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

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.

• Hint : Taylor expansion.
– jvc
May 18, 2020 at 19:45
• Hint: Mean Value Theorem makes this automatic. May 18, 2020 at 19:45
• Use the mean value theorem. May 18, 2020 at 19:45

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 $$x \in (0,1]$$.

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

• This approach requires that the derivative is integrable. May 18, 2020 at 20:08
• Well it is integrable in the Lebesgue sense since $\lvert f'(x) \rvert \le 1$, but you are correct, I suppose that means this approach can only guarantee that $\lvert f(x) \rvert \le 1$ for almost every $x \in [0,1]$. Continuity should cover the holes in all but the most pathological examples, but yes, at this point the MVT provides a much easier solution. May 18, 2020 at 20:59

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