Let $\mathbb{D}:=\{z \in \mathbb{C}|~|z|<1\}$. Consider $f,~g \in H^{\infty}(\mathbb{D}):=\{f\in\text{ Hol}(\mathbb{D}): \|f\|_{\infty}:= \sup_{z \in \mathbb{D}} |f(z)| < \infty\}$ such that $f^3=g^2$. Does there exist an analytic function $h \in H^{\infty}(\mathbb{D})$ such that $f=h^2$, and $g=h^3$?

## 2 Answers

Every bounded analytic function $h$ in the disk has the representation $$h(z)=B(z)\exp(-P(z)),$$ where $B$ is a Blaschke product and $P$ has positive imaginary part. Applying this to $h=f^3=g^2$, we conclude that every factor in the Blaschke product must occur $6n$ times. Therefore the Blaschke product $B$ has a 6-th root $B_0$ which is also a Blaschke product, and $h_0(z)=B_0(z)\exp(-P(z)/6)$ satisfies $h=h_0^6$, so $f=c_3h^2$ and $g=c_2h^2$, where $c_k$ are some $k$-th roots of unity. Multiplying $h_0$ on an appropriate $6$-th root of unity we obtain the requested function.

This holds in general in any domain $U$, while of course if $f=h^2$ is bounded, then $h$ bounded, so the extra condition is automatically satisfied (similarly note that $f$ bounded is equivalent to $g$ bounded, so only one of the two is needed).

Wlog let's assume $f, g$ not identically zero, so they have isolated zeroes since we are in a domain $U$. Since $f^3=g^2$ it follows that any zero of $f$ of multiplicity $k \ge 1$ is a zero of $g^2$ of multiplicity $3k$, hence $k$ even and as a zero of $g$ it has multiplicity $3k/2 >k$. This means that $h=g/f$ is analytic in $U$ so $g^2=h^2f^2=f^3$ hence $f=h^2$ outside of its isolated zeroes and by continuity there too, so in all $U$ and then $g=hf=h^3$ and we are done!