Conjecture: Every analytic function on the closed disk is conformally a polynomial.

Here is my conjecture, any proof, counter-example, or intuitions?

If $f$ is analytic on $\text{cl}(\mathbb{D})$ (that is, analytic on some open set containing $\text{cl}(\mathbb{D})$), then there is some injective analytic function $\phi:\mathbb{D}\to\mathbb{C}$ and polynomial $p$ such that $f=p\circ\phi$ on $\mathbb{D}$.

PS: I would then say that $f$ and $p$ are conformally equivalent. Is there a better term for this relationship?

Note: $\text{cl}(\mathbb{D})$ could be replaced by any simply connected compact set. If we wanted to replace “simply connected” with “finitely connected”, then we would have to replace “polynomial” with “rational function”.

Solutions Collecting From Web of "Conjecture: Every analytic function on the closed disk is conformally a polynomial."

Yes, it is true that you can write $f=p\circ\phi$, for a polynomial $p$ and an injective analytic $\phi\colon\mathbb{D}\to\mathbb{C}$. In fact, you can replace $\mathbb{D}$ by any compact subset of $\mathbb{C}$ and the results still holds, where $p$ is instead required just to be a rational function.

First, by expanding as a power series and truncating after $n$ terms, there exists a sequence of polynomials $p_n$ converging uniformly to $f$ on an open set containing $\mathbb{D}$. This is almost enough, applying the following.

Lemma 1: Let $f,\{p_n\}_{n=1,2,\cdots}$ be non-constant analytic functions on an open set $U\subseteq\mathbb{C}$ such that $p_n\to f$ uniformly on $U$. Let $K\subseteq U$ be compact and suppose that, for each $k\ge1$ and $z\in K$ with $f^\prime(z)=\cdots=f^{(k)}(z)=0$ then we also have $p_n(z)=f(z)$ and $p_n^\prime(z)=\cdots=p_n^{(k)}(z)=0$.
Then, for all large enough $n$, there exists ono-to-one analytic $\phi_n\colon K\to U$ with $f=p_n\circ\phi_n$.

This is enough to obtain the required result, so long as $f^\prime$ does not vanish on $\mathbb{D}$, as the condition on the derivatives on $f$ trivially holds. If $f^\prime$ does vanish somewhere on $\mathbb{D}$, then we can still obtain the required result by applying the following consequence of Lemma 1.

Lemma 2: Let $f,\{p_n\}_{n=1,2,\cdots}$ be non-constant analytic functions on an open set $U\subseteq\mathbb{C}$ such that $p_n\to f$ uniformly on $U$. Then, for any compact $K\subseteq U$, for large enough $n$ there exists polynomials $q_n$ and ono-to-one analytic $\phi_n\colon K\to U$ such that $f=(p_n-q_n)\circ\phi_n$.

The polynomials $q_n$ will be of uniformly bounded degree with coefficients tending to zero as $n\to\infty$, although we do not require this here. Once Lemma 2 has been established, our required result follows emmediately by using truncated power series approximations for the polynomials $p_n$. In fact, if $\mathbb{D}$ is replaced by any compact $K$ then the result will still hold where $p$ is instead required to be just be a rational functions. This is because we can uniformly approximate $f$ by a sequence of rational functions $p_n$ on some open set containing $K$, by Runge’s theorem. Then, Lemma 2 applies to get the required result.

I’ll now give proofs of the Lemmas. Here, I make repeated use of the fact that if analytic functions then their derivates to all orders converge uniformly on compacts (a consequence of Cauchy’s integral formula, as explained by David Speyer in his answer).

Proof of Lemma 1:
First, restricting $U$ a small enough open set containing $K$, we can w.l.o.g. suppose that for every $z\in U$ and $k\ge1$ with $f^\prime(z)=\cdots=f^{(k)}(z)=0$ then $p_n(z)=f(z)$ and $p_n^\prime(z)=\cdots=p_n^{(k)}(z)=0$. I’ll start by proving the result locally; for each $z_0\in U$ there exists an open neighbourhood $V$ of $z_0$ and a sequence of analytic functions $\phi_n\colon V\to U$ with $\phi_n(z)\to z$ uniformly on $V$ and $f=p_n\circ\phi_n$ on $V$ for all large $n$.

