# When a sequence of coefficients converges to the coefficients of a rational function $R$, when does the sequence $R_n$ converge uniformly to $R$?

Let $R$ be a rational function of degree $d$ mapping the Riemann sphere to itself:$$R(z) = \frac{a_d z^d + a_{d-1} z^{d-1} + \cdots + a_0}{b_d z^d + b_{d-1} z^{d-1} + \cdots + b_0}$$ where $a_d$ and $b_d$ are not both zero. And suppose that a sequence of coefficients $\{(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n\}$ converges to the coefficients of $R$ in $\mathbb{C}^{2d+2}$. Let $R_n$ be the rational function with coefficients $(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n$.

Is it a theorem that: The sequence $\{R_n\}$ converges uniformly to $R$ on the Riemann sphere if and only if for all $n$ sufficiently large the degree of $R_n$ matches the degree of $R$?

I ask because I am reading Beardon's Iteration of Rational Functions and in section 2.8 he introduces a mapping $\Psi:\mathcal{R}\to\mathbb{C}\mathbb{P}^{2d+1}$, which takes a rational function to its coefficients. Beardon omits the details, but claims that $\Psi$ is a homeomorphism of $\mathcal{R}_d$ onto $\Psi(\mathcal{R}_d)$, where $\mathcal{R}_d$ is the space of rational functions having degree exactly $d$.

My thought is, if for $n$ sufficiently large $R_n$ has degree $d$, then $(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n \in \Psi(\mathcal{R}_d)$. Now $\Psi^{-1}$ is defined and continuous at $(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n$, because $\Psi$ is a homeomorphism of $\mathcal{R}_d$ onto $\Psi(\mathcal{R}_d)$. And (up to multiplication by a constant) $R_n = \Psi^{-1}((a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n)$. Since $\Psi^{-1}$ is continuous on the sequence of coefficients and at the limit of the sequence of coefficients, $R_n$ converges to $R$. But I would like to know:

Is the convergence always uniform under these circumstances?

Conversely, if for all $n$ there exists $R_{m>n}$ in the sequence with degree not equal to $d$, then $\{R_n\}$ cannot converge to $R$ because (Beardon also states that) the spaces $\{\mathcal{R}_0, \mathcal{R}_1, \ldots, \mathcal{R}_d\}$ are the connected components of the space of rational functions with degree at most $d$.

The reason that I think this might be a theorem is that Beardon introduces a metric $\rho(R,S) = \sup_{z\in\mathbb{C}_\infty} \sigma_0 (R(z),S(z))$, $\sigma_0$ the chordal metric on $\mathbb{C}_{\infty}$, just a couple of pages earlier than the point where he introduces the mapping $\Psi$. It is tempting to read "$\Psi$ is a homeomorphism of $\mathcal{R}_d$ onto $\Psi(\mathcal{R}_d)$" as "$\Psi$ is a homeomorphism of the metric space $(\mathcal{R}_d,\rho)$ onto $\Psi(\mathcal{R}_d)$." If that is the correct reading then $R_n$ converges to $R$ under the metric $\rho$ and the convergence is clearly uniform. But I am not sure that this is the correct reading since Beardon is only explicitly treating $\mathcal{R}_d$ and $\mathbb{C}\mathbb{P}^{2d+1}$ as topological spaces at the point where he introduces $\Psi$.

Note: I have edited the question in response to Margaret Friedland's answer, in an attempt to clarify my assumptions (and hopefully my source of confusion). I am a little worried that the reason my point of confusion was unclear is that it is actually trivial. I apologize in advance if that is the case.

