Consider the Hardy space $H^p, 0<p\leq\infty$ ([defined here](http://math.stackexchange.com/questions/2586370/if-a-function-belongs-to-the-hardy-space)).


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](https://mathoverflow.net/questions/291160/regarding-representation-of-an-outer-function). So can anyone tell how the above statement is true?



Regarding outer functions again
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.


