Showing $f(x,y)$ satisfies the Lipschitz condition with respect to $y$ In order to prove that a certain ODE has a unique solution, I want to show that the function $f(x,y)=x\sin |y|-(x^2-2) \arctan y$ satisfies the Lipschitz condition with respect to $y$.
My attempt: Let $D=(\alpha,\beta)\times(-\infty,\infty)$ , $\alpha<\beta$. We need to show that there exists $L$ such that $|f(x,y_1)-f(x,y_2)|\leq L |y_1-y_2|$ holds for all $(x,y_1)$ and $(x,y_2)$ in $D$.
$$|f(x,y_1)-f(x,y_2)|=|x\sin |y_1|-(x^2-2)\arctan y_1-x\sin|y_2|+(x^2-2)\arctan y_2|\\\leq |x| |\sin|y_1|-\sin|y_2||+|x^2-2| |\arctan y_1 - \arctan y_2|\leq \\ \leq 2 |x| \left |\sin \left(\frac{|y_1|-|y_2|}{2} \right) \right |\left |\cos \left(\frac{|y_1|+|y_2|}{2} \right) \right |+|x^2-2| |\arctan y_1 - \arctan y_2|\leq \\ \leq |x| |y_1 - y_2| + |x^2-2| |\arctan y_1 - \arctan y_2|\leq |x| |y_1 - y_2| + |x^2-2| |\arctan y_1 - \arctan y_2|$$
The problem is with the $|x^2-2| |\arctan y_1 - \arctan y_2|$ term. I can't think of a way to get a proper bound for it, that is something like $|x^2-2| |\arctan y_1 - \arctan y_2|\leq f(x) |y_1-y_2|$. Any suggestions?
 A: In general, it is better to divide the problem into simpler parts. I think these two statements will help:


*

*If $f$ and $g$ are Lipschitz, then $h=\alpha f+\beta g$ with $\alpha,\beta\in\mathbb R$ is also Lipschitz. This is evident from the triangle inequality:
$$\begin{align}
\lVert h(x_1)-h(x_2)\rVert&= \lVert\alpha f(x_1)+\beta g(x_1)-\left(\alpha f(x_2)+\beta g(x_2)\right)\rVert \\
 & = \lVert\alpha\left(f(x_1)-f(x_2)\right)+\beta\left(g(x_1)-g(x_2)\right)\rVert \\ 
 & \le |\alpha|\cdot\lVert f(x_1)-f(x_2)\rVert+|\beta|\cdot\lVert g(x_1)-g(x_2)\rVert
\end{align}$$

*If $g$ is analytical and $\lVert g'(x)\rVert\le L$ for all $x$, then $g$ is Lipschitz (see here, the first sentence).


To show that $f(x,y)=x\sin{|y|}-(x^2-2)\arctan{y}$ is Lipschitz with respect to $y$, first note that since we are only concerned about $y$, then $x$ can be set fixed. We also know that the $\arctan{y}$ is analytical on $\mathbb R$ and its derivative, $\frac{1}{1+y^2}$ is always positive and has a maximum at $y=0$.
It seems we only need to show that $\sin{|y|}$ is Lipshitz. The sine is an odd function, i.e. $\sin{|y|}=|\sin{y}|$. In this other post on MathSE, it has been shown that $|\sin|$ is Lipschitz.
A: Alternative method:
Since $\arctan $ is differentiable, by MVT there exist $y_0$ between $y_1,y_2$ such that
$$
\left|\frac{\arctan y_1 - \arctan y_2}{y_1 - y_2} \right| \ \leq \ \arctan'y_0 \ = \ \frac{1}{1+y_0^2} \ \leq \ 1
$$
