Outer function

Remarks

$F$ is holomorphic and has no zeros, and $\log \abs{F}$ is the Poisson integral of $\log \varphi$. Its radial limits satisfy $\abs{F^*} = \varphi$ almost everywhere. For $0 < p \le \infty$, $F$ lies in the Hardy space $H^p$ if and only if $\varphi \in L^p$ (Lp space).

Example: $F(z) = 2 + z$ is outer. It is holomorphic and has no zeros on a neighbourhood of the closed disc, so $\log \abs{F}$ is harmonic and continuous up to the circle. Hence $\log \abs{F}$ equals the Poisson integral of $\log \varphi$, where $\varphi = \abs{2 + e^{it}}$. So $F$ and the outer function with modulus $\varphi$ have the same modulus and no zeros; their quotient is a holomorphic function of constant modulus $1$, hence a constant.

Non-example: $f(z) = z$ is not outer. Its boundary modulus is $\varphi = 1$, and the outer functions with this modulus are the constants of modulus $1$, which differ from $z$.