Proof of Vanishing of the Derivative at an Interior Local Extremum

By An Open Interval is an Interval All of Whose Points Are Interior the set $(p,q)$ is an interval and every point of it, in particular $c$, is an interior point, so differentiability at $c$ is meaningful. Write $L=g'(c)$ and let $|\cdot|$ be the absolute value, so that $d_{\mathbb{R}}(x,y)=|x-y|$ by The Absolute Value Metric on the Real Line. Claim numbers refer to Elementary Order Arithmetic in an Ordered Field and Properties of the Absolute Value in an Ordered Field as indicated.

Case A: $g$ has a local maximum at $c$ relative to $(p,q)$. By that definition there is $\delta_{0}$ with $0<\delta_{0}$ such that every $y\in(p,q)$ with $|y-c|<\delta_{0}$ satisfies $g(y)\le g(c)$.

Suppose, for contradiction, that $L\ne0$. Since $\le$ compares any two elements, either $0<L$ or $L<0$.

Subcase $0<L$. Differentiability at $c$ applied with $\varepsilon=L$ gives $\delta_{1}$ with $0<\delta_{1}$ such that every $k$ with $0<|k|<\delta_{1}$ and $c+k\in(p,q)$ satisfies $|Q_{k}-L|<L$, where $Q_{k}$ denotes the difference quotient $\bigl(g(c+k)-g(c)\bigr)/k$. By claim 9 of Properties of the Absolute Value in an Ordered Field this gives $-L<Q_{k}-L$, hence $0<Q_{k}$ by claim 1 of Elementary Order Arithmetic in an Ordered Field.

Since $c\in(p,q)$ we have $c<q$, so $0<q-c$ by claim 1. Using claim 9 of Elementary Order Arithmetic in an Ordered Field twice, choose $\mu$ with $0<\mu$, $\mu\le\delta_{0}$, $\mu\le\delta_{1}$ and $\mu\le q-c$, and put $k=\mu\cdot2^{-1}$, so that $0<k$ and $k<\mu$ by claim 8. Then $k<q-c$ by claim 2, so $c+k<q$ by claim 1; and $p<c<c+k$ gives $p<c+k$ by claim 2. Hence $c+k\in(p,q)$. Also $|k|=k<\mu$, so $0<|k|<\delta_{1}$ and $|k|<\delta_{0}$, both by claim 2.

Therefore $0<Q_{k}$, and $0<k$, so claim 5 gives $0<Q_{k}k=g(c+k)-g(c)$, whence $g(c)<g(c+k)$ by claim 1. But $c+k\in(p,q)$ and $|(c+k)-c|=|k|<\delta_{0}$, so the local maximum property gives $g(c+k)\le g(c)$; with $g(c)<g(c+k)$ and claim 2 this yields $g(c)<g(c)$, contradicting the irreflexivity of the strict order.

Subcase $L<0$. Then $0<-L$ by claim 4. Differentiability at $c$ applied with $\varepsilon=-L$ gives $\delta_{1}$ with $0<\delta_{1}$ such that every admissible $k$ satisfies $|Q_{k}-L|<-L$, hence $Q_{k}-L<-L$ by claim 9 of Properties of the Absolute Value in an Ordered Field, hence $Q_{k}<0$ by claim 1.

Since $p<c$ we have $0<c-p$ by claim 1. Choose $\mu$ with $0<\mu$, $\mu\le\delta_{0}$, $\mu\le\delta_{1}$ and $\mu\le c-p$ by claim 9, put $\nu=\mu\cdot2^{-1}$ and $k=-\nu$, so $0<\nu<\mu$ by claim 8 and $k<0$ by claim 4. Then $\nu<c-p$ by claim 2 gives $p<c-\nu=c+k$ by claim 1, and $c+k<c<q$ gives $c+k<q$ by claim 2, so $c+k\in(p,q)$. Also $|k|=|-\nu|=\nu<\mu$ by claim 2 of Properties of the Absolute Value in an Ordered Field, so $0<|k|<\delta_{1}$ and $|k|<\delta_{0}$.

Therefore $Q_{k}<0$ and $k<0$, so $0<-Q_{k}$ and $0<-k$ by claim 4, and claim 5 gives $0<(-Q_{k})(-k)=Q_{k}k=g(c+k)-g(c)$, whence $g(c)<g(c+k)$ by claim 1. As before this contradicts the local maximum property.

Both subcases are impossible, so $L=0$.

Case B: $g$ has a local minimum at $c$ relative to $(p,q)$. Let $0$ also denote the constant function on $(p,q)$ with value $0$; its difference quotient at $c$ is $0$ for every admissible $k$, so it is differentiable at $c$ with derivative $0$. By Derivative of a Sum and of a Difference the function $-g=0-g$ is differentiable at $c$ with $(-g)'(c)=0-L=-L$.

By the definition of a local minimum there is $\delta_{0}$ with $0<\delta_{0}$ such that every $y\in(p,q)$ with $|y-c|<\delta_{0}$ satisfies $g(c)\le g(y)$; claim 4 of Elementary Order Arithmetic in an Ordered Field turns this into $(-g)(y)\le(-g)(c)$, so $-g$ has a local maximum at $c$ relative to $(p,q)$ with the same $\delta_{0}$. Case A applied to $-g$ gives $-L=0$, hence $L=0$.