Proof of Vanishing of the Derivative at a Local Extremum Strictly Inside an Interval

Since $u<v$, the interval $I$ contains two distinct points. Let $L=f'(x)$, the derivative of $f$ at $x$ as in Derivative of a Real Function on an Interval (ii), unique by Uniqueness of the Derivative of a Real Function on an Interval (i). For $t\in I$ with $t\ne x$ write

$$q(t)=\frac{f(t)-f(x)}{t-x}.$$

Step 1: if $L\ne0$, there are $t_{-},t_{+}\in I$ with $t_{-}<x<t_{+}$, $|t_{-}-x|<\delta$ and $|t_{+}-x|<\delta$, such that $q(t_{-})$ and $q(t_{+})$ are both positive if $0<L$ and both negative if $L<0$. Suppose $L\ne0$. Then $|L|>0$ by Rules of Arithmetic and Order in an Ordered Field (x). First, Derivative of a Real Function on an Interval (i) with $\varepsilon=|L|$ gives $\delta_{1}>0$ such that every $t\in I$ with $0<|t-x|<\delta_{1}$ satisfies $|q(t)-L|<|L|$. For such $t$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ix) gives $-|L|<q(t)-L<|L|$. By Absolute Value in an Ordered Field (i), $|L|=L$ if $0<L$, and $|L|=-L$ if $L<0$. So if $0<L$ the left inequality gives $0<q(t)$, and if $L<0$ the right inequality gives $q(t)<0$. Next let

$$\eta_{+}=\min\{\delta,\delta_{1},v-x\},\qquad \eta_{-}=\min\{\delta,\delta_{1},x-u\},$$

both positive since $\delta$, $\delta_{1}$, $v-x$ and $x-u$ are, the last two as $x<v$ and $u<x$ (Difference and Quotient Identities in ℝ (i)), and put $t_{+}=x+\eta_{+}/2$ and $t_{-}=x-\eta_{-}/2$. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (vii), $0<\eta_{\pm}/2<\eta_{\pm}$, so $x<t_{+}<x+\eta_{+}\le v$ and $u\le x-\eta_{-}<t_{-}<x$, whence $x<t_{+}<v$ and $u<t_{-}<x$. As $u,x,v\in I$, Basic Facts about Intervals of the Real Line (iii) gives $t_{+},t_{-}\in I$. Finally $t_{+}-x=(x+\eta_{+}/2)-x=\eta_{+}/2$ and $t_{-}-x=-(x-t_{-})=-\bigl(x-(x-\eta_{-}/2)\bigr)=-(\eta_{-}/2)$ (Difference and Quotient Identities in ℝ (i)); as $0<\eta_{\pm}/2$, $|t_{\pm}-x|=\eta_{\pm}/2$ by Absolute Value in an Ordered Field (i) and Rules of Arithmetic and Order in an Ordered Field (x) for $|-w|=|w|$. So $0<|t_{\pm}-x|=\eta_{\pm}/2<\eta_{\pm}\le\min\{\delta,\delta_{1}\}$, so $|t_{\pm}-x|<\delta$, and $q(t_{\pm})$ has the sign stated, by the choice of $\delta_{1}$.

Step 2: signs of $q$ from the sign of $f(t)-f(x)$. Let $t\in I$, $t\ne x$, and put $w=f(t)-f(x)$, so that $q(t)=w\,(t-x)^{-1}$ (Difference and Quotient Identities in ℝ (ii)). If $f(t)\le f(x)$, then $w\le0$, by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (ii) with $-f(x)$ added to both sides; if $f(x)\le f(t)$, then $0\le w$, by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (i). If $x<t$, then $0<t-x$ (Difference and Quotient Identities in ℝ (i)) and $0<(t-x)^{-1}$ by Rules of Arithmetic and Order in an Ordered Field (viii); so by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (iv), $q(t)\le0$ when $f(t)\le f(x)$ and $0\le q(t)$ when $f(x)\le f(t)$. If $t<x$, then $0<x-t=-(t-x)$ (Difference and Quotient Identities in ℝ (i)), so $t-x<0$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (iii), and $(t-x)^{-1}<0$ (Difference and Quotient Identities in ℝ (iii)); so by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (v), $0\le q(t)$ when $f(t)\le f(x)$ and $q(t)\le0$ when $f(x)\le f(t)$.

Proof of (i). Suppose $f(t)\le f(x)$ for every $t\in I$ with $|t-x|<\delta$, and suppose $L\ne0$. Take $t_{-},t_{+}$ as in Step 1. Since $|t_{\pm}-x|<\delta$, Step 2 gives $q(t_{+})\le0$ and $0\le q(t_{-})$. If $0<L$, Step 1 gives $0<q(t_{+})$, contradicting $q(t_{+})\le0$; if $L<0$, Step 1 gives $q(t_{-})<0$, contradicting $0\le q(t_{-})$. Hence $L=0$, that is $f'(x)=0$.

Proof of (ii). Suppose $f(x)\le f(t)$ for every $t\in I$ with $|t-x|<\delta$, and suppose $L\ne0$. Take $t_{-},t_{+}$ as in Step 1. Step 2 gives $0\le q(t_{+})$ and $q(t_{-})\le0$. If $0<L$, Step 1 gives $0<q(t_{-})$, contradicting $q(t_{-})\le0$; if $L<0$, Step 1 gives $q(t_{+})<0$, contradicting $0\le q(t_{+})$. Hence $f'(x)=L=0$.