The problem is as follow:
Let $g \in C^{1}(\mathbb{R}), \, f\in C(\mathbb{R})$. Prove that the function $F: D\subseteq \mathbb{R}^2 \to \mathbb{R}^2$, $F(x,y) = ( -f(x) y-g(x) , y)$ is Lipschitz, where $D$ is a compact.
My attempt:
Let $(x,y) , (a,b) \in D$. Then we have $$ ||F(x,y)-F(a,b)|| = || (y-b,-f(x)y-g(x)+f(a)b +g(a) \,)|| \leq \sqrt{(y-b)^2+( |f(x)y - f(a)b|+ |g(x)-g(a)|)^2}. $$ Since $g$ is continuously differentiable, we have $|g^{'}(s)| \leq A$ on the compact set $p_Y(D)$, the projection of $D$ onto $y$. Hence the term $|g(x)-g(a)| \leq A |x-a|$.
However, I have no idea how to deal with the term $|f(x)y- f(a)b|$ inside the square root. Can anyone provide a solution on how to show the Lipschitz condition? Thanks.
This is false. Suppose $f$ is continuous but not Lipschitz. For a counter-example take $g \equiv 0$ and consider $\|F(x,1)-F(x',1)\|=|f(x)-f(x')|$.