$\displaystyle |(2-\sin(x))-(2-\sin(y))|\leq \frac{1}{2};\quad\quad k|x-y|\leq \frac{\pi}{3}$

But what if we take the $\lim_{x \to y}$? $|\sin(x)-\sin(y)|=|x-y|$

So we won't be able to make $|g(x)-g(y)|\leq k|x-y|$ with $k\in (0,1)$?

  • $\begingroup$ What is your objection to the assertion that $|g(x)-g(y)|\leqslant\cos(\pi/6)\,|x-y|$ for every $x$ and $y$ in $[\pi/6,\pi/2]$? Please be specific. $\endgroup$
    – Did
    Sep 27, 2016 at 20:48
  • $\begingroup$ Well if you take the $lim_{x\to y}$ $|g(x)-g(y)|=|x-y|$ so therefore the assertion is incorrect right? $\endgroup$
    – Peter
    Sep 27, 2016 at 21:30
  • $\begingroup$ I fail to understand: $\lim\limits_{x\to y}|g(x)-g(y)|=|x-y|$ since the RHS should not depend on $(x,y)$, being the limit of the LHS. For example, the derivative of a function $h$ at some point $x_0$ is $h'(x_0)=\lim\limits_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}$ and $h'(x_0)$ does not depend on $x$ (and which $x$ would that be?). Finally, since your function $g$ is continuous, $\lim\limits_{x\to y}|g(x)-g(y)|=0$. Please explain what you mean. $\endgroup$
    – Did
    Sep 27, 2016 at 21:56
  • $\begingroup$ @Did $lim_{x\to y}|g(x)-g(y)|\geq cos(\frac{\pi}{6})\lim_{x\to y}|x-y|$ Therefore, $|g(x)-g(y)|\nleq k|x-y| \forall x,y\in (0,1)$ $\endgroup$
    – Peter
    Sep 27, 2016 at 22:00
  • $\begingroup$ Sure, both are zero hence this (true) remark is offtopic. Note also that $A\geqslant B$ does not contradict $A\leqslant B$. (And the "Therefore" part you added is plain wrong... which is good because now we have an idea of where you went astray: yes, $|g(x)-g(y)|\leqslant\cos(\pi/6)\,|x-y|$ for every $x$, $y$ in $[\pi/6,\pi/2]$, why do you think otherwise? And why do you mention the interval $(0,1)$, which is quite irrelevant here?) $\endgroup$
    – Did
    Sep 27, 2016 at 22:02

1 Answer 1


We have $g'(x) = -\cos(x)$.

So, on the interval $[\pi/6,\pi/2]$, we have $$ |g'(x)| \le \cos(\pi/6) < 1 $$ and, consequently, for every pair $(x_{1}, x_{2})$ of points in that interval, $$ \left|g(x_2) - g(x_1)\right| = \left| \int_{x_1}^{x_2} g'(x) \, dx \right| \leq \cos(\pi/6) \; |x_2 - x_1|. $$

  • $\begingroup$ Shouldn't $|g(x)-g(y)|\leq k|x-y|$ $\forall x,y\in [\frac{\pi}{6},\frac{\pi}{2}]$, which it isn't for $\lim_{x\to y}$? $\endgroup$
    – Peter
    Sep 27, 2016 at 21:23
  • $\begingroup$ u can use g(x)-g(y)=(x-y)g'(c) $\endgroup$ Sep 27, 2016 at 21:31
  • $\begingroup$ yes but $|g(x)-g(y)|\geq g'(c)|x-y|$ when you take the $\lim_{x\to y}$ $\endgroup$
    – Peter
    Sep 27, 2016 at 21:42

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