Let $f:\left[ 0,1\right] \rightarrow \mathbb{R}$ be a continuous function such that $f\left( 0\right) =0$ and $\left| f'\left( x\right) \right| \leq 1$ for all $x$ in $\left( 0,1\right)$. Prove that $-x\leq f\left( x\right) \leq x$ for all $x$ in $\left[ 0,1\right]$.

Proof-trying. We will use the mean value theorem. By the mean value theorem and assumption, there is a $c$ in $\left( a,b\right)$ such that

$\left| f'\left( c\right) \right| =\left| \dfrac {f\left( 1\right) -f\left( 0\right) } {1-0}\right| =\left| f\left( 1\right) \right| \leq 1 $. Hence, $-1\leq f\left( 1\right) \leq 1$. So, what should I do?

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    $\begingroup$ It's equivalent to proving $$\left|\frac {f\left(x\right) -f\left( 0\right) } {x-0}\right| \le 1.$$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '17 at 22:14
  • $\begingroup$ @GNUSupporter by the mean value theorem, can we say that there is a $c$ in $(a,b)$ $f\left( c\right) =\dfrac {f\left( x\right) -f\left( 0\right) } {x-0} $? $\endgroup$ – PozcuKushimotoStreet Mar 12 '17 at 22:23
  • $\begingroup$ No, because you missed a ' on the LHS. $$f'(c) = \frac{f(x)-f(0)}{x-0}$$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '17 at 22:30
  • $\begingroup$ @GNUSupporter Then, why did you say $\left| \dfrac {f\left( x\right) -f\left( 0\right) } {x-0}\right| \leq 1$? $\endgroup$ – PozcuKushimotoStreet Mar 12 '17 at 22:31
  • $\begingroup$ Due to your edit: $\left| f'\left( x\right) \right| \leq 1\forall x \in (0,1)$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '17 at 22:33

In response to OP's comments, I convert my comments into an answer.

It suffices to show that $$\left|\frac {f\left(x\right) -f\left( 0\right) } {x-0}\right| \le 1 \quad\forall x\in[0,1].$$

Since we have the continuity of $f$ on the closed interval $[0,1]$ and the differentiability on the open interval $(0,1)$, we can apply the Mean Value Theorem to conclude that for each $x \in [0,1]$, there exists $c \in (0,x)$ so that $$f'(c) = \frac{f(x)-f(0)}{x-0}.$$

Since $c \in (0,x) \subseteq (0,1)$, $|f'(c)| \le 1$, so we're done.

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    $\begingroup$ Very courteous of you to do for the OP. (+1) $\endgroup$ – Chris Mar 12 '17 at 23:00
  • $\begingroup$ @Chris Thanks for your comment and +1 $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '17 at 23:01
  • $\begingroup$ You showed that $\left| f'\left( c\right) \right| \leq 1$ but you didn't show $\left| f'\left( x\right) \right| \leq x$ for all $x$ in $\left[ 0,1\right]$? $\endgroup$ – PozcuKushimotoStreet Mar 12 '17 at 23:31
  • $\begingroup$ @Kahler Why do I need this? The question asks for $$-x\leq f\left( x\right) \leq x\iff|f(x)|\le x\quad\forall x \in [0,1].$$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '17 at 23:39
  • $\begingroup$ @GNUSupporter Yes, the question asks for this. But, you only showed $\left| f\left( c\right) \right| \leq 1$ for $c\in(0,x)$. Is it enough? Why? I couldn't be convinced. $\endgroup$ – PozcuKushimotoStreet Mar 12 '17 at 23:47

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