If $f'(x)\not = 1$ for all real numbers $x$, then $f$ has at most one fixed point

Question

A number $a$ is called a fixed point of a function $f$ if $f(a)=a$.Prove that if $f'(x)\not = 1$ for all real numbers $x$, then $f$ has at most one fixed point.

This is an exercise in Stewart's calculus textbook. Do we need to suppose $f$ is continuous? Or the assumption, $f'(x) \not = 1$ for all real numbers $x$, guarantees that $f$ is continuous? What if $f'$ is not defined at some points? For example, when $f$ is an increasing piecewise function which has infinitely many intersections with the line $y=x$, $f$ has infinitely many fixed points and $f'(x) \not =1$ for all real $x$.

Answer 1

Suppose $f(x)$ has two fixed points $a$ and $b$:

$f(a) = a, \; f(b) = b; \tag 1$

we may without loss of generality assume that

$a < b; \tag 2$

then by the mean value theorem there exists $c \in (a, b)$ with

$b - a = f(b) - f(a) = f'(c)(b - a), \tag 3$

and we thus find

$f'(c) = \dfrac{b - a}{b - a} = 1, \tag 4$

which contradicts

$f'(x) \ne 1, \forall x \in \Bbb R; \tag 5$

therefore, $f(x)$ has at most one fixed point.

A little commentary on our OP user398843's closing questions: the assertion (5) precludes the possibility that $f'(x)$ is not defined for some $x \in \Bbb R$; saying $f'(x) \ne 1$ means that $f'(x)$ exists, that is, is defined, and that it's value is not equal to one; the statement is meaningless if $f'(x)$ is undefined. Since differentiability at a point implies continuity at that point, $f(x)$ is continuous. The key lies in the title, "$f'(x) \ne 1$ for all real numbers $x$ . . ."

Answer 2

Let $x,y$ be two distinct fixed points of a differentiable function $f:(a,b)\to\Bbb R$ with $x,y\in(a,b)$. Then by a Lagrange MVT we get $$1=\frac{x-y}{x-y}=\frac{f(x)-f(y)}{x-y}=f'(c)$$ for some $c$ between $x$ and $y$. A contradiction.

Answer 3

If $f$ has two fixed points, $a$ and $b$, then $\frac{f(b)-f(a)}{b-a}=1$. Now apply the mean value theorem.

Answer 4

Let $$g(x)=f(x)-x$$

assume there are two fixed points $a$ and $b$ for $f$.

$g$ is differentiable at $[a,b]$ and $g(a)=g(b)=0$, thus by Rolle's theorem

there exists $c$ such that $$g'(c)=0=f'(c)-1$$

this contradicts the hypothese $f'(c)\ne 1$.

Answer 5

$f'(x)\not =1$ , $x \in \mathbb{R}.$

1) $f'(x)>1$, $x\in \mathbb{R}$, or

2)$f'(x)<1$, $x \in \mathbb{R}.$

$F(x):=f(x)-x;$

1)Let $f'(x)>1$.

$F'(x)=f'(x)-1>0.$

$F$ is strictly increasing, i.e.

$F$ has at most one zero.

2) Let $f'(x) <1$.

$F'(x)=f'(x)-1 <0$, $F$ is strictly decreasing, has at most one zero.

3) See comment by Jens Schwaiger.Thanks .

Assume $f'(x) >1$ for some $x$, and $f'(y)<1$ for some $y$, say, $x <y$, then by Darboux' Intermediate Value Property, there is a $z \in (x,y)$ s.t. $f'(z)=1$. Contradiction.