On continuity of roots of a polynomial depending on a real parameter

Question

Problem

Suppose $f^{(t)}(z)=a_0^{(t)}+\dotsb+a_{n-1}^{(t)}z^{n-1}+z^n\in\mathbb C[z]$ for all $t\in\mathbb R$, where $a_0^{(t)},\dotsc,a_{n-1}^{(t)}\colon\mathbb R\to\mathbb C$ are continuous on $t$. Is it true that there is always a complex-valued continuous function $\phi^{(t)}\colon\mathbb R\to\mathbb C$ such that $f^{(t)}(\phi^{(t)})=0\;(\forall t\in\mathbb R)$?

Topologically,

Let $S=\big\{\,(a_0,\dotsc,a_{n-1},z)\in\mathbb C^{n+1}\,\big\vert\,a_0+a_1z+\dotsb+a_{n-1}z^{n-1}+z^n=0\,\big\}$ and $\pi\colon S\to\mathbb C^n,(a_0,\dotsc,a_{n-1},z)\mapsto(a_0,\dotsc,a_{n-1})$. Is it true that $\pi$ has path lifting property (i.e. for each continuous map $p\colon[0,1]\to\mathbb C^n$, there exists a continuous map $\tilde p\colon[0,1]\to S$ such that $p=\pi\circ\tilde p$? (Sorry, I cannot find a good reference for that term. In my definition, there's no assumption of uniqueness.)

Or algebraically,

Let $R=\mathcal C(\mathbb R,\mathbb C)$ denote the ring of complexed-valued real continuous functions. Is it true that any monic polynomial over $R$ has a root in $R$?

Discussion

It's certainly true that there is a function $\phi_{t_0}^{(t)}$ continuous at $t=t_0$ such that $f(\phi_{t_0}^{(t)})=0\;(\forall t\in\mathbb R)$, no matter whether $t$ is a real or complex parameter, or a parameter from some Hausdorff space. It follows directly from, say, Rouché's theorem. For an elementary proof, see Michael Artin's Algebra, proposition 5.2.1(b). A sharper proposition of the original problem, i.e. replacing the real parameter $t\in\mathbb R$ with a complex parameter $w\in\mathbb C$, is generally demonstrably false, i.e. there could be no continuous function to be a root of the polynomial. Here's a simple counterexample: $f^{(w)}(z)=z^2-w$. Note that there's a branch point at $w=0$, and if we draw an arbitrary circle around the origin in the $w$-plane, we'll see that a root shouldn't be continuous on the whole $w$-plane $\mathbb C$, for otherwise letting $w$ travel the circle would lead to a contradiction.

It seems true when $t$ is a real parameter, since the dimension is lower. I have no idea how to attack this. I expect your grateful ideas or hints. Thanks!

Postscript

There's an old post related, inequivalent but informative and interesting.

Answer 1

Let $Pol_n$ denote the space of degree $n$ monic polynomials. You have the natural continuous map $R: Pol_n \to Q= {\mathbb C}^n/S_n$ sending each polynomial to its set of (unordered) roots; here $S_n$ is the permutation group on $n$ letters. Let $q: {\mathbb C}^n\to Q$ denote the quotient map. The key observation is that the map $q$ has the path-lifting property (no uniqueness of the lift is assumed, only existence). This is a special case of the path-lifting property for orbifold-coverings, Lemma 4.1.3 here, or Lemma 2 in

M. Armstrong, The fundamental group of the orbit space of a discontinuous group. Proc. Cambridge Philos. Soc. 64 (1968) 299–301.

Note that Armstrong works even in greater degree of generality, namely with proper (but not, in general, free) discrete group actions on locally compact metrizable topological spaces.

Now, we can prove the path-lifting property you are asking for. Take a map $f: {\mathbb R}\to Pol_n$, compose it with $R$. The result is a map $$ g: {\mathbb R}\to Q. $$ Applying the above path-lifting property to $g$ we obtain a lift $$ \tilde g: {\mathbb R}\to {\mathbb C}^n. $$ Let $\tilde g_1$ denote the first component of this lift (the "1st root" of the polynomial $f(t)$). Then the map $$ \tilde f: t\mapsto (f(t), \tilde g_1(t))\in Pol_n \times {\mathbb C} $$ is the lift you want.

Edit: Armstrong's proof depends on Theorem 2 (path lifting property for "open light maps") in

E.E. Floyd, "Some characterizations of interior maps". Ann. of Math. 51 (1950), 571-575.

Armstrong simply observes that quotient maps like $q$ in your case, are "light and open": Open is clear (since it is a quotient map), "light" follows from the fact that point preimages are discrete (finite in your case).

Hence, Floyd's theorem applies. (One needs a tiny compactness argument since Floyd assumes that the domain and the range are compact, but the path-lifting property is a purely local issue, so it works for locally compact spaces.)

Floyd's paper is available (for free) via Jstor, once you open a free account with them.

Answer 2

Isn't a consequence of the continuity of the roots respect of the coefficients?

http://www.ams.org/journals/proc/1987-100-02/S0002-9939-1987-0884486-8/S0002-9939-1987-0884486-8.pdf

Possible problem: the "space of the roots" isn't ${\Bbb C}$, but a quotient.

Answer 3

There are two reasons that $\pi$ doesn't have the path-lifting property.

First, if $a_n$ is allowed to vanish, one of the $n$ complex roots (or more, if more coefficients vanish simultaneously) will go off to infinity. For example, consider the polynomial $P_t(z) = tz + 1$ : when $t \neq 0$, it has a unique root $z = -1/t$, but when $t = 0$, $z$ diverges to infinity. This problem can be solved by either restricting $a_n$ to nonzero values, or extending the range of $z$ to the Riemann sphere $\Bbb C \cup \{ \infty \}$

Secondly, the path may not be unique. This happens when you pass through a polynomial that has a double (or triple or more ...) root. For example, if $P_t(z) = z^2 + t$, if you start at $t= -1$ and $z(-1)= 1$, you have a unique path only up to $t=0$. Then you have two choices : you can have (for $t > 0$) either $z(t) = +\sqrt t i $, either $z(t) = - \sqrt t i$. You can solve this problem by restricting your polynomials to polynomials whose discriminant doesn't vanish.

If you stick with polynomials with nonzero discriminant, essentially nothing bad can happen. You should be able to apply some version of the inverse function theorem to a simply connected neighbourhood of $P$ to show that the roots of $P$ vary continuously in the coefficients and unambiguously on that neighbourhood.

Answer 4

I think I have an operator-theoretic formulation of your problem:

Given an $n \times n$ matrix-valued function $B(t), \ t\in \mathbb{R}$ whose entries are continuous, complex-valued functions on $\mathbb{R}$, is there a continuous function $\lambda(t), \ t \in \mathbb{R}$ such that $\lambda(t)$ is an eigenvalue of $B(t)$ for all $t \in \mathbb{R}$?

This is equivalent to your analytic question as every characteristic polynomial of such $B(t)$ is of the form of your function $f^{(t)}$, and every polynomial of the RHS of your equation can be interpreted as a characteristic polynomial of a companion matrix. This formulation of the question seems to be answered in an older question: Eigenvalues of matrix with entries that are continuous functions.

Further results in that direction can be found in T. Kato, Perturbation theory of linear operators, Chapter Two, §5.