If a Sequence of Polynomials Converge to Another Polynomial Then the Roots Also Converge.

Proposition 5.2.1 in Artin states that:

THEOREM. Let $p_k(t)\in \mathbf C[t]$ be a sequence of monic polynomials of degree $\leq n$, and let $p(t)\in \mathbf C[t]$ be another monic polynomial of degree $n$. Let $\alpha_{k,1},\ldots,\alpha_{k,n}$ and $\alpha_1,\ldots,\alpha_n$ be the roots of these polynomials. If $$\lim_{k\to\infty}p_k=p$$ then the roots $\alpha_{k,\nu}$ of $p_k(t)$ can be numbered in such a way that $$\lim_{k\to\infty}\alpha_{k,\nu}=\alpha_{\nu}, \quad \forall \nu=1,\ldots,n.$$

There is a simple proof given in Artin. I approached the problem using the inverse function theorem and for that I required a more stringent hypothesis to prove the theorem. Under this new stringent hypothesis, I was able to say something more(see the boxed part i the Lemma below) about the roots of the polynomials $p_k(t)$.

LEMMA. Let $p_k(t)\in\mathbf R[t]$ be a sequence of monic polynomials of degree no more than $n$, and let $p(t)\in\mathbf R[t]$ be another monic polynomial of degree $n$. Let $\alpha_{k,1},\ldots,\alpha_{k,n}$ and $\alpha_1,\ldots,\alpha_n$ denote the roots of the polynomials $p_k(t)$ and $p(t)$ over the field $\mathbf C$. Assume that $\alpha_1,\ldots,\alpha_n$ are real and pairwise distinct. Then if $$\lim_{n\to\infty}p_k\to p$$ There exists $K\in\mathbf N$ such that $$\boxed{k\geq K \quad \Rightarrow \quad \alpha_{k,1},\ldots,\alpha_{k,n}\text{ are real and pairwise distinct.}}$$ Further, $\alpha_{k,1},\ldots,\alpha_{k,n}$ can be numbered in such a way that $$\lim_{k\to\infty}\alpha_{k,\nu}=\alpha_{\nu}, \quad \forall \nu=1,\ldots,n.$$

(The Problem) The proof of this is presented below. I am curious as to how to adjust for the case when the roots of $p(t)$ are not all distinct (This leads to a singular point of the function $f$ defined below not allowing to use the inverse function theorem) and also how to cover the complex case using this approach.

PROOF. For each $r\in\{1,\ldots,n\}$, define $S_r:\mathbf R^n\to\mathbf R$ as $$S_r(x_1\ldots,x_n)=(-1)^r\left[\sum_{1\leq i_1<\cdots<i_r\leq n}\left(\prod_{j=1}^l x_{i_j}\right)\right]$$ ($S_r$ is the $r$-th symmetric function on the letters $x_1,\ldots,x_n$). Now define a function $f:\mathbf R^n\to\mathbf R^n$ as $$f(\mathbf x)=(S_1(\mathbf x),\ldots,S_n(\mathbf x)), \quad \forall \mathbf x\in \mathbf R^n$$ where $\mathbf x=(x_1,\ldots,x_n)$. It can be shown that: \begin{equation*} |\det(Df(x_1,\ldots,x_n))|=\left|\prod_{1\leq i<j\leq n}(x_i-x_j)\right|, \quad \forall (x_1,\ldots,x_n)\in\mathbf R^n \tag{1} \end{equation*} So by hypothesis, $\det(Df(\alpha_1,\ldots,\alpha_n))\neq 0$. So by the Inverse Function Theorem, there exists a neighborhood $U$ of $(\alpha_1,\ldots,\alpha_n)$ in $\mathbf R^n$ such that $f|_U:U\to f(U)$ is a smooth diffeomorphism. Write $V=f(U)$ and say $g:V\to U$ be the inverse of $f|_U$. Let $$[t^r](p_k(t))=a_r^k,\quad \forall k\in\mathbf N, 0\leq r\leq n-1$$ and $$[t^r](p(t))=a^r, \quad \forall 0\leq r\leq n-1$$ where $[t^r]f(t)$ denotes the coefficient of the $r$-th power of $t$ in the expression of $f(t)$. For each $k\in\mathbf N$ write $\mathbf a^k=(a_{n-1}^k,\ldots,a_0^k)$ and also write $\mathbf a=(a_{n-1},\ldots,a_0)$. Note that $\lim_{k\to\infty}p_k=p$ says nothing but $\lim_{k\to\infty}\mathbf a^k=\mathbf a$. Thus there exists $K\in\mathbf N$ such that $$k\geq K \quad \Rightarrow \quad \mathbf a^k\in V$$ We claim that for each $k\geq K$, $p_k(t)$ has $n$-distinct pairwise distinct real roots. To see this fix $k\geq K$ and let $g(\mathbf a^k)=(\beta_{k,1},\ldots,\beta_{k,n})=\boldsymbol \beta_k$. Then Since $\boldsymbol \beta_k\in U$, we have $\det Df(\boldsymbol\beta_k)\neq 0$. Thus by (1) $\beta_{k,1},\ldots,\beta_{k,n}$ are pairwise distinct. Also since $S_r(\boldsymbol \beta_{k})=a^k_{n-r}=[t^{n-r}](p_k(t))$,we see that $\beta_{k,1},\ldots,\beta_{k,n}$ are all the roots of $p_k(t)$. So our claim is proved. Now we will show that the roots $\alpha_{k,1},\ldots,\alpha_{k,n}$ of $p_k(t)$ can be numbered in such a way that the $\lim_{k\to\infty}\alpha_{k,\nu}=\alpha_{\nu}$. But this is clear since by continuity of $g$, we have $$\lim_{k\to\infty}\beta_{k,\nu}=\alpha_\nu, \quad \forall 1\leq\nu\leq n$$ Noting that $\beta_{k,\nu}$'s are a permutation of $\alpha_{k,\nu}$'s, we are done.

• What chapter is this in? It's from "Algebra" correct? – ClassicStyle Sep 30 '15 at 17:03
• Yes. It is prop. 5.2.1 so it must be chapter 5. I do not have the book right now. It was from the second edition. – caffeinemachine Sep 30 '15 at 18:55