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Except in trivial cases, your rational function $\bf R$ has a pole somewhere, and generally the poles of $\bf R_n$ will be different. How can it converge uniformly in a neighbourhood of the pole? Or are you thinking of rational functions as mappings of the Riemann sphere to itself? – Robert Israel Jul 24 '11 at 20:36
I am thinking of rational functions as mapping the Riemann sphere to itself. I forgot to state that in the question. I will edit the question to clarify the setting. – Aaron Golden Jul 25 '11 at 3:51
I think you need to clean up your assumptions a little bit, especially at the beginning of the question, and things will become clearer. E.g. Are you considering the coefficients in projective space or not? Do you assume that the limit has degree exactly d or at most d? Margaret Friedland's answer suggests that the answer may change depending on the specific assumptions. – Thierry Zell Jul 26 '11 at 14:36
Thierry, do my assumptions seem more clear now? I am tempted to completely rewrite the question to try to simplify/clarify it, but I don't know if that is appropriate since there are already comments and an answer that depend on the original version of the question for context. – Aaron Golden Jul 28 '11 at 1:55

Since you reformulated your question, it deserves a new answer. Now, the theorem in the gray box in the new version is true, but what does it really say? It says that the map $\Psi^{-1}: \Psi(\mathcal{R}_d) \mapsto \mathcal{R}_d$ is continuous! (The topology on $\mathcal{R}_d$ is inherited from $\mathbb{P}^{2d+2}$; on $\Psi(\mathcal{R}_d)$ we have, as you state, the topology of uniform convergence with respect to e.g. chordal metric). So what you are really asking is: how to prove that $\Psi^{-1}$ is continuous in these topologies? This is not so hard if you are used to working with homogeneous coordinates on projective spaces, but there is a subtle point involved. Here is the argument: Let $[r:Z]=[a_0:...:a_d:b_0:...:b_d:Z_0:Z_1] \in \mathbb{P}^{2d+2}\times \mathbb{P}^1$ and define $p([r:Z] \in \mathbb{P}^1$ as $p([r:Z]=[a_0{Z_0}^d+a_1{Z_0}^{d-1}Z_1+...+a_d{Z_1}^d:b_0{Z_0}^d+b_1{Z_0}^{d-1}Z_1+...:b_d{Z_1}^d]$. Then the map $p$ is continuous (being given by homogeneous polynomials in homogeneous coordinates), so as $[r_n:Z^{(n)}] \to [r,Z]$, we have $p[r_n:Z^{(n)}] \to p[r:Z]$. Note that $p[r:Z]$ is the value of the rational function $R$ with the coefficients $r$ at the point $[Z_0:Z_1]\in \mathbb{P}^1$. Note also that the continuity of $p$ means that $R_n$ converge continuously to $R$ on $\mathbb{P}^1$. On a compact space , continuous convergence of a sequence of continuous functions is equivalent to its uniform convergence.

The subtlety is the following: the domain of the map $p$ is $\mathcal{R}_d \times \mathbb{P}^1$, not the whole product $\mathbb{P}^{2d+2}\times \mathbb{P}^1$. (The ratio of two polynomials is not defined at their common zeros. We saw this in Beardon's counterexample: the limit expression $z/z$ (or $[Z_1:Z_1]$) in the limit is not defined at the point $0=[1:0]$.) So the assumption that all $R_n$ and $R$ have the same degree $d$ guarantees that there are no factors in common for each numerator and denominator and we do not consider fractional expressions that are not rational functions on $\mathbb{P}^1$.

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Thank you very much for this new answer. The relationship between the sequence of coefficients and the sequence of rational functions is clear when everything is spelled out explicitly with homogeneous coordinates, as you've done here. – Aaron Golden Aug 11 '11 at 9:45
I am glad it helps. Homogeneous coordinates are very useful indeed. And even if your interest is in one-dimensional complex dynamics, it pays to have some insight into higher-dimensional setting. – Margaret Friedland Aug 11 '11 at 13:44

The equivalence you state in your first question cannot be a theorem, because the "if" part is false (take $R_n(z):=(n+1)z, R_n(\infty)=\infty), n \geq 1$ and $R(z)=z$; the degrees certainly match, but $R_n$ does not converge to $R$, even pointwise). The "only if" part is true and can be proved using Rouche's theorem (Beardon does include some detail on this).

