Solution of A Function with Small Derivative Has Exactly One Fixed Point

Step 1: $f$ is continuous on $\mathbb{R}$. Every $x\in\mathbb{R}$ is an interior point of the interval $\mathbb{R}$, and $f$ is differentiable there by hypothesis, so $f$ is continuous at $x$ by The Increment Form of the Derivative, and Differentiability Implies Continuity. Hence $f$ is continuous on $\mathbb{R}$.

Step 2: the auxiliary function $g$. Define $g:\mathbb{R}\to\mathbb{R}$ by

$$g(x)=f(x)+(-1)\,x .$$

The identity map $\iota(x)=x$ is continuous on $\mathbb{R}$ by The Real Line: Standing Notation and Background for Calculus on Intervals (v), so $g$ is continuous on $\mathbb{R}$ by clauses 2, 4 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space. By clause 1 of Derivative of a Polynomial Function on the Real Line the map $\iota$ is differentiable at every $x\in\mathbb{R}$ with derivative $1$, so by clauses 2 and 3 of Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (sum and constant multiple) $g$ is differentiable at every $x\in\mathbb{R}$ with

$$g'(x)=f'(x)+(-1)\cdot 1=f'(x)-1 .$$

By Properties of the Absolute Value in an Ordered Field, $|f'(x)|\le\tfrac12$ gives $f'(x)\le\tfrac12$, hence

$$g'(x)\le\tfrac12-1=-\tfrac12<0\qquad\text{for every }x\in\mathbb{R} .$$

Step 3: at most one fixed point. Since $g$ is continuous on the interval $\mathbb{R}$ and differentiable at every interior point of it with $g'(x)<0$ there, clause 4 of The Sign and Size of the Derivative: Monotonicity, Constancy and Lipschitz Bounds says $g$ is strictly decreasing on $\mathbb{R}$. If $c\ne c'$, then either $c<c'$ or $c'<c$, and in either case $g(c)\ne g(c')$. So at most one real number $c$ satisfies $g(c)=0$. Since $g(c)=0$ holds exactly when $f(c)=c$, there is at most one fixed point.

Step 4: increments of $g$. Let $v\in\mathbb{R}$ with $0<v$. By clauses 1 and 2 of Restriction Stability of Continuity and of the Derivative the restriction of $g$ to $[0,v]$ is continuous on $[0,v]$ and differentiable at every point of $(0,v)$, with the same derivative. By Mean Value Theorem on a Closed Real Interval and on Any Interval there is $\xi\in(0,v)$ with $g(v)-g(0)=g'(\xi)\,v$. Since $g'(\xi)\le-\tfrac12$ and $0<v$, the order arithmetic of Elementary Order Arithmetic in an Ordered Field gives $g'(\xi)\,v\le-\tfrac{v}{2}$, so

$$g(v)\le g(0)-\tfrac{v}{2} .$$

Similarly, let $u\in\mathbb{R}$ with $u<0$. The restriction of $g$ to $[u,0]$ is likewise continuous on $[u,0]$ and differentiable at every point of $(u,0)$ with the same derivative, by clauses 1 and 2 of Restriction Stability of Continuity and of the Derivative. Applying Mean Value Theorem on a Closed Real Interval and on Any Interval to it gives $\eta\in(u,0)$ with $g(0)-g(u)=g'(\eta)\,(0-u)$. Since $g'(\eta)\le-\tfrac12$ and $0<-u$, we get $g(0)-g(u)\le-\tfrac{-u}{2}=\tfrac{u}{2}$, that is,

$$g(0)-\tfrac{u}{2}\le g(u) .$$

Step 5: existence of a fixed point. Put

$$v=\max\{1,\ 2\,g(0)+1\},\qquad u=\min\{-1,\ 2\,g(0)-1\} .$$

Then $0<1\le v$ and $2g(0)<2g(0)+1\le v$, so $g(0)<\tfrac{v}{2}$ and hence $g(0)-\tfrac{v}{2}<0$; with Step 4 this gives $g(v)<0$. Likewise $u\le-1<0$ and $u\le 2g(0)-1<2g(0)$, so $\tfrac{u}{2}<g(0)$ and hence $0<g(0)-\tfrac{u}{2}$; with Step 4 this gives $0<g(u)$.

Thus $u<0<v$, so $u<v$, and

$$g(v)\le 0\le g(u) .$$

The restriction of $g$ to $[u,v]$ is continuous on $[u,v]$ by clause 1 of Restriction Stability of Continuity and of the Derivative. By Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals applied on $[u,v]$ with the value $0$, there is $c\in[u,v]$ with $g(c)=0$, that is, $f(c)=c$.

By Steps 3 and 5 there is exactly one real number $c$ with $f(c)=c$.