Global Lipschitz behavior of given function

Question

Is the function $$ f(x) = sin(x)sgn(x) $$ globally lipschitz? The textbook solutions says so but I've some doubts since $$\frac{\partial f}{\partial x} = 1 \quad\forall x > 0^+ \quad ;\quad = -1 \quad\forall x < 0^-$$ Thus it's not continuous at x = $0$.

Answer 1

Suppose $x>0$ and $y<0$. Then $\frac {|f(x)-f(y)|} {|x-y|}=\frac {|\sin (x)+\sin (y)|} {|x-y|}$. Since $|\sin x| \leq |x|$ and $|\sin y| \leq |y|$ we get $\frac {|f(x)-f(y)|} {|x-y|} \leq \frac {|x|} {|x-y|} + \frac {|y|} {|x-y|}=\frac {x-y} {|x-y|}=1$. The same inequality holds when $x$ and $y$ have the same sign (by MVT) so $|f(x)-f(y)| \leq |x-y|$ for all $x$ and $y$. PS: Lipschtiz functions need not have continuous derivatives.