Theorem: If Polynomials Converge, Roots Also Converge

[caffeinemachine]

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.

 

[jvicens]

Hello,

Thank you for sharing your approach to the problem and providing a proof for the lemma. Your use of the inverse function theorem is interesting and provides a different perspective on the problem. However, as you mentioned, your approach requires a more stringent hypothesis and does not cover the case when the roots of $p(t)$ are not all distinct.

To address this issue, we can modify your approach by using the fundamental theorem of algebra, which states that every non-constant polynomial with complex coefficients has at least one complex root. This allows us to extend the result to cover the case when the roots of $p(t)$ are not all distinct. In particular, we can use the fundamental theorem of algebra to prove that the roots of $p_k(t)$ converge to the roots of $p(t)$ as $k$ tends to infinity. This can be done by considering the polynomial $p_k(t)-p(t)$ and showing that it has at least one root.

Furthermore, to cover the complex case, we can use the fact that the inverse function theorem holds for smooth functions in both real and complex variables. This allows us to extend your approach to the complex case by considering the functions $f:\mathbf C^n\to\mathbf C^n$ and $g:\mathbf C^n\to\mathbf C^n$, where $f$ and $g$ are defined in a similar manner as in your proof. We can then use the same argument as in your proof to show that the roots of $p_k(t)$ converge to the roots of $p(t)$ in the complex plane.

Overall, your approach is a creative and interesting way to approach the problem. By incorporating the fundamental theorem of algebra and extending the inverse function theorem to complex variables, we can address the issues you mentioned and extend the result to cover the cases you mentioned.