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 = $0$.
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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.
Kavi Rama Murthy
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So, continuous derivatives is not a necessary condition (but it is a sufficient one?) Since, Khalil's book mentions that if $ f(t,x) ;\quad \frac{\partial f}{\partial x} (t,x)$ continuous on [a,b] x $R^n \implies f$ is globally lipschitz in x on $[a,b]xR^n$ iff $\frac{\partial f}{\partial x} $ uniformly bounded $on [a,b]xR^n$ – db18 May 03 '18 at 14:25
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Even differentiability at every point is not necessary: $|x|$ is Lipschitz but not differentiable at 0. – Kavi Rama Murthy May 05 '18 at 00:00