Uniqueness of the Fixed Point when $f'(x) < 1$ at Fixed Points

Question

For $f : [a,b] \rightarrow [a,b]$, show that if $f$ is continuous and differentiable and $f'(x) < 1$ at points such that $f(x) = x$, then $f$ has a unique fixed point.

The proof of the existence of the fixed point is straightforward, but to get uniqueness, we can no longer directly apply Mean Value Theorem when the assumption on the derivative is more restrictive. I am trying to show that if there are two fixed points, then there must be a fixed point between them with derivative greater than 1.

Answer 1

Hint: define $g(x)=f(x)-x$. So you have to show that if $g'(x)<0$ whenever $g(x)=0$, then $g(x)$ has a unique zero. Now suppose for a contradiction that there exist $\alpha<\beta$ such that $g(\alpha)=g\beta)=0$. $g'(\alpha)<0$ means that there exists $\xi\in(\alpha,\beta)$ with $g(\xi)<0$; and $g'(\beta)<0$ means that there exists $\eta\in(\xi,\beta)$ such that $g(\eta)>0$.

Answer 2

Here is one argument:

Claim: Let $f:[0,1]\to [0,1]$ be continuous. If for any $ x\in \operatorname{Fix}(f)$ the derivative $f'(x)$ exists and $f'(x)<1$, then $\# \operatorname{Fix}(f)=1$.

(Here $\operatorname{Fix}(f)=\{x\in[0,1]\,|\, f(x)=x\}$ is the set of fixed points of $f$.)

Proof: First we claim that $f$ has at least one fixed point. Indeed, if $f(0)=0$ or $f(1)=1$, we're done. Otherwise $0< f(0)$ and $f(1)< 1$. Since $f$ is continuous by the intermediate value theorem $f$ ought to have a fixed point in $]0,1[$.

Observe that by the hypothesis on the fixed points we have:

\begin{align*} \forall x\in \operatorname{Fix}(f), \exists \delta_x\in\mathbb{R}_{>0},\,\,& \forall z\in [0,1]\,\,\cap\,\,]x,x+\delta_x[: f(z)< z\text{ and }\\ &\forall z\in [0,1]\,\,\cap\,\,]x-\delta_x,x[: z< f(z). \quad\quad(\dagger) \end{align*}

(The first line is by unfolding the definition of the right limit for $f'(x)$ and the second line is by unfolding the definition of the left limit for $f'(x)$. Heuristically $f'(x)<1$ means that the graph of $f$, compared to the diagonal, is tilted up for $f(x-\epsilon)$ and tilted down for $f(x+\epsilon)$. It's clear that we don't really need the two limits to coincide; only that they both be less than $1$.)

In particular $\operatorname{Fix}(f)$ is discrete. By the continuity of $f$ it's also closed, hence compact, whence finite. Let us put $L_x=[0,1]\,\,\cap\,\,]x-\delta_x,x[$ and $R_x=[0,1]\,\,\cap\,\,]x,x+\delta_x[$ for $x\in\operatorname{Fix}(f)$.

Next we claim that $f$ has not more than one fixed point. As we observed that $f$ has finitely many fixed points, it suffices to show that if $f$ has two distinct fixed points, then it has infinitely many fixed points. Let $x,y\in\operatorname{Fix}(f)$ be such that $x< y$. By $(\dagger)$, since $R_x\,\,\cap\,\, ]x,y[\neq\emptyset\neq L_y\,\,\cap\,\, ]x,y[$, there are $a,b\in]x,y[$ such that

$$a<b,\quad f(a)<a,\quad b<f(b).$$

Again by the intermediate value theorem $f$ has a fixed point in $]a,b[$. By induction $f$ has infinitely many fixed points, a contradiction.