Proving that the set of $a$ for which the roots of $P(X)-a$ are all real is an interval

Question

I found this problem on a problem sheet about polynomials:

Let $P \in \mathbb{R}[X]$, $S_a = \{t \in \mathbb{C} \mid P(t) = a\}$, and $I = \{a \in \mathbb{R} \mid S_a \subset \mathbb{R}\}$. Show that $I$ is an interval.

I can see intuitively why this must be true: if we take a polyomial $P$ such that all the roots of $P'$ are real and simple, then the graph of $P$ shows that this interval is exactly the interval between the smallest maximum of $P$ and the largest minimum of $P$. I tried to show that this holds, but I couldn't.

Answer 1

$\def\R{\mathbb{R}}\def\ge{\geqslant}\def\emptyset{\varnothing}\def\le{\leqslant}\def\paren#1{\left(#1\right)}\def\={\mathrel{\hphantom{=}}}$Lemma: For any $Q(x) \in \R[x] \setminus \{0\}$ and $a \in \R^*$, $x^3 Q(x) - a$ does not split in $\R[x]$.

Proof: Suppose$$ x^3 Q(x) - a = c\prod\limits_{k = 1}^n (x - x_k), $$ where $c \in \R^*$, $x_1, \cdots, x_n \in \R$. Comparing the coefficients of the lowest three terms yields:\begin{gather*} \begin{cases} \prod\limits_{k = 1}^n x_k \ne 0\\ \sum\limits_{k = 1}^n \prod\limits_{j \ne k} x_j = 0\\ \sum\limits_{k_1 < k_2} \prod\limits_{j \ne k_1, k_2} x_j = 0 \end{cases}, \end{gather*} thus$$ \sum_{k = 1}^n \frac{1}{x_k^2} = \paren{ \sum_{k = 1}^n \frac{1}{x_k} }^2 - 2 \sum_{k_1 < k_2} \frac{1}{ x_{k_1} x_{k_2} } = \paren{ \frac{ \sum\limits_{k = 1}^n \prod\limits_{j \ne k} x_j }{ \prod\limits_{k = 1}^n x_k } }^2 - \frac{ 2\sum\limits_{k_1 < k_2} \prod\limits_{j \ne k_1, k_2} x_j }{ \prod\limits_{k = 1}^n x_k } = 0, $$ a contradiction.

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Now return to the problem.

Case 1: $P'$ has at least one real multiple root.

Suppose $x_0 \in \R$ is a multiple root of $P'$, then ${(x - x_0)}^3 \mid P(x) - P(x_0)$. For any $a \in \R \setminus \{P(x_0)\}$, note that$$ P(x) - a = {(x - x_0)}^3 \cdot \frac{ P(x) - P(x_0) }{ {(x - x_0)}^3 } - (a - P(x_0)), $$ thus the lemma shows that $a \notin I$. Therefore, $I = \emptyset$ or $\{P(x_0)\}$.

Case 2: All real roots of $P'$ are simple.

Suppose the real roots of $P'$ are $x_1 < \cdots < x_m$. (In case $P'$ has no real root, set $m = 0$.) Define $x_0 = -\infty$, $x_{m + 1} = +\infty$, and $I_k = (x_k, x_{k + 1})$ ($0 \le k \le m$).

For any fixed $a \in \R$, note that $P - a$ is strictly monotonic on each of $\overline{I_0}, \cdots, \overline{I_m}$. For any $0 \le k \le m$:

• If $P - a$ has a root in $I_k$, it must be simple.
• If $P(x_k) = a$, then $x_k$ is a real root of multiplicity $2$ of $P - a$ since $x_k$ is a simple root of $P'$.

Therefore, the number of real roots of $P - a$ is:\begin{align*} &\= \sum_{k = 0}^m \delta(P - a \text{ has a root in } I_k) + 2\sum_{k = 1}^m \delta(P(x_k) = a)\\ &= \sum_{k = 1}^{m - 1} \delta(P - a \text{ has a root in } I_k) + \delta(P - a \text{ has a root in } I_0)\\ &\={} + \delta(P - a \text{ has a root in } I_m) + \sum_{k = 0}^{m - 1} \delta(P(x_{k + 1}) = a) + \sum_{k = 1}^m \delta(P(x_k) = a)\\ &= \sum_{k = 0}^m \delta(P - a \text{ has a root in } \overline{I_k})\\ &= |\{ 0 \le k \le m \mid a \in P(\overline{I_k}) \}|. \tag{1} \end{align*}

Note that $m \le \deg(P') \le \deg(P) - 1$, so $(1) \le \deg(P)$. Hence\begin{align*} &\mathrel{\hphantom{\iff}} (1) = \deg(P)\\ &\iff (m = n - 1) \land (\forall 0 \le k \le m\colon\ a \in P(\overline{I_k}))\\ &\iff (m = n - 1) \land \paren{ a \in \bigcap_{k = 0}^m P(\overline{I_k}) }. \end{align*}

Therefore, $I = \emptyset$ or $\bigcap\limits_{k = 0}^m P(\overline{I_k})$, the latter of which, being the intersection of intervals, is an interval or an empty set.

Answer 2

Let $n$ be the degree of $P$. The case $n\le 1$ is trivial, so suppose that $n\ge 2$. Let $I'$ consists of values $a$ which $P$ attains exactly $n$ times. Clearly, $I'\subset I$.

Let $y_1\le \dots\le y_k$ be the distinct roots of $P'$. Moreover, put $y_0=-\infty$ and $y_{k+1}=+\infty$. Then for each natural $i\le k+1$, $P'$ has no roots at the interval $(y_{i-1},y_i)$, so the polynomial $P$ is monotone on this interval and thus $P$ has at most one root at $[y_{i-1},y_i]$. Let $I_i$ be the open interval between $P(y_{i-1})$ and $P(y_i)$. Since $k\le n-1$, we see that $I'$ is empty, if $k<n-1$, and $I'=\bigcap_{i=1}^n I_i$, if $k=n-1$.

Let now $a\in I\setminus I'$ be any real number. Then there exists $y\in X$ such that $y$ is a root of $P(x)-a$ of multiplicity at least two. That is, $P'(y)=0$. Put $Y=\{y_1, \dots,y_k\}$ and $Y'=\{y\in Y:P(y)-a=0\}$. Note that no two consecutive elements of $Y$ belong to $Y'$. Thus the real roots of $P(x)-a$ are comprised of elements of $Y'$ and the elements of $(y_{i-1},y_i)$ for some $i$. Since in the latter case none of the endpoints $y_{i-1}$ and $y_i$ belong to $Y'$, there are at most $k+1-2|Y'|$ such roots. Let $s$ be the sum of the multiplicities of the roots of $P(x)-a$ from $Y'$. Then $k\le n-1-s+2|Y'|$, so $P(x)-a$ has at most $s+ n-1-s+2|Y'|+1-2|Y'|=n$ real roots counted with multiplicity, and the equality is possible only when for any natural $i\le k+1$ such that $y_{i-1},y_i\not\in Y'$ then the interval $(y_{i-1},y_i)$ contains a root of $P(x)-a$, so $a\in I_i$. For any remaining natural $i\le k+1$, we have that either $y_{i-1}\in Y'$ or $y_i\in Y'$, so anyway $a\in \overline{I_i}$.

Put $I''=\bigcap_{i=1}^{k+1} \overline{I_i}$. Then $I\subset I''$. When $|I''|\le 1$ then the set $I$ is either empty or one-point. So suppose that $|I''|\ge 2$.

To be continued...