Proof of The Sign and Size of the Derivative: Monotonicity, Constancy and Lipschitz Bounds

The mean value step. Let $x,y\in I$ with $x<y$. The hypotheses on $f$ are those of Mean Value Theorem on a Closed Real Interval and on Any Interval (ii), which gives $\xi\in I$ with $x<\xi<y$ and

$$f(y)-f(x)=f'(\xi)\,(y-x).$$

Since $x,y\in I$ and $x<\xi<y$, $\xi$ is an inner point of $I$, so each hypothesis below on the values $f'(x')$ at inner points $x'$ applies to $f'(\xi)$. Moreover $0<y-x$ (Difference and Quotient Identities in ℝ (i)).

Proof of (i). Let $x,y\in I$ with $x\le y$. If $x=y$ then $f(x)\le f(y)$ trivially. If $x<y$, take $\xi$ as in the mean value step; $0\le f'(\xi)$ and $0\le y-x$ give $0\le f'(\xi)\,(y-x)=f(y)-f(x)$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (iv), so $f(x)\le f(y)$ by Rules of Arithmetic and Order in an Ordered Field (iv).

Proof of (ii). Let $x,y\in I$ with $x<y$ and take $\xi$ as in the mean value step. Then $0<f'(\xi)$ and $0<y-x$ give $0<f'(\xi)\,(y-x)=f(y)-f(x)$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (iv), so $f(x)<f(y)$ (Difference and Quotient Identities in ℝ (i)).

Proof of (iii). Let $x,y\in I$ with $x\le y$; the case $x=y$ is trivial. If $x<y$, take $\xi$ as in the mean value step. From $f'(\xi)\le0$, Rules of Arithmetic and Order in an Ordered Field (v) gives $0\le-f'(\xi)$, so $0\le(-f'(\xi))\,(y-x)=-(f(y)-f(x))=f(x)-f(y)$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (iv), Rules of Arithmetic and Order in an Ordered Field (ii) for $(-u)\,v=-(u\,v)$, and Difference and Quotient Identities in ℝ (i) for the last equality, and $f(y)\le f(x)$ by Rules of Arithmetic and Order in an Ordered Field (iv).

Proof of (iv). Let $x,y\in I$ with $x<y$ and take $\xi$ as in the mean value step. From $f'(\xi)<0$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (iii) gives $0<-f'(\xi)$, so $0<(-f'(\xi))\,(y-x)=-(f(y)-f(x))=f(x)-f(y)$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (iv), Rules of Arithmetic and Order in an Ordered Field (ii) for $(-u)\,v=-(u\,v)$, and Difference and Quotient Identities in ℝ (i) for the last equality; and $f(y)<f(x)$ (Difference and Quotient Identities in ℝ (i)).

Proof of (v). If $f'(x')=0$ at every inner point $x'$, then both $0\le f'(x')$ and $f'(x')\le0$ hold there, so (i) and (iii) apply. Let $x,y\in I$. The order of $\mathbb{R}$ is total, so $x\le y$ or $y\le x$; by symmetry suppose $x\le y$. Then $f(x)\le f(y)$ by (i) and $f(y)\le f(x)$ by (iii), so $f(x)=f(y)$ by antisymmetry of $\le$.

Proof of (vi). Let $x,y\in I$. If $x=y$, both sides of the inequality are $0$, as $f(x)-f(x)=0$ and $x-x=0$ (Difference and Quotient Identities in ℝ (i)) and $|0|=0$ by Rules of Arithmetic and Order in an Ordered Field (x). Suppose $x<y$ and take $\xi$ as in the mean value step. By Rules of Arithmetic and Order in an Ordered Field (x), $|f(y)-f(x)|=|f'(\xi)|\,|y-x|$ and $0\le|y-x|$; since $|f'(\xi)|\le M$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (v) gives $|f'(\xi)|\,|y-x|\le M\,|y-x|$. So $|f(y)-f(x)|\le M\,|y-x|$. Finally suppose $y<x$. The case just proved, with the roles of $x$ and $y$ exchanged, gives $|f(x)-f(y)|\le M\,|x-y|$; and $|f(x)-f(y)|=|f(y)-f(x)|$ and $|x-y|=|y-x|$, since $f(x)-f(y)=-(f(y)-f(x))$ and $x-y=-(y-x)$ (Difference and Quotient Identities in ℝ (i)) and $|-z|=|z|$ for $z\in\mathbb{R}$ by Rules of Arithmetic and Order in an Ordered Field (x). So $|f(y)-f(x)|\le M\,|y-x|$ in every case, the order of $\mathbb{R}$ being total.