Is a differentiable non constant real function with continuous derivative strictly monotone on a interval?

Question

Let $f : [0,1] \to \mathbb{R_+}$ be a differentiable non constant real function with continuous derivative

My question:

Is it true that $\exists \lambda \in \mathbb{R_+}$ such that $f$ is strictly monotone on $[0, \lambda)$

Thanks

Answer 1

Consider $$f(x)=\begin{cases}x+x^5\sin\frac{1}{x^3} &\text{ if }x\in(0,1],\\ 0 &\text{ if }x=0.\end{cases}$$

The factor $x^5$ comes for continuous differentiablity of $f$. Note also that $f'$ is continuous as $f'(0)=1$ and $f'(x)=1+5x^4\sin\big(\frac{1}{x^3}\big)-3x\cos\big(\frac{1}{x^3}\big)$ for $x>0$.

Answer 2

Example 1.

Let $f(x)=2+x^3\sin 1/x$ when $x\ne 0,$ and $f(0)=2.$

For $0<x\le 1$ we have $f(x)\ge 2-|x^3\sin 1/x|\ge 2-|x^3|\ge 1.$

For $x\ne 0$ we have $f'(x)=3x^2\sin 1/x-x\cos 1/x,$ and $|f'(x)| \le |3x^2|+|-x|$ so $\lim_{0\ne x\to 0}f'(x)=0.$ And for $x\ne 0$ we have $|\frac {f(x)-f(0)}{x-0}|=|x^2\sin 1/x|\le |x^2|,$ so $f'(0)=0.$ So $f'$ is continuous.

If $1\ge\lambda>0$ take $n\in \Bbb N$ large enough that $0<v_n=1/(2n\pi+\pi/2)<u_n=1/(2n\pi)<\lambda.$ Then $f'(u_n)<0<f'(v_n )$ so $f$ cannot be monotonic on $[0,\lambda).$

Example 2.

Let $g(1/n)=(-1)^n /n^2$ for $n\in\Bbb N.$ For each $n\in \Bbb N$ let $g(x)$ be linear for $x\in [1/(n+1),\, 1/n].$ Let $g(0)=0.$

For $x\in [0,1]$ let $f(x)=10^{10}+\int_0^xg(t)dt.$