2
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.