Polynomials: $f_n \to p$ and $p$ has distinct real roots implies $f_n$ eventually has real roots.

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 $$f_n(x)=x^2+\frac{1}{n}.$$

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."

• 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. – Dionel Jaime May 5 at 16:36
• @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. – Dzoooks May 5 at 17:08

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 $$y_n$$.