Is there a characterization of the nonconstant entire functions $f$ that satisfy $|f(z)|=1$ for all $|z|=1$?

Clearly, $f(z)=z^n$ works for all $n$. Also, it's not difficult to show that if $f$ is such an entire function, then $f$ must vanish somewhere inside the unit disk. What else can be said about those functions?

Thank you


The only such functions are $f(z)=az^n$ with $|a|=1$. Let $f$ be such a function and define $$g(z)=\overline{f(\overline{z}^{-1})}^{-1}.$$ Due to the two complex conjugations, $g$ is meromorphic on $\mathbb{C}-\{0\}$. Also $g(z)=f(z)$ when $|z|=1$. By analytic continuation, $f(z)=g(z)$ for all nonzero $z$. Now $g$ has a pole of order $k$ at $\infty$ where $k$ is the order of $f$ at zero. Hence $f$ is a polynomial of degree $k$. As $k$ is also the order at $0$ then $f(z)=az^k$; clearly $|a|=1$.

  • $\begingroup$ I think You implicitly assumed that $f(z) = 0\iff z = 0$ ($g$ has a pole in $\bar{a}^{-1}$ if $f(a) = 0$), but in this case the claim follows almost immediately from maximum modulus principle. $\endgroup$ – ifk Aug 27 '10 at 9:31
  • $\begingroup$ Pierre-Yves, an entire function with a pole (i.e., not an essential singularity) at $\infty$ s a polynomial. $\endgroup$ – Robin Chapman Aug 28 '10 at 13:13
  • $\begingroup$ I deleted my previous question by mistake. It was: why is f a polynomial? Thanks for your answer, but now I don't understand why f is entire. (The function g is meremorphic, not holomorphic.) $\endgroup$ – Pierre-Yves Gaillard Aug 28 '10 at 13:19
  • 1
    $\begingroup$ Pierre-Yves, $f$ is entire, since that is the hypothesis! $\endgroup$ – Robin Chapman Aug 28 '10 at 13:59
  • 3
    $\begingroup$ This must be from the Book. $\endgroup$ – timur Dec 24 '10 at 6:14

If a function $f$ is holomorphic in a neighborhood of the closed disk and has modulus 1 on the circle, then $f$ is a finite Blaschke product. You can prove this by taking all of the zeros inside the disk counted according to multiplicity, dividing by corresponding holomorphic automorphisms of the disk that have those zeros, and showing that the result is constant. (This quotient and its reciprocal are analytic and bounded by 1 on the disk...) Thus, your $f$ is a finite Blaschke product, but it also has no poles in the plane. The only such functions are constant multiples of power functions, as ifk already said.

  • $\begingroup$ Nice answer. In fact, my proof was (again) flawed, but one can modify it to give alternative proof of this formula with finite Blaschke product. $\endgroup$ – ifk Aug 27 '10 at 10:15
  • $\begingroup$ Nice, the solution is easy once you use Blaschke products! $\endgroup$ – Malik Younsi Aug 27 '10 at 12:22

Edit: How frustrating, this is still not correct. It is not hard to make this proof work by generalizing to the functions holomorphic on unit disc and continuous on its closure, but such approach will be nothing else than proving the theorem cited by Jonas Meyer (see his answer).
Second attempt:

  1. Note that $f$ has finitely many zeros in $\mathbb{D}$ (identity principle + compactness).

  2. Observe that if $F\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}}),$ the only zero of $F$ is in zero, and $|F|$ is constant on unit circle, then $F(z) = e^{it}z^{k}$ where $k = ord_{0}F$.
    (apply the maximum principle for $F(z)/z^{k}$ and $z^{k}/F(z)$)

  3. Let $z_{1},\ldots, z_{n}$ be all pairwise distinct zeros of $f$ in $\mathbb{D}$. By induction on $n$ we will show that $n = 1$ and $z_{1} = 0$.

Let $n = 1$.
We are taking $h \in Aut(\mathbb{D})$ s.t. $h(0) = z_{1}$ and $g:= f\circ h\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}})$.
Then $|g|$ is constant on unit circle, $0$ is the only zero of $g$ in $\mathbb{D}$. So $g(z) = e^{it}z^{k}$.
Therefore $f(z) = e^{it}\cdot \left(\frac{z - z_{1}}{1 - \bar{z_{1}}z}\right)^{k}$ in $\mathbb{D}$.
From identity principle for meromorphic functions this equality holds in $\mathbb{C}$, and gives contradiction unless $z_{1} = 0$ (otherwise $f$ is not entire).

