\right)$ is located on the line $s=i\omega$:
$$sup\left|f\left(s\right)\right|\in\left\{\left|f\left(s\right)\right||s=i\omega\right\}=\left\{0\right\}{\Rightarrow}sup\left|f\left(s\right)\right|=0\Rightarrow\left|f\left(s\right)\right|=0{\Rightarrow}f\equiv0$$
I'm not sure how correct and airtight the above "proof" is, is it complete and true?