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, $\Omega\subseteq \mathbb{C}\setminus\mathbb{D}$?

  • $\begingroup$ 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. $\endgroup$
    – user752802
    Feb 28, 2020 at 17:05
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    $\begingroup$ 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. $\endgroup$ Feb 28, 2020 at 17:15
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    $\begingroup$ 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 $\endgroup$
    – Conrad
    Feb 28, 2020 at 21:11

1 Answer 1


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!

  • $\begingroup$ 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! $\endgroup$ Feb 28, 2020 at 22:10
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    $\begingroup$ 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 $f(z)-f(-z)=2z$ $\endgroup$
    – Conrad
    Feb 28, 2020 at 22:58
  • $\begingroup$ 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 $z\to z^2$. $\endgroup$ Mar 10, 2020 at 21:07
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    $\begingroup$ @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 $\endgroup$
    – Conrad
    Mar 10, 2020 at 22:13

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