A nonzero best polynomial error must alternate at n+2 extrema

Put $e=f-p$. It is continuous by [F3] and has norm $E$ by [F1]. Since $I$ is compact, [F3] makes $\mathrm{\mid }e\mathrm{\mid }$ attain $E$, so its active-extrema set $S=\{x\in I:\mathrm{\mid }e(x)\mathrm{\mid }=E\}$ is nonempty. Suppose there are not $n+2$ ordered alternating points in $S$. Applying [F2] gives a polynomial $h$ of degree at most $n$ with $e(x)h(x)>0$ for every $x\in S$. In particular, $h$ is nonzero on $S$.

The set $S$ is compact because it is a closed subset of $I$. The continuous function $\mathrm{\mid }h\mathrm{\mid }$ therefore has a positive minimum $c$ on $S$ by [F3]. The set $U=\{x\in I:e(x)h(x)>0,\mathrm{\mid }e(x)\mathrm{\mid }>E\mathrm{/}2,\mathrm{\mid }h(x)\mathrm{\mid }>c\mathrm{/}2\}$ is a relatively open neighborhood of $S$, by continuity and step 1.1.

If $I\setminus U$ is nonempty, it is compact and contains no point of $S$, so [F3] gives $\eta :=E-{\max }_{x\in I\setminus U}\mathrm{\mid }e(x)\mathrm{\mid }>0$. If it is empty, set $\eta :=E$; the outside bound is then vacuous.

The polynomial $h$ is continuous and has finite norm. Choose $\epsilon >0$ with $\epsilon \mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty },I}<\min (E\mathrm{/}2,\eta )$. On $U$, $e$ and $h$ have the same sign, $\mathrm{\mid }e\mathrm{\mid }>E\mathrm{/}2$, and $\epsilon \mathrm{\mid }h\mathrm{\mid }<E\mathrm{/}2$, so $\mathrm{\mid }e-\epsilon h\mathrm{\mid }=\mathrm{\mid }e\mathrm{\mid }-\epsilon \mathrm{\mid }h\mathrm{\mid }<E-\epsilon c\mathrm{/}2$. On $I\setminus U$, when nonempty, $\mathrm{\mid }e-\epsilon h\mathrm{\mid }\le E-\eta +\epsilon \mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty },I}<E$. Thus $\mathrm{\parallel }f-(p+\epsilon h){\mathrm{\parallel }}_{\mathrm{\infty },I}<E$, while $p+\epsilon h\in {\mathcal{P}}_n(I)$, contradicting that $p$ is best. Hence the required $n+2$ alternating extrema exist.