Determine $M$ in an inequality of absolute value

Question

Given $a < \dfrac{\pi} {2}$. If $M < 1$ with $|\cos x - \cos y| \leq M|x-y|$ for every $x, y \in [0,a]$, then compute the value of $M$

I have no idea to approach this problem (come from real analysis olympiad in my state) , but my friend said that we should use intermediate value theorem in real analysis. Too bad I don't have much experience to apply that theorem. Could you help me? Regards.

user · Accepted Answer · 2018-03-20T13:26:27.680

1

HINT

Note that if $f$ is differentiable in the interval $I$

$f $ is Lipschitz $\iff f'(x)$ is bounded in $I$

and

$$M=\sup\{|f'(x)|, x\in I\}$$

See the related Is a function Lipschitz if and only if its derivative is bounded? and Prove that if $f$ is differentiable on $[a,b]$ and $f$ is Lipschitz, then $f$ has a bounded derivative.

edited Mar 20 '18 at 13:26

answered Mar 20 '18 at 13:12

user

154,566

$f'(x) = -\sin x$ and the supremum of its derivative with a given interval is zero. So, $M = 0$? – Shane Dizzy Sukardy Mar 20 '18 at 13:25
@ShaneDizzySukardy ops sorry there was a typo, it is |f'(x)| of course! – user Mar 20 '18 at 13:26
$M = 1$ that's it? – Shane Dizzy Sukardy Mar 20 '18 at 13:28
@ShaneDizzySukardy Yes it is, note that we can derive directly the result by MVT since $|\cos x - \cos y| = |f'(c)||x-y|$ and from $|f'(c)|\le M$ we obtain $|\cos x - \cos y| \le M |x-y|$ – user Mar 20 '18 at 13:36
But it says that $M < 1$, does not it contradict? – Shane Dizzy Sukardy Mar 20 '18 at 13:39
@ShaneDizzySukardy Oh yes sorry of course, with this condition sup$(f'(x))=\sin a$. – user Mar 20 '18 at 13:43

Determine $M$ in an inequality of absolute value

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