# $f(z) = z + f(z^2)$ outside the unit disk?

The function $$f(z) = \sum_{n=0}^\infty z^{2^n}$$ which satisfies the functional equation $$f(z) = z + f(z^2)$$ is a classic example of a function analytic in $$\mathbb{D} = \{z:|z|<1\}$$ that cannot be analytically extended beyond the boundary anywhere.

My question is, are there any other analytic solutions to $$f(z) = z +f(z^2)$$ defined on a different domain, $$\Omega\subseteq \mathbb{C}\setminus\mathbb{D}$$?

• If $f$ is analytic on a disc $U$ outside $\mathbb{D}$, to satisfy the functional equation it should also be defined on $U^2$, and $U^4,...,U^{2^n},...$. So if would be defined on an annulus around $\infty$. Therefore, it would admit a convergent series expansion of the form $a_0+a_1/z+a_2/z^2+...$. Putting this in the functional equation gives $0=a_1=a_2=...$. Then $f$ must be constant. But $a_0=z+a_0$ wouldn't be satisfied.
– user752802
Feb 28, 2020 at 17:05
• Being defined on an annulus around infinity only would give us that it has a convergent Laurent series on the annulus right? I can't see how you get the positive terms to be 0 unless you know it's analytic on a punctured neighborhood of infinity, which doesn't seem to follow from the assumptions, e.g. if $U = \{ 2 < z < 3\}$, then $U^{2^n}$ are all disjoint if I'm not mistaken. Feb 28, 2020 at 17:15
• This is a good question - I incline to think the answer is no and I agree that if one can define it outside a disc centered at the origin, the Laurent series method work Feb 28, 2020 at 21:11

Assuming the problem is not vacuous (obviously, one can take a domain $$U$$ s.t $$U^2 \cap U= \emptyset$$ and any analytic $$f$$ on $$U$$ and declare the condition satisfied trivially because there is no $$z^2 \in U$$ when $$z \in U$$) the answer is no.

Assume by contradiction such an $$f$$ exists on a domain $$U$$ exterior to the unit disc and that the formula gives the analytic continuation of $$f$$ to $$U^2, U^4,...$$ (where $$U^2$$ open is the image of $$U$$ under $$z \to z^2$$, etc) and let $$W$$ the open set (could be disconnected) that is the union of $$U, U^2..$$ so $$f$$ is analytic on $$W$$

The fundamental fact we will use is that if $$z, -z \in W$$ then $$f(z)-f(-z)=2z$$ hence $$|f(z)| +|f(-z)| \ge 2|z|$$

Now, picking $$a \in U$$, there is a disc centered at $$a$$ contained in $$U$$, which means there is a root of unity $$\alpha$$ of some order $$2^k$$ and $$r>1$$ st $$r\alpha \in U$$ (pick a ray through the origin passing through the above disc and get a close enough ray with an appropriate angle...). This means that some positive number $$R >1$$ is in $$W$$.

But now $$W$$ is open so there is a small closed disc $$\bar V$$ centered at $$R$$ included in $$W$$ and again picking appropriate rays we find that $$Re^{\pi 2^{-N-1}i}$$ is in $$V$$ for all high enough $$N$$ (or notice that $$Re^{\pi 2^{-N-1}i} \to R, N \to \infty)$$. Since $$f$$ is continuous on $$\bar V$$, $$|f|$$ attains a maximum $$M$$ there.

Now $$-R^{2^{N+1}}$$ is in $$W$$, so $$f(R^{2^{N+1}})-f(-R^{2^{N+1}})=2R^{2^{N+1}}$$

$$f(R)=R+R^2+...R^{2^n}+f(R^{2^{n+1}})$$ for any $$n \ge 0$$, hence $$|f(R^{2^{N+1}})| \le (N+2)R^{2^{N}}+M$$ and similarly for $$|f(-R^{2^{N+1}})|$$, which gives $$2R^{2^{N+1}}\le (2N+4)R^{2^{N}}+2M$$ and that is impossible for high enough $$N$$. Done!

• I suspected there was a way to prove it using the growth rate, I just couldn't quite see how to do it. $f(z) - f(-z) = 2z$ is a neat trick! Feb 28, 2020 at 22:10
• I tried to construct a counterexample by taking $U=D(3,1)$ and then $W$ as above is a disjoint union of sets since each successive power is in annulus bounded by circles with radiuses consecutive powers of powers of two and the obstruction came precisely when both $z,-z$ are in one such set as then $f(z^2)$ must agree - from there it was clear that leads to a contradiction by taking large enough modulus - we also can show that $f(-R^{2^N})$ cannot be real for high $N$ since the largest term in the iteration is purely imaginary and dominates the rest and then use directly $f(z)-f(-z)=2z$ Feb 28, 2020 at 22:58
• If I'm understanding correctly, your proof extends to show non-existence of continuous (not just analytic) solutions to $f(z) = z + f(z^2)$ on any set $W\subset\{z\mid |z|>1\}$ that is closed under $z\to z^2$. Mar 10, 2020 at 21:07
• @Dark looks like it which makes sense since $W$ can be a disjointed union of open sets, so basically the only connection is given by $z^2$ coming from the previous component and forcing some equalities on $f$ that eventually lead to contradiction Mar 10, 2020 at 22:13