# Showing a function satisfies a Lipschitz condition

Have I got this right -

$$f(t,y) = 1 + t \sin(ty),\quad 0 \leq t \leq 2.$$

Here's as far as I have gotten -

$|f(t, u) - f(t,y)|$

$= |1 + t\sin(tu) - 1 - t\sin(tv)|$

$= t\cdot |\sin(tu) - \sin(tv)|$

$= t\cdot|\sin(tu) - \sin(tv)|\leq t\cdot|tu - tv|$

$= t^2|u-v|$

Is the inequality allowed?

So the function is Lipschitz with $L = 4$. It's the dropping the $\sin$ part I am not sure about.

Your step would be allowed if you knew $|\sin(x) - \sin(y)| \le |x-y|$ to be true, i.e. if you knew that the function $\sin$ is $1$-Lipschitz.

• After googling a bit I read that $sin$ is Lipschtitz with $L = 2$. So does that mean I should have a final answer in my question of $L = 8$ instead of $L = 4$?
– csss
Nov 17, 2012 at 17:07
• Curious to know which website indicate Lipschitz = 2 for the sine.
– Did
Nov 17, 2012 at 17:20
• I found it here - uk.answers.yahoo.com/question/index?qid=20090514030319AANZ1Uy
– csss
Nov 17, 2012 at 17:33
• @Did Yahoo! Answers is a veritable fountain of wisdom. "The sine function takes values from 1 to -1 so no matter what inputs you throw at this beast, the output will never vary by more then 2! So if you take your filter to be 2 you are guaranteed that | f(x) - f(y) | ≤ 2. Thus, sin(x) is a Lipschitz function."
– user53153
Feb 24, 2013 at 17:54
• Just to be sure everybody gets it right (yes, I have seen worse on MSE): @5pm's first sentence is a JOKE and Y!A quote is absurd.
– Did
Feb 24, 2013 at 18:31