Proof of Fermat's theorem on stationary points

Proof

We treat a local maximum first, using one-sided sequences of difference quotients.

1. Suppose $f$ has a local maximum at $a$. Then there is $\delta > 0$ with $f(x) \le f(a)$ for all $x \in I$ with $\abs{x - a} < \delta$.
2. Since $I$ is an open interval containing $a$, we may shrink $\delta$ so that $(a - \delta, a + \delta) \subseteq I$.
3. For $h \ne 0$ with $a + h \in I$, write $q(h) = \frac{f(a + h) - f(a)}{h}$. By the definition of the derivative, $q(h) \to f'(a)$ as $h \to 0$.
4. For $0 < h < \delta$, the numerator of $q(h)$ is $\le 0$ and $h > 0$. So $q(h) \le 0$.
5. For $-\delta < h < 0$, the numerator is $\le 0$ and $h < 0$. So $q(h) \ge 0$.
6. Put $h_n = \frac{\delta}{n + 2}$ for $n \in \N$. Then $0 < h_n < \delta$. Also $h_n \to 0$, by this lemma and the epsilon-N form.
7. By the sequential characterization of limits, $q(h_n) \to f'(a)$. By step 4 and limits preserve non-strict inequalities, compared with the constant sequence $0$, $f'(a) \le 0$.
8. The sequence $-h_n$ also tends to $0$, with $-\delta < -h_n < 0$. In the same way, step 5 gives $f'(a) \ge 0$.
9. So $f'(a) = 0$ at a local maximum.
10. Now suppose $f$ has a local minimum at $a$. Then $-f$ has a local maximum at $a$, since $f(x) \ge f(a)$ means $-f(x) \le -f(a)$.
11. By the constant multiple rule with $c = -1$, $-f$ is differentiable at $a$ with derivative $-f'(a)$.
12. Steps 1 to 9 for $-f$ give $-f'(a) = 0$. So $f'(a) = 0$.