Function $ f:[0,1] \rightarrow \mathbb{R}$ is continuous at $[0,1]$ and differentiable at $(0,1)$

Question

If $f(0) = 0$ and $|f'(x)| \leq 1$ $\forall x \in(0,1)$. Prove that $|f(x)| \leq 1$ $\forall x\in(0,1)$.

I tried to solve it via definition of continuity $(\forall\varepsilon>0)(\exists\delta>0)(\forall x\in A)(|x-a|<\delta\implies |f(x)-f(a)|<\varepsilon)$ for $a = 0$ and $\delta = 1$ and I got $|x|<1 \implies |f(x)|<\varepsilon$ but I don't know how to find $\varepsilon$ and how to use derivative to get the solution. Please help.

Answer 1

Using the mean value theorem, we can find $c \in (0,x)$ such that: $$f(x)-f(0) = f'(c)(x-0) \Rightarrow f(x) = f'(c)x$$ for every $x \in (0,1]$. Because $x \in (0,1]$, we have $|x| \le 1$ and $|f'(c)| \le 1$ by hypothesis. Thus: $$|f(x)| = |f'(c)||x| \le 1$$ for every $x \in (0,1]$.

Answer 2

There are a few suggestions in the comments; you can also just use the fundamental theorem of calculus: $$\lvert f(x) \rvert = \lvert f(x) - f(0) \rvert = \left \lvert \int^x_0 f'(t)dt \right \rvert \le \int^x_0 \lvert f'(t)\rvert dt \le \int^x_0 1 dt = x \le 1.$$

Answer 3

Assume there is an $x \in (0,1)$ s.t. $|f(x)| >1$.

Then

$|\dfrac{f(x)-f(0)}{x}| =|f'(s)|$, where $s \in (0,x)$.

$1<\dfrac{|f(x)|}{x}= |f'(s)|$, a contradiction