If $n > 1$, then we're reasoning by contradiction.
W.l.o.g. $z_{n}\neq 0$.
We are taking $h \in Aut(\mathbb{D})$ s.t. $h(0) = z_{n}$ and $g:= f\circ h\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}})$, and then $h^{-1}(z_{1}),\ldots, h^{-1}(z_{n-1}), 0$ are the all zeros of $g$ in $\mathbb{D}$.
Let $k = ord_{0}g$, and $G(z) = g(z)/z^{k}\in \mathcal{O}(\mathbb{D})\cap \mathcal{C}(\bar{\mathbb{D}})$.
$G$ has $n-1$ zeros on $\mathbb{D}$, $|G(z)| = 1$ on unit circle, thus from inductive assumption $n-1 = 1$ or $G$ is constant (we cannot use inductive assumption since in general $G\not\in \mathcal{O}(\mathbb{C})$!).
If $G$ is non-constant then $n-1 = 1$ and $G(z) = e^{it}z^{l}, \ l > 0$ - a contradiction with $ord_{0}g = k$. Otherwise $n - 1 = 0$ - a contradiction.

Finally 1. and 3. proves that all entire functions with desired property are of the form $f(z) = e^{it}z^{k}$.

The following reasoning is not correct (see the Malik Younsi's answer):

This should be the comment to the Leandro's answer, although I cannot add comments yet.

W.l.o.g. $f$ is non-constant. Then by maximum modulus $f$ has zero inside $\mathbb{D}$. Denote this zero by $a$.

Composing $f$ with an automorphism $h$ of disc s.t. $h(0) = a$ we obtain that $g = f\circ h$ satisfies $g(0)= 0$ and moreover $|g(z)|\le |z|^{k}$ inside $\mathbb{D}$, where $k$ is the order of zero of $g$ in $0$ (indeed $g(z)/z^{k}$ is entire and we may use the maximum modulus principle).

By Schwarz lemma there exists $\theta\in \mathbb{R}$ satisfying $g(z) = e^{i\theta}z^{k}$.

From classification of automorphisms of the disc, we have $h^{-1}(z) = e^{i\omega}\frac{z-a}{1-\bar{a}z}$ for some $\omega\in \mathbb{R}$, and from this the full classification easily follows.

  • $\begingroup$ Hi ifk, this is in fact a complement to my partial answer :), nice! $\endgroup$ – Leandro Aug 26 '10 at 21:02

Partial answer.

If $|f(z)|=1$ for all $|z|=1$ and $f$ is entire function, then $f:\mathbb D \to \mathbb D$, by the Maximum Modulus Theorem.

Using Schwarz-Lemma we can characterize, at least, the conformal ones.
( See Functions of one Complex Variable, Conway 2ed pag. 131 )


For your $f$ we know there are no zeros on the boundary of the circle, and only a finite number of zeros inside. So you can form the Blaschke product $b$ for the non-zero zeros of $f$ inside the circle, and $f(z)/(z^{n}b(z))$ is non-vanishing and holomorphic on an open neighborhood of the closed unit disk for some natural number $n$, and has modulus $1$ on the unit circle. Of course, it may be that $f$ has no non-zero zeros inside the disk, in which case, define $b\equiv 1$.

So $\ln|f(z)/(z^{n}b(z))|$ is harmonic on an open neighborhood of the closed unit disk, and it vanishes on the unit circle. So this harmonic function is identically $0$, which means $|f(z)/(z^{n}b(z))|\equiv 1$ inside the disk. Now, by the maximum modulus principle for holomorphic functions, $f(z)=Cz^{n}b(z)$ for some unimodular constant $C$, and for all $|z| \le 1$. By the identity theorem, this must be true for all $z$ where $b(z)$ is finite, which means that the Blaschke factor $b$ must not be present in the factoring because $f$ is entire. Therefore, $f=Cz^{n}$ for some constant $C$ with $|C|=1$ and for some $n \ge 0$.


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