Showing that the function $f(x,y)=x\sin y+y\cos x$ is Lipschitz I wanted to show that  $f(x,y)=x\sin y+y\cos x$ sastisfy Lipschitz conditions. but I can't separate it to $L|y_1-y_2|$. 

According to my lecturer, the Lipschitz condition should be $$|f(x,y_1)-f(x,y_2)|\le  L|y2-y1|$$ 
I was able show that   $x^2+y^2$ in the rectangle $|x|\le a$, $|y|\le b$  satisfies the Lipschitz condition, with my $L=2b$. But I had problem showing this for $f(x,y)=x\sin y+y\cos x $.
 A: It seems that you want to prove that $f$ is Lipschitz with respect to $y$.  When $f$ is viewed as a global function $f:\>{\mathbb R}^2\to{\mathbb R}$ the statement is wrong, because
$$\left|{f\bigl(x,{\pi\over 2}\bigr)-f(x,0)\over {\pi\over2}}\right|\geq |x|-{\pi\over2}$$
assumes arbitrarily large values. The function $f$ is, however, locally Lipschitz with respect to $y$ in ${\mathbb R}^2$. This means that any point $(x_0,y_0)$ has a neighborhood $W$ whithin which the Lipschitz condition is fulfilled. In order to see this it is sufficient to note that $f\in C^1({\mathbb R}^2)$, but maybe you want a selfcontained proof.
From
$${\partial f(x,y)\over\partial y}= x\cos y+\cos x$$
it follows that $\bigl|{\partial f(x,y)\over\partial y}\bigr|\leq |x|+1$. Therefore any point $(x_0,y_0)$ is the center of a quite large window $W$ such that for a suitable $M$ one has
$$\left|{\partial f(x,y)\over\partial y}\right|\leq M\qquad\forall (x,y)\in W\ .$$
By means of the MVT we then conclude that
$$\bigl|f(x,y_1)-f(x,y_2)\bigr|\leq M\>|y_1-y_2|$$
for all $(x,y_1)$, $(x,y_2)\in W$.
A: Since $\frac {\partial f} {\partial y} =x \cos y+ \cos x$, it follows that
$$\left| \frac {\partial f} {\partial y} \right| = |x \cos y+ \cos x| \le |x \cos y|+|\cos x| = |x| \, |\cos y|+|\cos x| \le a+1$$
since $|x| \le a$, $|\cos y| \le 1$ and $|\sin y| \le 1$. This means that $|f(x,y_1) - f(x,y_2)| \le (a+1) \, |y_1 - y_2|$.
