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While reading this paper (top of page 2), I was confused by the following statement. (I will state it in a generality.)

Recall that a polynomial is hyperbolic if all its roots are real.

Let $p$ and $\{f_n\}_{n \geq 1}$ be polynomials of degree $d$. Suppose $\lim_{n \to \infty} f_n(x) = p(x)$ where the limit is taken pointwise with respect to $x$. Prove that if $p$ is hyperbolic and has distinct roots, then there exists an $N$ such that for all $n \geq N$, the polynomials $f_n$ are hyperbolic.

It is false if $p(x)$ does not have distinct roots, for example $f_n(x)=x^2+\frac{1}{n}.$

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We assume the limit converges uniformly in compact subsets of $\mathbb{R}$. Pick $[a,b]$ large enough so that it contains the $d$ roots of $p$. Note that $p(x)$ crosses the $x$-axis $d$ times because its roots are distinct. We can pick $\epsilon$ small enough so that for $n \geq N_{\epsilon}$ and $|f_n(x)-p(x)| < \epsilon$ for $x \in [a,b]$, the function $f_n$ crosses the $x$-axis at least $d$ times (near the roots of $p$). Since deg$(f_n)=d$, this implies $f_n$ has $d$ distinct roots for $n$ greater than some $N_{\epsilon}$.


We cannot weaken "$p$ has distinct roots" to "$p$ has odd multiplicity roots" because of $f_n(x)=x\left(x^2+\frac{1}{n}\right).$

Since $f_n$ and $p$ are uniformly continuous on $[a,b]$, we get uniform convergence automatically, so this is not an "extra assumption."

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  • $\begingroup$ While I think it's intuitively clear that the $f_n$ crosses the $x$- axis at least $d$ times nears the roots of $p$, I think it requires some amount of justification. $\endgroup$ – Dionel Jaime May 5 at 16:36
  • $\begingroup$ @DionelJaime The $f_n$ are continuous, so intermediate value theorem + details. It's very easy to see. I don't want to clutter my answer with the details, but I'm happy to send them to you if you'd like. You may also view robjohn's answer. $\endgroup$ – Dzoooks May 5 at 17:08
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Let $\{x_j\}_{j=1}^n$ be the roots of $p$ (where $x_j\lt x_{j+1}$). Let $y_0=x_1-1$, $y_n=x_n+1$ and $y_j=\frac{x_j+x_{j+1}}2$ for $1\le j\le n-1$.

There is an $N$ so that for $k\ge N$, $\left|\,f_k(y_j)-p(y_j)\,\right|\le\frac12\left|\,p(y_j)\,\right|$ for $0\le j\le n$. This means the sign of $f_k(y_j)$ is the same as that of $p(y_j)$.

Then, for $k\ge N$, $f_k$ changes sign $n$ times between $y_0$ and $y_n$.

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