Let $f$ be continuous in $[0,2]$ and differentiable in $(0,2)$ such that $|f'(x)|\leqslant1$ and $f(0)=1=f(2)$. Prove that $f(x)\geqslant0$.

Question

Let $f$ be a continuous function in $[0,2]$ and differentiable in $(0,2)$, such that $|f'(x)|\leqslant1$ for all $x\in(0,2)$ and also $f(0)=1=f(2)$.

Prove that $f(x)\geqslant0$ for all $x\in(0,2)$.

My approach is to use Intermediate Mean value theorem and then we get, that there exists a point $c\in(0,2)$ such that $|f'(c)|=0$, so $c$ is local extreme. If $c$ is local minimum than the proof is straight forward, bit I have difficulties proving if $c$ is local maximum.

I also need to be sure that I can continue in this way or I need another way to prove the statement.

Thanks a lot!

Answer 1

For any $\,x\in(0,1]\,,\,$ by applying Intermediate Mean Value Theorem to the function $\,f\,$ on the interval $\,[0,x]\,,\,$ we get that there exists $\,c_x\in(0,x)\subseteq(0,1)\,$ such that $\;f(x)-f(0)=f’(c_x)\cdot\!x\;,\;\;$ consequently ,

$f(x)=f(0)+f’(c_x)\cdot\!x\geqslant1-|f’(c_x)|\!\cdot\!x\geqslant1-x\geqslant0\,.$

So, $\;f(x)\geqslant0\;$ for any $\,x\in(0,1]\,.$

Analogously, for any $\,x\in(1,2)\,,\,$ by applying Intermediate Mean Value Theorem to the function $\,f\,$ on the interval $\,[x,2]\,,\,$ we get that there exists $\,d_x\in(x,2)\subseteq(1,2)\,$ such that $\;f(2)-f(x)=f’(d_x)\cdot\!(2-x)\;,\;\;$ consequently ,

$\begin{align}f(x)&=f(2)-f’(d_x)\cdot\!(2-x)\geqslant1-|f’(d_x)|\!\cdot\!(2-x)\geqslant\\[3pt]&\geqslant1-(2-x)=x-1>0\,.\end{align}$

So, $\;f(x)>0\;$ for any $\,x\in(1,2)\,.$

Hence, $\;f(x)\geqslant0\;\;$ for any $\,x\in(0,2)\,.$

Answer 2

If there is a $a \in (0,1]$ such that $f(a)$ is negative: Then

1. Letting $y_0$ be such thst $f(a)=y_0$, [with $y_0$ negative] conclude that the function $h(x) = f(x) - \frac{(1-y_0)x}{a}$ $=$ $f(x) - \frac{1+|y_0|}{a}$ satisfies $h(0)=h(a)=0$, and is continuous and differentiable on $(0,a)$.

2. And thus by the Mean Value Theorem, there is a point $c \in (0,a)$ such that $h'(c)=0$.

3. But $0=h'(c)= f'(c) - \frac{1+|y_0|}{a}$, and so the equation $h'(c)=0$ implies the equation $f'(c) = \frac{1+|y_0|}{a}$. However, as $a \le 1$ and $y_0 <0$, this gives $|f'(c)| = \Big|\frac{1+|y_0|}{a}\Big| > 1$.

4. So conclude that there cannot be an $a \in (0,1]$ such that $f(a)$ is negative, given the conditions put upon $f$.

Can you now use the above reasoning, using $f(2)=1$ instead of $f(0)=1$, to conclude that there cannot be an $a \in [1,2)$ such that $f(a)$ is negative, given the conditions put upon $f$.