# 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.

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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 '12 at 17:07