Showing that the function $f(t,y) = 1 + t \sin ty ,\quad 0 \leq t \leq 2$ satisfies the Lipschitz condition

Question

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

Here is as far as I have gotten - \begin{align*} |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|\\ & \le t\cdot |tu - tv|\\ & = t^2|u-v| \end{align*}

Is the inequality allowed?

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

Answer 1

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.