The convergence of rational functions in the sense of coefficients does not imply uniform convergence: let $R_n(z)=(z+1/n)/z, R(z)=1$; the degree drops for the limit function (this example is right there in Beardon).\

Edit: And of course all $R_n$ have a pole at $0$, while $R$ does not (as Robert Israel suggested) so this is the true reason why uniform convergence does not hold. Still, $(1/n,1,0,1)$ converge to $(0,1,0,1)$.

The map between rational functions and their coefficients that Beardon introduces is not the same as the map you wrote above, if by $\mathcal{R}$ you denote the space of all rational functions of degree at most $d$. Your map cannot have a continuous inverse, since the projective space is connected and the space $\mathcal{R}$ is a union of non-trivial connected components $\mathcal{R}_j$ , of functions of degree exactly $j$, $j=0,...,d$.

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Thank you for the answer. I see now that my question was unclear. I mean to require that: (1) $R$ has degree exactly $d$. (2) $R_n$ is the sequence of functions associated with a sequence of coefficients converging to the coefficients of $R$ in $\mathbb{C}^{2d+2}$. And (3) For sufficiently large $n$ the degree of $R_n$ is exactly $d$. Then your first sequence is not a counterexample because the coefficients of $R_n$ do not converge to the coefficients of $R$, and in the second example, as you and Beardon both point out, the degree drops in the limit, violating (3). – Aaron Golden Jul 26 '11 at 16:40

After rereading section 2.8 in Iteration of Rational Functions I am convinced that when Beardon writes "$\Psi$ is a homeomorphism of $\mathcal R_d$ onto $\Psi(\mathcal R_d)$," he means $\mathcal R_d$ to be the space of rational functions with degree exactly $d$, with the topology induced by the metric: $$\rho(R,S) = \sup_\{z\in\mathbb C_\infty\} \sigma(R(z), S(z))$$ where $\sigma$ is the spherical metric on $\mathbb{C}_\infty$ (but Beardon also says that he could equally well have used the chordal instead of the spherical metric because the two differ only by a bounded factor, and I'll make $\sigma$ the chordal metric in what follows). Beardon explicitly treats $\mathcal{R}$ as a metric space (with metric $\rho$) in his proof that the degree function, $\deg:\mathcal{R}\to\mathbb{Z}$, is continuous, and to show that $\mathcal R_d$ is an open subset of $\mathcal R$. He couldn't possibly mean any other topology by the time he introduces $\Psi$. So

If $R$ is a rational function of degree exactly $d$ mapping $\mathbb{C}_\infty$ to itself and $R_n$ converges to $R$ in the sense of coefficients, and for all $n$ sufficiently large $\deg{R_n} = d$, then $R_n$ converges to $R$ uniformly on $\mathbb{C}_\infty$ under the spherical (or equivalently the chordal) metric.

Proof: Let $\Theta = \Psi(R)$ be the coefficients of the limit function $R$, and let $\Theta_n = \Psi(R_n)$. Since $R_n$ converges to $R$ in the sense of coefficients $\Theta_n$ converges to $\Theta$. Since both $R$ and $R_n$ (for sufficiently large $n$) have degree exactly $d$, $\Theta\in\Psi(\mathcal{R}_d)$ and $\Theta_n\in\Psi(\mathcal{R}_d)$ for sufficiently large $n$.

Now, $\Psi$ is a homeomorphism of the topological space $\mathcal{R}_d$ with the topology induced by the metric $\rho$ onto $\Psi(\mathcal{R}_d)$. So $\Psi^{-1}$ is defined and continuous at $\Theta$ and at $\Theta_n$ for large $n$, and $R = \Psi^{-1}(\Theta)$ and $R_n = \Psi^{-1}(\Theta_n)$, so $R_n$ converges to $R$ with respect to the metric $\rho$. Since $\rho$ is a metric of uniform convergence on $\mathbb{C}_\infty$, the convergence of $R_n$ is uniform on $\mathbb{C}_\infty$. QED.

