Proof of The Sign of the Derivative and Monotonicity

Step 1: an increment formula. Let $u,v\in I$ with $u<v$. We claim there is $\xi$ with $u<\xi<v$ such that

$$f(v)-f(u)=f'(\xi)\,(v-u) .$$

Since $u,v\in I$ and $u<v$, we have $[u,v]\subseteq I$ by The Real Line: Standing Notation and Background for Calculus on Intervals (iv). Let $g=f|_{[u,v]}$ be the restriction of $f$ to $[u,v]$. By clause 1 of Restriction Stability of Continuity and of the Derivative, $g$ is continuous on $[u,v]$, since $f$ is continuous on $I$.

Let $x\in(u,v)$. Then $u<x<v$ with $u,v\in[u,v]$, so $x$ is an interior point of $[u,v]$. By clause 2 of Restriction Stability of Continuity and of the Derivative applied to the intervals $[u,v]\subseteq I$, the point $x$ is also an interior point of $I$; hence $f$ is differentiable at $x$ by hypothesis, and the same clause gives that $g$ is differentiable at $x$ with $g'(x)=f'(x)$.

So $u<v$, $g$ is continuous on $[u,v]$, and $g$ is differentiable at every point of $(u,v)$. By Mean Value Theorem on a Closed Real Interval and on Any Interval there is $\xi\in(u,v)$ with

$$g'(\xi)=\frac{g(v)-g(u)}{v-u} .$$

Since $g(u)=f(u)$, $g(v)=f(v)$ and $g'(\xi)=f'(\xi)$, multiplying by $v-u$ gives the claimed identity. Note also that $\xi$ is an interior point of $I$, and that $0<v-u$ because $u<v$.

Step 2: the four cases. Throughout we use the sign rules for products and sums of Elementary Order Arithmetic in an Ordered Field.

1. Suppose $0\le f'(x)$ for every interior point $x$ of $I$, and let $x,y\in I$ with $x\le y$. If $x=y$ then $f(x)\le f(y)$. If $x<y$, Step 1 gives an interior point $\xi$ of $I$ with $f(y)-f(x)=f'(\xi)(y-x)$; since $0\le f'(\xi)$ and $0<y-x$, the product is nonnegative, so $0\le f(y)-f(x)$ and hence $f(x)\le f(y)$. Therefore $f$ is nondecreasing on $I$ in the sense of Monotone Real Function (i).

2. Suppose $0<f'(x)$ for every interior point $x$ of $I$, and let $x,y\in I$ with $x<y$. Step 1 gives an interior point $\xi$ of $I$ with $f(y)-f(x)=f'(\xi)(y-x)$; since $0<f'(\xi)$ and $0<y-x$, the product is positive, so $0<f(y)-f(x)$ and hence $f(x)<f(y)$. Therefore $f$ is strictly increasing on $I$ in the sense of Monotone Real Function (iii).

3. Suppose $f'(x)\le 0$ for every interior point $x$ of $I$, and let $x,y\in I$ with $x\le y$. If $x=y$ then $f(y)\le f(x)$. If $x<y$, Step 1 gives $\xi$ with $f(y)-f(x)=f'(\xi)(y-x)$; since $f'(\xi)\le 0$ and $0<y-x$, the product is nonpositive, so $f(y)-f(x)\le 0$ and hence $f(y)\le f(x)$. Therefore $f$ is nonincreasing on $I$ in the sense of Monotone Real Function (ii).

4. Suppose $f'(x)<0$ for every interior point $x$ of $I$, and let $x,y\in I$ with $x<y$. Step 1 gives $\xi$ with $f(y)-f(x)=f'(\xi)(y-x)$; since $f'(\xi)<0$ and $0<y-x$, the product is negative, so $f(y)-f(x)<0$ and hence $f(y)<f(x)$. Therefore $f$ is strictly decreasing on $I$ in the sense of Monotone Real Function (iv).