5
$\begingroup$

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 $g=h^3$?

$\endgroup$

2 Answers 2

6
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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!

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.