IMC 2007 · Day 2 · P12

For the proof, we need the following

Lemma 1. For any polynomial $g$, denote by $d(g)$ the minimum distance of any two of its real zeros ($d(g) = \infty$ if $g$ has at most one real zero). Assume that $g$ and $g + g'$ both are of degree $k \ge 2$ and have $k$ distinct real zeros. Then $d(g + g') \ge d(g)$.

Proof of Lemma 1: Let $x_1 < x_2 < \dots < x_k$ be the roots of $g$. Suppose $a, b$ are roots of $g + g'$ satisfying $0 < b - a < d(g)$. Then, $a, b$ cannot be roots of $g$, and $$\tag{1} \frac{g'(a)}{g(a)} = \frac{g'(b)}{g(b)} = -1.$$ Since $\frac{g'}{g}$ is strictly decreasing between consecutive zeros of $g$, we must have $a < x_j < b$ for some $j$.

For all $i = 1, 2, \dots, k-1$ we have $x_{i+1} - x_i > b - a$, hence $a - x_i > b - x_{i+1}$. If $i < j$, both sides of this inequality are negative; if $i \ge j$, both sides are positive. In any case, $\frac{1}{a - x_i} < \frac{1}{b - x_{i+1}}$, and hence $$\frac{g'(a)}{g(a)} = \sum_{i=1}^{k-1} \frac{1}{a - x_i} + \underbrace{\frac{1}{a - x_k}}_{< 0} < \sum_{i=1}^{k-1} \frac{1}{b - x_{i+1}} + \underbrace{\frac{1}{b - x_1}}_{> 0} = \frac{g'(b)}{g(b)}$$ This contradicts (1).

Now we turn to the proof of the stated problem. Denote by $m$ the degree of $f$. We will prove by induction on $m$ that $f_n$ has $m$ distinct real zeros for sufficiently large $n$. The cases $m = 0, 1$ are trivial; so we assume $m \ge 2$. Without loss of generality we can assume that $f$ is monic. By induction, the result holds for $f'$, and by ignoring the first few terms we can assume that $f_n'$ has $m - 1$ distinct real zeros for all $n$. Let us denote these zeros by $x_1^{(n)} > x_2^{(n)} > \dots > x_{m-1}^{(n)}$. Then $f_n$ has minima in $x_1^{(n)}, x_3^{(n)}, x_5^{(n)}, \dots$, and maxima in $x_2^{(n)}, x_4^{(n)}, x_6^{(n)}, \dots$. Note that in the interval $(x_{i+1}^{(n)}, x_i^{(n)})$, the function $f_{n+1}' = f_n' + f_n''$ must have a zero (this follows by applying Rolle's theorem to the function $e^x f_n'(x)$); the same is true for the interval $(-\infty, x_{m-1}^{(n)})$. Hence, in each of these $m - 1$ intervals, $f_{n+1}'$ has exactly one zero. This shows that $$\tag{2} x_1^{(n)} > x_1^{(n+1)} > x_2^{(n)} > x_2^{(n+1)} > x_3^{(n)} > x_3^{(n+1)} > \dots$$

Lemma 2. We have $\lim\limits_{n \to \infty} f_n(x_j^{(n)}) = -\infty$ if $j$ is odd, and $\lim\limits_{n \to \infty} f_n(x_j^{(n)}) = +\infty$ if $j$ is even.

Lemma 2 immediately implies the result: For sufficiently large $n$, the values of all maxima of $f_n$ are positive, and the values of all minima of $f_n$ are negative; this implies that $f_n$ has $m$ distinct zeros.

Proof of Lemma 2: Let $d = \min \{ d(f'), 1 \}$; then by Lemma 1, $d(f_n') \ge d$ for all $n$. Define $\varepsilon = \frac{(m-1) d^{m-1}}{m^{m-1}}$; we will show that $$\tag{3} f_{n+1}(x_j^{(n+1)}) \ge f_n(x_j^{(n)}) + \varepsilon \quad \text{for } j \text{ even.}$$ (The corresponding result for odd $j$ can be shown similarly.) Do to so, write $f = f_n$, $b = x_j^{(n)}$, and choose $a$ satisfying $d \le b - a \le 1$ such that $f'$ has no zero inside $(a, b)$. Define $\xi$ by the relation $b - \xi = \frac{1}{m}(b - a)$; then $\xi \in (a, b)$. We show that $f(\xi) + f'(\xi) \ge f(b) + \varepsilon$.

Notice, that $$\begin{align*} \frac{f''(\xi)}{f'(\xi)} &= \sum_{i=1}^{m-1} \frac{1}{\xi - x_i^{(n)}} \\ &= \underbrace{\sum_{i<j} \frac{1}{\xi - x_i^{(n)}}}_{< \frac{1}{\xi - a}} + \frac{1}{\xi - b} + \underbrace{\sum_{i>j} \frac{1}{\xi - x_i^{(n)}}}_{< 0} \\ &< (m-1) \frac{1}{\xi - a} + \frac{1}{\xi - b} = 0. \end{align*}$$ The last equality holds by definition of $\xi$. Since $f'$ is positive and $\frac{f''}{f'}$ is decreasing in $(a, b)$, we have that $f''$ is negative on $(\xi, b)$. Therefore, $$f(b) - f(\xi) = \int_{\xi}^{b} f'(t)\,dt \le \int_{\xi}^{b} f'(\xi)\,dt = (b - \xi) f'(\xi)$$ Hence, $$\begin{align*} f(\xi) + f'(\xi) &\ge f(b) - (b - \xi) f'(\xi) + f'(\xi) \\ &= f(b) + (1 - (\xi - b)) f'(\xi)$$

\\ &= f(b) + \left( 1 - \tfrac{1}{m}(b - a) \right) f'(\xi) \\ &\ge f(b) + \left( 1 - \tfrac{1}{m} \right) f'(\xi). \end{align*} Together with $$f'(\xi) = |f'(\xi)| = m \prod_{i=1}^{m-1} \underbrace{\left| \xi - x_i^{(n)} \right|}_{\ge |\xi - b|} \ge m |\xi - b|^{m-1} \ge \frac{d^{m-1}}{m^{m-2}}$$ we get $$f(\xi) + f'(\xi) \ge f(b) + \varepsilon.$$ Together with (2) this shows (3). This finishes the proof of Lemma 2.