## Issues with poles of $R$ and $R_n$

Some of the responses to this question have indicated that the convergence cannot be uniform in the neighborhood of a pole of $R$ when the $R_n$ do not share that pole. It is true that if $z$ is a pole of $R$ and no pole of $R_n$ approaches $z$, then $R_n$ does not converge to $R$ at $z$. For if $R(z) = \infty$ and $\left| R_n(z)\right| < M < \infty$ for all $n$, then as $n\to\infty$, using the chordal metric: $$\sigma(R_n(z), R(z)) = \sigma(R_n(z), \infty) = \frac{2}{(1 + \left| R_n(z) \right|^2)^{1/2}} > \frac{2}{(1 + M^2)^{1/2}} \not\to 0.$$ And the same calculation works if all the $R_n$ share a pole not shared by $R$. Margaret Friedland's answer gives an explicit example of this. If $R_n(z) = (z + 1/n)/z$, $R(z) = 1$, then $\sigma(R_n(0), R(0)) = \sigma(\infty, 1) = 2 / \sqrt{2}$.

But what if the poles of $R_n$ do not match those of $R$, but approach them? Then $R_n$ may converge uniformly to $R$ on the sphere. We can show this directly with an example. Let $R(z) = 1/z$ and $R_n(z) = 1/(z - 1/n)$. Then, under the chordal metric, for $z\neq 0$: $$\sigma(R_n(z),R(z)) = \frac{2 \left| 1/(z - 1/n) - 1/z \right|}{(1 + \left| 1/(z-1/n)\right|^2)^{1/2} (1 + \left| 1/z \right|^2)^{1/2}}$$ simplifies (try running it through WolframAlpha) to: $$2 / ((|z|^2 + 1)^{1/2} (|n z - 1|^2 + n^2)^{1/2})$$ which is less than or equal to $2/n$ for $n>0$. And when $z=0$ we have $$\sigma(R_n(0),\infty) = \sigma(-n,\infty) = 2/\sqrt{1 + |n|^2} \leq 2/n.$$ So for any given $\epsilon>0$, choose $n>2/\epsilon$, then $\rho(R_n, R) = \sup_{z\in\mathbb{C}_\infty} \sigma(R_n(z), R(z)) \leq 2/n < \epsilon$.

The discussion of above indicates that the requirement that $\deg R_n = \deg R$ for all $n$ sufficiently large is actually a requirement that all the poles of $R_n$ converge to the poles of $R$. (Beardon even hints at this, it turns out. He mentions "sliding" the roots and poles of one function to another, when the degrees match, as a way to show that each $\mathcal{R}_d$ is connected.)

## Summary

1. Convergence of the coefficients, on its own, is not even a sufficient condition for point-wise convergence of $R_n$ to $R$ on the sphere. In Margaret Friedland's example, $R_n(z) = (z + 1/n)/z$, $R(z) = 1$ and the coefficients $(1/n, 1; 0, 1)$ converge to $(0, 1; 0, 1)$, but $R_n(0)$ is bound away from $R(0)$ for all $n$.

2. In general if the poles of $R_n$ fail to converge to the poles of $R$ then $R_n$ does not converge to $R$ at some pole of $R$.

3. If the poles of $R_n$ do converge to the poles of $R$ then it is possible that $R_n$ converges uniformly to $R$ on the sphere (for example when $R(z) = 1/z$ and $R_n(z) = 1/(z - 1/n)$).

4. If the coefficients converge and we require that $\deg R_n$ matches $\deg R$ for all $n$ sufficiently large, then $R_n$ converges uniformly to $R$ on the sphere (based on the discussion of Beardon's mapping, $\Psi$).

5. (2) and (4) together yield: If the coefficients of $R_n$ converge to the coefficients of $R$ and $\deg R_n$ matches $\deg R$ for all $n$ sufficiently large, then all the poles of $R_n$ converge to poles of $R$.

I don't think that any of these points contradict Margaret Friedland's answer and I hope that this answer makes my original question and thinking a little more clear. I would still like to have a direct proof for item (4). It is unsatisfying to rely on Beardon's $\Psi$ mapping, even assuming that I have interpreted the text correctly, since Beardon's discussion of $\Psi$ is so light on details.

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