Suppose that $f^\prime(z_0)\not=0$. By rescaling $f$ and $p_n$ if required, we can suppose that $f^\prime(z_0)=1$. For some $r > 0$ the closed ball $\bar B_r(z_0)$ is contained in $U$ and $\Re[f^\prime] > 1/2$ on $\bar B_r(z_0)$. Then, by uniform convergence, $\Re[p_n^\prime] > 1/2$ on $\bar B_r(z_0)$ for large $n$. This implies that $\Re[(p(y)-p(x))/(y-x)] > 1/2$ on the closed ball, so $p_n$ is one-to-one there with derivative $\lVert p_n^\prime\rVert\ge1/2$. This implies that $p_n(B_r(z_0))$ contains $B_{r/2}(p_n(z_0))$, so $p_n(B_r(z_0))\supseteq B_{r/3}(f(z_0))$ for large $n$ and there is a unique (analytic) $p_n^{-1}\colon B_{r/3}(f(z_0))\to B_r(z_0)$ with $p_n\circ p_n^{-1}(z)=z$ (by the inverse function theorem). Choosing an open neighborhood $V$ of $z_0$ such that $f(V)\subseteq B_{r/3}(f(z_0))$, then the analytic functions $\phi_n\colon V\to U$, $\phi_n(z)=p_n^{-1}\circ f(z)$ satisfy the required properties.

Now suppose that $f^\prime(z_0)=0$. Then, there is a $k\ge0$ with $f^\prime(z_0)=\cdots=f^{(k)}(z_0)=0$ and $f^{(k+1)}(z_0)\not=0$. Subtracting a constant from $f$ and $p_n$ if necessary, we can suppose that $f(z_0)=0$. Then, $f(z)=(z-z_0)^kg(z)$ for an analytic function $g$ on $U$ with $g(z_0)\not=0$. By assumption, $p_n(z_0)=\cdots=p_n^{(k)}(z_0)=0$, so we can decompose $p_n(z)=(z-z_0)^kq_n(z)$ where $q_n\to f$ uniformly on $U$. Then, on some neighborhood of $z_0$, $g$ is nonzero and $q_n$ are nonzero for large $n$. Hence, we can take $k$’th roots to obtain analytic functions $\tilde f(z)=(z-z_0)g(z)^{1/k}$ and $\tilde p_n(z)=(z-z_0)q_n(z)^{1/k}$. Furthermore, $\tilde p_n\to\tilde f$ uniformly (so long as we take consistent $k$’th roots for $g$ and $q_n$). We also have $\tilde f^\prime(z_0)=g(z_0)^{1/k}\not=0$. So, by the first case above, there exists an open neighborhood $V$ of $z_0$ and analytic $\phi_n\colon V\to U$ with $\phi_n(z)\to z$ uniformly and $\tilde f=\tilde p_n\circ\phi_n$. Taking $k$’th powers gives $f=p_n\circ\phi_n$.

Now, by compactness of $K$ and the fact that the analytic functions $\phi_n$ exist locally as shown above, there exists a finite cover $\lbrace B_1,\ldots,B_m\rbrace$ of $K$, where $B_i$ are open balls in $U$, and analytic functions $\phi_{in}\colon B_i\to U$ satisfying $f=p_n\circ\phi_{in}$ on $B_i$ and $\phi_{in}(z)\to z$ uniformly as $n\to\infty$.

