# Functional equation of bounded analytic functions

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 $$g=h^3$$?

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!