A polynomial with the sign pattern of the active extrema

Continuity and attainment in [F2] show that $S_{+}\cup S_{-}$ is nonempty and compact: the maximum of $\mathrm{\mid }e\mathrm{\mid }$ is attained and $E>0$, while each level set is closed in the compact interval [F1]. If one of $S_{+}$ or $S_{-}$ is empty, take $h=1$ or $h=-1$ according to the sign of the nonempty set; this has degree zero and the required sign.

Suppose both sets are nonempty. Their product is a closed bounded subset of ${\mathbb{R}}^2$. For $(x,y),(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\in {\mathbb{R}}^2$, the reverse and ordinary triangle inequalities in [F3] give $\mid \mathrm{\mid }x-y\mathrm{\mid }-\mathrm{\mid }{x}^{\mathrm{\prime }}-y^{\mathrm{\prime }}\mathrm{\mid }\mid \le \mathrm{\mid }(x-y)-(x^{\mathrm{\prime }}-y^{\mathrm{\prime }})\mathrm{\mid }\le \mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }+\mathrm{\mid }y-y^{\mathrm{\prime }}\mathrm{\mid }\le 2\mathrm{\parallel }(x,y)-(x^{\mathrm{\prime }},y^{\mathrm{\prime }}){\mathrm{\parallel }}_2$, so the distance function is continuous. By [F2] it attains a minimum separation $\delta =\min \{\mathrm{\mid }x-y\mathrm{\mid }:x\in S_{+},y\in S_{-}\}$. This minimum is positive because a minimizing pair at distance zero would give a point in both disjoint sets. Starting at the leftmost point of $S_{+}\cup S_{-}$, repeatedly take the first point of the opposite-sign set to the right and the last current-sign point before it. These extrema exist because the level sets are compact, and each transition gives an open gap of length at least $\delta$ containing no point of $S_{+}\cup S_{-}$. The gaps are pairwise disjoint subintervals of $[a,b]$, so any $k$ transitions satisfy $k\delta \le b-a$; hence there are finitely many transitions, each with a distinct ordered gap.

If there were at least $n+1$ transitions, choosing one extremal point from each successive sign run would give $n+2$ ordered alternating points, contrary to the hypothesis. Thus the number $k$ of transitions is at most $n$. Choose one point ${\xi }_j$ in each transition gap and set $h(x)=c{\prod }_{j=1}^k(x-{\xi }_j)$, where $c\in \{-1,1\}$ is chosen so that $h$ has the sign of $e$ on the first run. The polynomial has degree $k\le n$, has no zero on $S_{+}\cup S_{-}$, and changes sign exactly when the extremal sign changes. Hence $e(x)h(x)>0$ on the entire extremal set.