definition of Smirnov class

Question

The Smirnov class in the unit disc $\mathbb{D}$ can be characterized as space $$ N^{+}=\{ \phi/\psi\; :\; \phi ,\psi\in H^{\infty},\; \psi\;\; outer\} $$ where $H^{\infty}$ denotes the algebra of bounded analytic functions on $\mathbb{D}$.

My question is what the meaning of outer is inside of the brace?

Answer 1

A function $\psi\in H^1(\mathbb{D})$ is called outer if it has the integral representation $$ \psi(z) = \alpha \exp\left(\frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{e^{it} + z }{e^{it}-z} k(t)\,dt\right) $$ for some constant $\alpha$ with $|\alpha|=1$ and some real-valued $k\in L^1[-\pi, \pi]$. The main features of outer function:

• $\psi$ does not vanish in the unit disk
• It also "does not vanish too much on the boundary". Precise meaning: $|\psi|$ has boundary limit values a.e. on the boundary, and $\log|\psi|$ is integrable (because it's $k$). So for example $\exp(1/(1-z))$ is not an outer function: it tends to $0$ too fast as $z\to 1$.

The main purpose of these is the inner-outer factorization: every $H^1$ function is the product of an inner function (modulus $1$ a.e. on the boundary) and an outer function, in an essentially unique way; you can only move a unimodular constant from one factor to the other.

Also, the reciprocal of an outer function has integral representation $$ 1/\psi(z) = \alpha^{-1} \exp\left(-\frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{e^{it} + z }{e^{it}-z} k(t)\,dt\right) $$ which makes it somewhat manageable, even though $1/\psi$ need not be in a Hardy class. This is why having an outer functions in the denominator, as is the case in your formula, still makes a reasonable function class.