For whenever $B_i,B_j$ have non-empty intersection then, as $f$ is non-constant, its derivative will be non-zero at some point $z_0\in B_i\cap B_j$ and, wlog, we can suppose that $f^{\prime}(z_0)=1$. Then, by uniform convergence, there is an open neighborhood $B^\prime$ of $z_0$ on which $\Re[p_n^\prime]\ge1/2$ for all large $n$ so, as with the argument above, $p_n$ is one-to-one on this neighborhood. As $p_n\circ\phi_{in}=p_n\circ\phi_{jn}$ this implies that $\phi_{in}=\phi_{jn}$ in a neighborhood of $z_0$ whenever $n$ is large enough that $\phi_{in}(z_0)$ and $\phi_{jn}(z_0)$ are in $B^\prime$. By analytic continuation, this gives $\phi_{in}=\phi_{jn}$ on $B_i\cap B_j$. Therefore, with $V=\cup_iB_i$, we can define analytic $\phi\colon V\to U$ by setting $\phi=\phi_{in}$ on $B_i$. Then, $f=p_n\circ\phi_n$ and $\phi_n(z)\to z$ uniformly.

It only remains to show that $\phi_n$ is one-to-one for large $n$. By uniform convergence of $\phi_n^\prime$ to $1$ on any closed ball $\bar B$ in $V$, we have $\Re[\phi^\prime_n]\ge1/2$ on $\bar B$ for large $n$, so $\phi_n$ is one-to-one on $\bar B$. So, letting $B_1^\prime,\ldots,B_r^\prime$ be open balls covering $K$ and whose closures are in $V$, $\phi_n$ is eventually one-to-one on each $B^\prime_i$. If we let $K^\prime\subseteq\bigcup_iB^\prime_i$ be a compact set with interior containing $K$, then, by compactness, there is an $\epsilon > 0$ such that any $x\not=y\in K^\prime$ with $\lVert x-y\rVert\le\epsilon$ are both contained in some $B^\prime_i$, so $\phi_{in}(x)\not=\phi_{in}(y)$ for large enough $n$. Alternatively, if $n$ islarge enough that $\Vert\phi_n(z)-z\rVert < \epsilon/2$ on $V$ then we have $\lVert\phi_n(x)-\phi_n(y)\rVert\ge\epsilon-\lVert x-y\rVert\gt0$ whenever $\lVert x-y\rVert\gt\epsilon$. So, $\phi_n$ is one-to-one for large $n$.

Proof of Lemma 2:
As explained in David Speyer’s answer, Lagrange interpolation gives a sequence of polynomials $q_n$ converging uniformly to $0$ on compacts (as they are of uniformly bounded degree with coefficients tending to zero) such that, for each $z\in K$ and $k\ge1$ with $f^\prime(z)=\cdots=f^{(k)}(z)=0$, then $p_n(z)-q_n(z)=f(z)$ and $p_n^\prime(z)-q_n^\prime(z)=\cdots=p_n^{(k)}(z)-q_n^{(k)}(z)=0$. Then, applying Lemma 1 with $p_n-q_n$ in place of $p_n$ gives the result.

This is an attempt to save George Lowther some time by removing the need for him to prove his Lemma 2.

Lemma Let $f$ be analytic and nonconstant on a neighborhood of the closed disc. Then there is a sequence of polynomials $r_n$, uniformly approaching $f$, such that, if $f'(z)=f”(z)=\cdots f^{(k)}(z)=0$, then $r_n(z)=f(z)$ and $r_n'(z) = \cdots = r^{(k)}_n(z)=0$.

With this Lemma, the result follows directly from his Lemma 1 (whose proof I am still eagerly awaiting.)

Proof: As in George’s answer, use Taylor series to build a sequence of polynomials $p_n$ uniformly converging to $f$.

Since $f$ is non-constant, $f’$ only has finitely many zeroes in the closed disc; say $z_1$, $z_2$, $\cdots$, $z_N$. Let $f'(z_i)=\cdots = f^{(k_i)}(z_i)=0$ for $1\leq{i}\leq{N}$, and let $K = \sum (k_i+1)$. By Lagrange interpolation, there is a unique polynomial $q_n(z)$ of degree $K-1$ such that $p_n(z_i) – q_n(z_i) = f(z_i)$ and $p_n^{(j)}(z_i) – q_n^{(j)}(z_i) = 0$ for $1 \leq j \leq k_i$. We will show that $p_n – q_n$ also uniformly approaches $f$; it is equivalent to show that $q_n$ uniformly approaches $0$.

