Is the set of all monic polynomials in $\mathbb{C}[X]$ with roots in a compact set $K$ compact?

Question

Let $K$ be a compact subset of $\mathbb{C}$ and $T$ a bounded operator in a Hilbert space $H$ (or an element of a unital Banach algebra $B$).

1. Does there exist a norm on $\mathbb{C}[X]$ such that the set of all monic polynomials in $\mathbb{C}[X]$ with roots in the compact set $K$ is compact ?

2. Is the set $\{P(T) : \text{P a monic polynomial in } \mathbb{C}[X] \text{ with roots in the compact set } K \}$ compact in $B(H)$ (or $B$)?

I tried to use the map $\phi$ from $(K^\mathbb{N})_0$ (finite sequences of elements of $K$) to $\mathbb{C}[X]$ (or $B(H)$) defined by $\phi((\rho_n)_{n=1}^m) = \prod_{n=1}^m (X-\rho_n)$ or ($\prod_{n=1}^m (T-\rho_n)$) but I failed to show that it is continuous or that $(K^\mathbb{N})_0$ is compact.

Answer 1

Since $K$ is compact the coefficients of the corresponding polynomials are bounded. On the other side if we consider a converging sequence of polynomials, if their coefficients are close then the corresponding roots are close, which would make the roots of the limit polynomial contained in $K$ due to the compactness of $K$.

EDIT: This would only work if the degree is fixed. Still I'll leave the answer as some might find it useful.

Answer 2

$\def\CC{\mathbb{C}}$Regarding (1): If you want "coefficient of $X^1$" to be a continuous function with respect to your norm, then no. All the polynomials $(X+1)^n$ have roots in the unit disc, so the coefficient of $X^1$ (namely $n$) is not bounded on this set, so this set cannot be compact.

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On the other hand, if $K$ is contained in the closed ball of radius $r$ around $0$, for some $r<1$, and if you include the $0$ polynomial in your set, then I can do it. Specifically, let $f_0 + f_1 X + \cdots + f_{n-1} X^{n-1} + f_n X^n$ be the coefficients of a polynomial in $\mathbb{C}[X]$. Choose a positive real number $s$ with $s<1-r$. My norm will be $$\sum_k s^{2k} |f_k|^2 = \frac{1}{2 \pi} \int_{\theta=0}^{2 \pi} |f(s e^{i \theta})|^2 d \theta. \qquad (\clubsuit)$$ It is important that I include the $0$ polynomial in my set: We have $\lim_{n \to \infty} X^n = 0$. I can't think of anything else that I would want as an accumulation point of the sequence $(X^n)$, so I think this is non-negotiable.

Note that the restriction of $(\clubsuit)$ to the polynomials of degree $\leq n$ gives the standard topology on $\CC^{n+1}$.

Let $S_n \subset \CC[X]$ be the set of degree $n$ monic polynomials with roots in $K$. The map $K^n \to \CC^n$ sending $(\rho_1, \rho_2, \ldots, \rho_n)$ to the coefficients of $\prod (X+\rho_i)$ is continuous, so each $S_n$, being the image of compact set under a continuous map, is compact. We want to show that $\bigcup_n S_n \cup \{ 0 \}$ is compact. To this end, it is enough to show that, if $f^i(X) \in S_{n_i}$ is any sequence of polynomials with $n_i \to \infty$, then $\lim f^i(X) =0$.

This is not hard at all. Let $f(X) = \prod (X+\rho_i)$ be a polynomial of degree $x$. If $z$ is on the circle $|z|=s$ and the $\rho_i$ are in the disc of radius $r$, then we have $$|f(z)|^2 \leq \prod_i (|z|+|\rho_i|)^2 = |s+r|^{2n}.$$ So the average value of $|f|^2$ on the circle of radius $s$ is bounded by $|s+r|^{2n}$. As we took $s<1-r$, this goes to $0$ as $n \to \infty$.

Answer 3

Let $\tilde K$ be the set of all monic polynomials in $\mathbb{C}[X]$ with roots in the set $K$. Let $r>1$ be any number. Put $B_r=\{z\in\mathcal C:|z|\le r\}$. Let $P\in \mathbb{C}[X]$ be any polynomial. Let $d$ be degree of $P$ and $M$ be the maximal absolute value of a coefficient of $P$. Pick any natural $n$ such that $r^n>M(d+1)$. Pick $Q(z)=z^{n+d}+P(z)$. Let $z\in\mathbb C\setminus B_r$ be any complex number. Then $$|Q(z)|\ge |z|^{n+d}- M(d+1)|z|^d=|z|^d(|z|^n-M(d+1))\ge |z|^d(r^n- M(d+1))>0.$$ So all roots of $Q$ are in $B_r$, that is $Q\in \widetilde{B_r}$ and $P\in \widetilde{B_r}-z^{n+d}$.

Suppose that $\widetilde{B_r}$ is compact. Then $\mathbb{C}[X]$ is covered by countably many compact sets. Suppose for a contradiction that $\mathbb{C}[X]$ is complete. By the Baire category theorem, $\mathbb{C}[X]$ contains a compact set with nonempty interior. So $\mathbb{C}[X]$ is a locally compact normed space. Then $\mathbb{C}[X]$ is finitely dimensional (see, for instance, the beginning of this my answer), a contradiction.

Maybe a similar construction can be done for Banach algebras.

Unfortunately, when $\mathbb{C}[X]$ then there is no contradiction similarly the case of complete $\mathbb C[X]$, because there exist $\sigma$-compact infinitely dimensional normed spaces, for instance, the subspace of $\ell^2$ consisting of the sequences which are eventually zero. (2)

2. Let $(B,\|\cdot\|)$ be a Banach algebra, $T\in B$, and $$\widetilde{K}=\{P(T):P \mbox{ is a monic polynomial in } \mathbb{C}[X] \text{ with roots in } K \}.$$

Then $\widetilde{K}$ is a topological semigroup with respect to the multiplication. Let $R$ be any element of $\widetilde{K}$. For each natural $n$ put $[R_n]=\{R^m:m\in\mathbb N,\,m\ge n\}$. Then $\bigcap_{n=1}^\infty\overline{[R_n]}$ is a compact group, see [Num] and [Sch], so it contains an idempotent $e$. So $e\in\widetilde{K}$.

Suppose that $\|e\|<1$. If $\|e\|>0$ then $\|e\|=\|e\|\cdot \|e\|<\|e\|$, which is impossible. So $e=0$. On the other hand, $e\in\widetilde{K}$ so either there exists a monic polynomial $P\in\mathbb{C}[X]$ with roots in $K$ such that $P(T)=e=0$ or we consider the zero of $\mathbb{C}[X]$ as a polynomial with roots in $K$.

References

[Num] Katsumi Numakura, On bicompact semigroups, Mathematical Journal of Okayama University 1 (1952) 99-108.

[Sch] Štefan Schwarz, K teorii Hausdorfovyh bicompaktnyh polugrupp *(On Hausdorff bicompact semigroups), Czechoslovak Mathematical Journal, 5:80 (1955) 1-23.