Example of a nowhere differentiable contraction mapping












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The Weierstrass function https://en.wikipedia.org/wiki/Weierstrass_function



is a pathological example of a continuous nowhere differentiable function.. Since a conttaction mapping is necessarily Lipschitz continuous (which is a stronger form of continuity), I was wondering if there exists a (pathological) example of a contraction mapping that is nowhere differentiable










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    3












    $begingroup$


    The Weierstrass function https://en.wikipedia.org/wiki/Weierstrass_function



    is a pathological example of a continuous nowhere differentiable function.. Since a conttaction mapping is necessarily Lipschitz continuous (which is a stronger form of continuity), I was wondering if there exists a (pathological) example of a contraction mapping that is nowhere differentiable










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      3












      3








      3





      $begingroup$


      The Weierstrass function https://en.wikipedia.org/wiki/Weierstrass_function



      is a pathological example of a continuous nowhere differentiable function.. Since a conttaction mapping is necessarily Lipschitz continuous (which is a stronger form of continuity), I was wondering if there exists a (pathological) example of a contraction mapping that is nowhere differentiable










      share|cite|improve this question









      $endgroup$




      The Weierstrass function https://en.wikipedia.org/wiki/Weierstrass_function



      is a pathological example of a continuous nowhere differentiable function.. Since a conttaction mapping is necessarily Lipschitz continuous (which is a stronger form of continuity), I was wondering if there exists a (pathological) example of a contraction mapping that is nowhere differentiable







      continuity lipschitz-functions contraction-operator






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      asked Dec 29 '18 at 7:39









      Somo S.Somo S.

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          No, by Rademacher's theorem a Lipschitz function $Bbb R^ntoBbb R^n$ is differentiable almost everywhere. A proof of this result can be found, for example, in Emmanuele DiBenedetto's Real Analysis, Theorem 21.1.






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            No, by Rademacher's theorem a Lipschitz function $Bbb R^ntoBbb R^n$ is differentiable almost everywhere. A proof of this result can be found, for example, in Emmanuele DiBenedetto's Real Analysis, Theorem 21.1.






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              $begingroup$

              No, by Rademacher's theorem a Lipschitz function $Bbb R^ntoBbb R^n$ is differentiable almost everywhere. A proof of this result can be found, for example, in Emmanuele DiBenedetto's Real Analysis, Theorem 21.1.






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                3





                $begingroup$

                No, by Rademacher's theorem a Lipschitz function $Bbb R^ntoBbb R^n$ is differentiable almost everywhere. A proof of this result can be found, for example, in Emmanuele DiBenedetto's Real Analysis, Theorem 21.1.






                share|cite|improve this answer











                $endgroup$



                No, by Rademacher's theorem a Lipschitz function $Bbb R^ntoBbb R^n$ is differentiable almost everywhere. A proof of this result can be found, for example, in Emmanuele DiBenedetto's Real Analysis, Theorem 21.1.







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                edited Dec 29 '18 at 13:03

























                answered Dec 29 '18 at 9:52









                Alessandro CodenottiAlessandro Codenotti

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