# 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".

• Essentially the question is whether the non-injectivity of $f$ can be captured by the polynomial. Since we know $f$ can only take on any particular value a finite number of times, this could be true, but I don't have a good idea off the top of my head. – Sharkos Jul 12 '13 at 20:34
• @Sharkos Agreed. That is an interesting way of characterizing the question. – Trevor J Richards Jul 13 '13 at 0:16
• @AlexanderGruber Thanks for the bounty. I will add my own if yours wears off without an answer. – Trevor J Richards Jul 13 '13 at 0:43
• No help yet on the notation "conformally equivalent". Does someone have a better term, or is this the one? – Trevor J Richards Jul 17 '13 at 19:35

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 $n$. QED

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. QED

• Let me just mention the following : If $\Omega$ is any open set in the plane and $f$ is holomorphic in $\Omega$, then we can always approximate $f$ locally uniformly on $\Omega$ by rational functions. This is an easy consequence of Runge's Theorem. – Malik Younsi Jul 14 '13 at 12:11
• As this took a bit longer to write up than I was anticipating, I'll have to come back later today and complete it by adding the proofs of Lemmas 1 and 2. They are not too involved though. For Lemma 1, it is just the inverse function theorem in the case that $f^\prime\not=0$. Where the first $k$ derivatives of $f$ vanish, the same idea applies to $f^{1/k}$. Lemma 2 is just a matter of subtracting out a polynomial term to ensure that the conditions of Lemma 1 are satisfied. – George Lowther Jul 14 '13 at 12:12
• @Malik: Thanks, I added this to the answer and linked to Runge's theorem on wikipedia. – George Lowther Jul 14 '13 at 12:16
• @GeorgeLowther You only need to finish Lemma 1; see my answer. – David E Speyer Jul 17 '13 at 3:18
• I'll add that, in Lemmas 1 and 2, $f$ is required to be nowhere constant (I.e., not constant on any open sets) Not just non-constant. As even very minor edits can take a long time with the general slowness I'm experiencing on this site, I'll leave that fix for now. – George Lowther Jul 18 '13 at 2:51

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 is the first time I have looked at Lagrange interpolation. Glancing at the Wikipedia page, it is a process for finding a poly. whose graph contains the points $(x_0,y_0),\ldots,(x_k,y_k)$, and does not seem to have control of the critical points of said polynomial. How do you guarantee that you have the desired critical points with the desired multiplicities? An explanation would be great but a link would suffice. Thanks – Trevor J Richards Jul 17 '13 at 17:43
• @Trevor Here is a link... the polynomial interpolation process which also matches some derivatives is usually called Hermite interpolation. – 40 votes Jul 20 '13 at 16:19

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.

• Thank you, this is very interesting. I was aware that this result has been proven for simply connected functions and polynomials, but did not know of the full strength above. – Trevor J Richards Jul 14 '13 at 2:06
• Note that the degree $n$ of the proper holomorphic mapping in the theorem is exactly the number of boundary components of $X$. If $X$ is simply connected, then $f$ is a biholomorphic map of $X$ onto $\mathbb{D}$ and in this case the result is trivial : one can take $R$ the identity and $\phi=f$. Which result are you referring to? – Malik Younsi Jul 14 '13 at 2:28
• You are right, I was thinking of a stronger result, that if $f:\mathbb{D}\to\mathbb{D}$ is precisely $n$-to-$1$ for some $n=1,2,\ldots$, then there is an injective map $\phi$ and a polynomial $p$ such that $f\equiv{p}\circ\phi$. As an application of Theorem 3.1 here. – Trevor J Richards Jul 14 '13 at 4:15

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.

• $f=1/(z+2)$ and $\phi= f, P=z$ indicates what usually happens - whatever behaviour $f$ has is imitated by $\phi$. The notion of the non-injective nature of $f$ is the important aspect. – Sharkos Jul 13 '13 at 9:25
• well, what I meant by "bad behaviour" is not a pole (here f is injective, so obviously the claim holds). I was thinking about many essential singularities along the boundary, where the function becomes highly non-injective. – leshik Jul 13 '13 at 17:36
• But $\phi$ is only injective on the open unit disc, which is isolated from non-analyticity of $f$ because the latter is analytic on an open set containing the closed unit disc, so this seems optimistic...? Just thinking out loud! – Sharkos Jul 13 '13 at 17:56

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.