Is this function $$x1_{\{x>0\}}(x)$$ Lipschitz? It's not differential so mean value cant be used here.
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Yes. You only need $|f(x)-f(y)| \le K |x-y|$, and $K=1$ will do.
Richard Martin
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please explain your reasoning, I cant see why. – Vaolter Nov 22 '18 at 14:45
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1@Vaolter: Have you sketched a graph of your function? – hmakholm left over Monica Nov 22 '18 at 14:47
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Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn – Richard Martin Nov 22 '18 at 14:48
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Yes. Or draw it. – Richard Martin Nov 22 '18 at 14:48
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@Richard Martin, Ok, thank you, I have checked it. – Vaolter Nov 22 '18 at 15:13