If $f(a)=b$ and $f(b)=a$. Prove that there exists at least one $c$ such that $|f'(c)|<1$. Also prove that ther exists some 'd' such that $|f'(d)|>1$

Question

Let $f:[a,b]\rightarrow[a,b]$ where $a<b$ be a non-linear differentiable function such that $f(a)=b$ and $f(b)=a$.

Prove that there exists at least one $c$ such that $|f'(c)|<1$ .Also prove that there exists at least one $d$ such that $|f'(d)|>1$ .

My Attempt

By LMVT we have $f'(e)=\frac{f(b)-f(a)}{b-a}=\frac{a-b}{b-a}=-1\Rightarrow |f'(e)|=1$ .

Since $f(x)$ is non-linear we can't have $f'(x)=-1$ for all $x$.

So we may have $f'(c)<1$ or $f'(c)>1$

But I was wondering if there is a more conclusive argument.

Answer 1

Derivatives have Intermediate Value Propery. If $|f'(x)| \ge 1$ for all $x$ then either $f'(x) \ge 1$ for all $x$ or $f'(x) \le -1$ for all $x$.

Suppose $f'(x) \ge 1$ for all $x$. Then $f(b)-f(a)=(b-a)f'(c)\ge b-a$ for some $c$. But then, $a-b \ge b-a$, a contradiction.

Suppose $f'(x) \le -1$ for all $x$. Let $x \le y$. Then $f(y)-f(x)=(y-x)f'(c)\le -(y-x)$ for some $c$. This gives $f(x)+x\ge f(y)+y$. In particular, $f(x)+x \ge f(b)+b=a+b$. In the other hand, $f(y)+y \le f(x)+x$ and this gives $f(y)+y \le f(a)+a=a+b$. Puttimg these together we get $f(t)+t=a+b$ for all $t$. But $f$ is given to be non-linear.

This proves the existence of $c$.

Existence of $d$: Suppose $|f'(x)|\le 1$ for all $x$. Let $g(x)=f(x)+x$. Then, $g'(x)=f'(x)+1\ge 0$. Thus, $g$ is increasing. This gives $g(x) \le g(b)=f(b)+b=a+b$ and $g(x) \ge g(a)=f(a)+a=b+a$. Hence, $g(x)=a+b$ for all $x$ which makes $f$ linear.

Answer 2

Suppose $f'(t)\geq -1$ for every $t\in [a, b]$. Then, if $a\leq x< y\leq b$, there exists $c\in (x,y)$ such that $$ \frac{f(y)-f(x)}{y-x}=f'(c)\geq -1. $$ Therefore, $$ f(y)-f(x)\geq x-y $$ for $x<y$, which implies that the function $g(t)=t+f(t)$ is non-decreasing. However, since $g(a)=a+b=g(b)$, $g$ is constant, which implies $f$ is linear. Therefore, if $f$ is non-linear, there is $d\in [a, b]$ such that $f'(d)<-1$.

Now proceed in an analogous way assuming $f'\leq -1$, get a contradiction and find $e$ such that $f'(e)>-1$. Since derivatives satisfy the Intermediate Value Property, you are done.

Answer 3

Consider $g(x)=f(x)+x-a-b.$ Then $g(a)=g(b)=0.$ As $f$ is not linear the function $g$ is not constantly equal $0.$ If $g'(x)\ge 0$ for all $x$ the function $g$ would be increasing. Thus there exist $x_0$ such that $g'(x_0)<0.$ Hence $f'(x_0)=g'(x_0)-1<-1.$ Similarly there is $x_1$ such that $g'(x_1)>0.$ Thus $f'(x_1)=g'(x_1)-1>-1.$ If $f'(x_1)<1$ we are done as $|f'(x_1)|<1.$ If $f'(x_1)\ge 1$ then by the intermediate value property of the derivative, there is $x_2$ between $x_0$ and $x_1$ such that $f'(x_2)=0.$

Answer 4

As $f$ is not linear, there is some point $x_0$ such that $f(x_0)\neq a+b-x_0$, which in particular means $x_0 \in (a,b)$.

Suppose $f(x_0) > a+b-x_0$. Then by the mean value theorem, there are $c,d$ such that $$f'(c) = \frac{f(x_0)-f(a)}{x_0-a} > \frac{a+b-x_0-b}{x_0-a} = -1 \\ f'(d) = \frac{f(b)-f(x_0)}{b-x_0} <\frac{a-(a+b-x_0)}{b-x_0} = -1$$ and to finish it suffices to note $f(x_0) \leq b$, so additionally $$f'(c) = \frac{f(x_0)-f(a)}{x_0-a} \leq \frac{b-b}{x_0-a} = 0.$$

The case of $f(x_0) < a+b-x_0$ is similar.