Lipschitz functions are $o(|x|)$?












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Consider a Lipschitz function from $mathbb{R}tomathbb{R}$. Can we say that $lim_{xtoinfty}frac{f(x)}{|x|}=0$. Can we also say that $f$ is differentiable. Continuity is quite evident. But linear order is hard to come by. Any hints? Thanks beforehand.






ordinary-differential-equations continuity lipschitz-functions







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    @user539887 thanks. edited the post
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    – vidyarthi
    Dec 10 '18 at 7:30
















0












$begingroup$


Consider a Lipschitz function from $mathbb{R}tomathbb{R}$. Can we say that $lim_{xtoinfty}frac{f(x)}{|x|}=0$. Can we also say that $f$ is differentiable. Continuity is quite evident. But linear order is hard to come by. Any hints? Thanks beforehand.






ordinary-differential-equations continuity lipschitz-functions







share|cite|improve this question











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  • 1




    $begingroup$
    @user539887 thanks. edited the post
    $endgroup$
    – vidyarthi
    Dec 10 '18 at 7:30














0












0








0





$begingroup$


Consider a Lipschitz function from $mathbb{R}tomathbb{R}$. Can we say that $lim_{xtoinfty}frac{f(x)}{|x|}=0$. Can we also say that $f$ is differentiable. Continuity is quite evident. But linear order is hard to come by. Any hints? Thanks beforehand.






ordinary-differential-equations continuity lipschitz-functions







share|cite|improve this question











$endgroup$




Consider a Lipschitz function from $mathbb{R}tomathbb{R}$. Can we say that $lim_{xtoinfty}frac{f(x)}{|x|}=0$. Can we also say that $f$ is differentiable. Continuity is quite evident. But linear order is hard to come by. Any hints? Thanks beforehand.






ordinary-differential-equations continuity lipschitz-functions




ordinary-differential-equations continuity lipschitz-functions






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edited Dec 10 '18 at 7:30







vidyarthi

















asked Dec 10 '18 at 7:20









vidyarthividyarthi

2,9541832




2,9541832








  • 1




    $begingroup$
    @user539887 thanks. edited the post
    $endgroup$
    – vidyarthi
    Dec 10 '18 at 7:30














  • 1




    $begingroup$
    @user539887 thanks. edited the post
    $endgroup$
    – vidyarthi
    Dec 10 '18 at 7:30








1




1




$begingroup$
@user539887 thanks. edited the post
$endgroup$
– vidyarthi
Dec 10 '18 at 7:30




$begingroup$
@user539887 thanks. edited the post
$endgroup$
– vidyarthi
Dec 10 '18 at 7:30










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$f(x)=x$ is a counterexample for the first part. Lipschitz functions need not be differentiable at every point but they are differentiable almost everywhere.






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    No, consider the function $f(x)=|x|$ which is Lipschitz in $mathbb{R}$ (with constant $1$), not differentiable at $0$, and
    $$lim_{xtopminfty}frac{f(x)}{|x|}=pm 1.$$






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    • $begingroup$
      @vidyarthi Are you still interested in this question?
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      – Robert Z
      Jan 16 at 17:34











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    2 Answers
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    2 Answers
    2






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    0












    $begingroup$

    $f(x)=x$ is a counterexample for the first part. Lipschitz functions need not be differentiable at every point but they are differentiable almost everywhere.






    share|cite|improve this answer









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      0












      $begingroup$

      $f(x)=x$ is a counterexample for the first part. Lipschitz functions need not be differentiable at every point but they are differentiable almost everywhere.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $f(x)=x$ is a counterexample for the first part. Lipschitz functions need not be differentiable at every point but they are differentiable almost everywhere.






        share|cite|improve this answer









        $endgroup$



        $f(x)=x$ is a counterexample for the first part. Lipschitz functions need not be differentiable at every point but they are differentiable almost everywhere.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 7:23









        Kavi Rama MurthyKavi Rama Murthy

        58.2k42161




        58.2k42161























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            $begingroup$

            No, consider the function $f(x)=|x|$ which is Lipschitz in $mathbb{R}$ (with constant $1$), not differentiable at $0$, and
            $$lim_{xtopminfty}frac{f(x)}{|x|}=pm 1.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @vidyarthi Are you still interested in this question?
              $endgroup$
              – Robert Z
              Jan 16 at 17:34
















            0












            $begingroup$

            No, consider the function $f(x)=|x|$ which is Lipschitz in $mathbb{R}$ (with constant $1$), not differentiable at $0$, and
            $$lim_{xtopminfty}frac{f(x)}{|x|}=pm 1.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @vidyarthi Are you still interested in this question?
              $endgroup$
              – Robert Z
              Jan 16 at 17:34














            0












            0








            0





            $begingroup$

            No, consider the function $f(x)=|x|$ which is Lipschitz in $mathbb{R}$ (with constant $1$), not differentiable at $0$, and
            $$lim_{xtopminfty}frac{f(x)}{|x|}=pm 1.$$






            share|cite|improve this answer











            $endgroup$



            No, consider the function $f(x)=|x|$ which is Lipschitz in $mathbb{R}$ (with constant $1$), not differentiable at $0$, and
            $$lim_{xtopminfty}frac{f(x)}{|x|}=pm 1.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 10 '18 at 8:06

























            answered Dec 10 '18 at 7:22









            Robert ZRobert Z

            96.3k1065136




            96.3k1065136












            • $begingroup$
              @vidyarthi Are you still interested in this question?
              $endgroup$
              – Robert Z
              Jan 16 at 17:34


















            • $begingroup$
              @vidyarthi Are you still interested in this question?
              $endgroup$
              – Robert Z
              Jan 16 at 17:34
















            $begingroup$
            @vidyarthi Are you still interested in this question?
            $endgroup$
            – Robert Z
            Jan 16 at 17:34




            $begingroup$
            @vidyarthi Are you still interested in this question?
            $endgroup$
            – Robert Z
            Jan 16 at 17:34


















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