8
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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. $\endgroup$
    – Dark Malthorp
    Feb 28, 2020 at 17:15
  • 1
    $\begingroup$ 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

Reset to default
8
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ 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! $\endgroup$
    – Dark Malthorp
    Feb 28, 2020 at 22:10
  • 1
    $\begingroup$ 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 $z\to z^2$. $\endgroup$
    – Dark Malthorp
    Mar 10, 2020 at 21:07
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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