Regarding outer functions again

Question

Consider the Hardy space $H^p, 0<p\leq\infty$ (defined here).

It is said that given any two outer functions $x_1$ and $x_2$ in $H^p$, there exists $a_1$ and $a_2$ in $H^\infty$ such that $a_1x_1=a_2x_2$ . And $a_1s$ and $a_2 s$ are outer functions for any outer function $s$.

The hint provided for the above is that every outer function $f$ can be written in the form $\frac{g}{h}$ where $g$ and $h$ are bounded outer functions. Earlier I thought that for an outer function $f$ the corresponding $g$ is the constant function 1. But that is not the case as seen here. So can anyone tell how the above statement is true?

Answer 1

If $f$ is outer, $1/f$ is outer. (More generally $f$ raised to any real - say non-zero to avoid constants - power is outer.) Any outer function has no zeros as those are factored out with Blaschke products, so we can talk about $\log f$ and any complex power of $f$ in the disk.

$1/f$ always belongs to the Nevanlinna class $N$ by the way, and if for example $\Re f>0$, than $\Re(1/f)>0$ so $1/f$ belongs to all Hardy classes of exponent less than $1$.

Applying the above factorization with bounded analytic functions for class $N$ for the ratio of any two outer functions, we get your answer.