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 $.
Best Answer
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 $(x,y_1)$, $(x,y_2)\in W$.