The coefficients of $q_n$ depend in a linear manner on the $K$ quantities $p_n^{(j)}(z_i) – f^{(j)}(z_i)$, for $0 \leq j \leq k_i$. The coefficients of this linear dependence do not depend on $n$. So it is enough to show that $p_n^{(j)}(z_i) – f^{(j)}(z_i)$ approaches $0$ as $n \to \infty$, for $0 \leq j \leq k_i$.

Let $\gamma_i$ be a small circle around $z_i$, not enclosing any other $z_{k}$. Then
$$ p_n^{(j)}(z_i) – f^{(j)}(z_i) = \frac{j!}{2 \pi i} \oint_{\gamma_i} \frac{p_n(z)- f(z)}{(z-z_i)^{j+1}} dz.$$
Since $p_n \to f$ uniformly, this integral approaches $0$ as $n \to \infty$.

This doesn’t answer your interesting question, but maybe it is relevant and perhaps interesting for you, especially after reading your note about replacing “simply connected” by “finitely connected”, provided “polynomials” is replaced by “rational functions”.

Let $X$ by an $n$-connected domain in the Riemann sphere. Suppose that $X$ is bounded by $n$ disjoint Jordan curves. A holomorphic function $f:X \rightarrow \mathbb{D}$ is called an $n$-to-$1$ proper holomorphic mapping if every point in $\mathbb{D}$ has exactly $n$ preimages in $X$ counting multiplicites, and if $f$ extends to a continuous function of $\overline{X}$ onto $\overline{\mathbb{D}}$ mapping each of the boundary curve of $X$ homeomorphically onto the unit circle.

In this paper, the authors prove the following :

Theorem (factorization theorem)

Every $n$-to-$1$ proper holomorphic mapping $f:X \rightarrow \mathbb{D}$ can be factorized, in $X$, as
$$f=R \circ \phi,$$
where $R$ is a rational function of degree $n$ and $\phi$ is a biholomorphic mapping of $X$ onto $R^{-1}(\mathbb{D})$.

Just a remark : this theorem does not appear explicitely in the paper, but this is hidden in the proof of Theorem $1.2$. The authors prove it for a special $n$-to-$1$ proper holomorphic mapping called the Ahlfors function, but the proof works for every $n$-to-$1$ proper holomorphic mapping.

This is not what you want, but maybe the proof can give some insight about how proving such factorization results.

Let me point out how one may look for a counterexample. Shifting and multiplying by the constants, one may assume that $f(z)=P(\phi(z)),$ $f(0)=\phi(0)=0$ and $\phi'(0)=1.$ Now, $\phi^{-1}(z)$ is well defined analytic injective function on some closed disc around the origin (by Koebe’s theorem this disc has radius $\ge 1/4$). So, $f(\phi^{-1}(z))=P(z)$ on the small closed disc around the origin. Now the right hand side is nice entire function so if we extend $\phi^{-1}(z)$ to its maximal domain of analyticity and take $f$ to have sufficiently bad behaviour along the boundary of the domain of analyticity, we should get a contradiction.

Another solution to the simply connected case (ie disk case) of this problem may be found in my paper on the arxiv here. The idea is the following.

  1. Take the level curves of your function $f$ (technically of $|f|$) and extend them outside of the domain of $f$ in such a way as to form a configuration of level curves corresponding to those of a polynomial.

  2. Find a polynomial $p$ which has this configuration of level curves. (This may be done as a result of my theorem in the final section of this paper, also on the arxiv.

  3. By a theorem in the same paper, any two functions having the same level curve configuration are conformally equivalent. Therefore if we restrict the domain of $p$ appropriately, it will be conformally equivalent to $f$.

Full details on the arxiv.