Is this function involving indicator function Lipschitz?











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Is this function $$x1_{{x>0}}(x)$$ Lipschitz? It's not differential so mean value cant be used here.










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    Is this function $$x1_{{x>0}}(x)$$ Lipschitz? It's not differential so mean value cant be used here.










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      Is this function $$x1_{{x>0}}(x)$$ Lipschitz? It's not differential so mean value cant be used here.










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      Is this function $$x1_{{x>0}}(x)$$ Lipschitz? It's not differential so mean value cant be used here.







      lipschitz-functions






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      asked Nov 22 at 14:30









      Vaolter

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          Yes. You only need $|f(x)-f(y)| le K |x-y|$, and $K=1$ will do.






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          • please explain your reasoning, I cant see why.
            – Vaolter
            Nov 22 at 14:45






          • 1




            @Vaolter: Have you sketched a graph of your function?
            – Henning Makholm
            Nov 22 at 14:47












          • Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
            – Richard Martin
            Nov 22 at 14:48










          • Yes. Or draw it.
            – Richard Martin
            Nov 22 at 14:48










          • @Richard Martin, Ok, thank you, I have checked it.
            – Vaolter
            Nov 22 at 15:13











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          Yes. You only need $|f(x)-f(y)| le K |x-y|$, and $K=1$ will do.






          share|cite|improve this answer





















          • please explain your reasoning, I cant see why.
            – Vaolter
            Nov 22 at 14:45






          • 1




            @Vaolter: Have you sketched a graph of your function?
            – Henning Makholm
            Nov 22 at 14:47












          • Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
            – Richard Martin
            Nov 22 at 14:48










          • Yes. Or draw it.
            – Richard Martin
            Nov 22 at 14:48










          • @Richard Martin, Ok, thank you, I have checked it.
            – Vaolter
            Nov 22 at 15:13















          up vote
          2
          down vote



          accepted










          Yes. You only need $|f(x)-f(y)| le K |x-y|$, and $K=1$ will do.






          share|cite|improve this answer





















          • please explain your reasoning, I cant see why.
            – Vaolter
            Nov 22 at 14:45






          • 1




            @Vaolter: Have you sketched a graph of your function?
            – Henning Makholm
            Nov 22 at 14:47












          • Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
            – Richard Martin
            Nov 22 at 14:48










          • Yes. Or draw it.
            – Richard Martin
            Nov 22 at 14:48










          • @Richard Martin, Ok, thank you, I have checked it.
            – Vaolter
            Nov 22 at 15:13













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Yes. You only need $|f(x)-f(y)| le K |x-y|$, and $K=1$ will do.






          share|cite|improve this answer












          Yes. You only need $|f(x)-f(y)| le K |x-y|$, and $K=1$ will do.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 14:41









          Richard Martin

          1,6328




          1,6328












          • please explain your reasoning, I cant see why.
            – Vaolter
            Nov 22 at 14:45






          • 1




            @Vaolter: Have you sketched a graph of your function?
            – Henning Makholm
            Nov 22 at 14:47












          • Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
            – Richard Martin
            Nov 22 at 14:48










          • Yes. Or draw it.
            – Richard Martin
            Nov 22 at 14:48










          • @Richard Martin, Ok, thank you, I have checked it.
            – Vaolter
            Nov 22 at 15:13


















          • please explain your reasoning, I cant see why.
            – Vaolter
            Nov 22 at 14:45






          • 1




            @Vaolter: Have you sketched a graph of your function?
            – Henning Makholm
            Nov 22 at 14:47












          • Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
            – Richard Martin
            Nov 22 at 14:48










          • Yes. Or draw it.
            – Richard Martin
            Nov 22 at 14:48










          • @Richard Martin, Ok, thank you, I have checked it.
            – Vaolter
            Nov 22 at 15:13
















          please explain your reasoning, I cant see why.
          – Vaolter
          Nov 22 at 14:45




          please explain your reasoning, I cant see why.
          – Vaolter
          Nov 22 at 14:45




          1




          1




          @Vaolter: Have you sketched a graph of your function?
          – Henning Makholm
          Nov 22 at 14:47






          @Vaolter: Have you sketched a graph of your function?
          – Henning Makholm
          Nov 22 at 14:47














          Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
          – Richard Martin
          Nov 22 at 14:48




          Evaluate both sides, taking cases (i) $x,y>0$, (ii) $x,y<0$ etc in turn
          – Richard Martin
          Nov 22 at 14:48












          Yes. Or draw it.
          – Richard Martin
          Nov 22 at 14:48




          Yes. Or draw it.
          – Richard Martin
          Nov 22 at 14:48












          @Richard Martin, Ok, thank you, I have checked it.
          – Vaolter
          Nov 22 at 15:13




          @Richard Martin, Ok, thank you, I have checked it.
          – Vaolter
          Nov 22 at 15:13


















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