If $f'$ is bounded on $mathbb{R}$, then is $f$ uniformly continuous?












1















Say $f : mathbb{R} to mathbb{R}$ is differentiable.



Show if $f'$ is bounded, then $f$ is uniformly continuous.





Try



For any $[a, b] subset mathbb{R}$$(a<b)$, I can show $f$ is uniformly continuous, as follows.




Say $x_0 in [a,b]$ is given. Fix $epsilon >0$.



Since $exists M >0$ s.t. $|f'(x)| le M$, we show $|y - x_0| < epsilon/M Rightarrow |f(y) - f(x_0)| < epsilon$.



Assume $exists y_0 in [a,b]$ s.t. $|y_0 - x_0| < epsilon/M$ and $|f(y_0) - f(x_0)| ge epsilon$.



Since $frac{f(y_0) - f(x_0)}{y_0 - x_0} > epsilon times M/epsilon = M$, by MVT $exists c in left[min{x_0, y_0} max{x_0, y_0}right]$ s.t. $f'(c) = frac{f(y_0) - f(x_0)}{y_0 - x_0}$. Observe $|f'(c)| > M$, which is a contradiction to $|f'(x)| le M$.




But I'm not sure I can generalize this fact on $[a,b]$ to $mathbb{R}$.






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  • Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
    – Peter Szilas
    Nov 29 '18 at 21:49
















1















Say $f : mathbb{R} to mathbb{R}$ is differentiable.



Show if $f'$ is bounded, then $f$ is uniformly continuous.





Try



For any $[a, b] subset mathbb{R}$$(a<b)$, I can show $f$ is uniformly continuous, as follows.




Say $x_0 in [a,b]$ is given. Fix $epsilon >0$.



Since $exists M >0$ s.t. $|f'(x)| le M$, we show $|y - x_0| < epsilon/M Rightarrow |f(y) - f(x_0)| < epsilon$.



Assume $exists y_0 in [a,b]$ s.t. $|y_0 - x_0| < epsilon/M$ and $|f(y_0) - f(x_0)| ge epsilon$.



Since $frac{f(y_0) - f(x_0)}{y_0 - x_0} > epsilon times M/epsilon = M$, by MVT $exists c in left[min{x_0, y_0} max{x_0, y_0}right]$ s.t. $f'(c) = frac{f(y_0) - f(x_0)}{y_0 - x_0}$. Observe $|f'(c)| > M$, which is a contradiction to $|f'(x)| le M$.




But I'm not sure I can generalize this fact on $[a,b]$ to $mathbb{R}$.






real-analysis







share|cite|improve this question






















  • Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
    – Peter Szilas
    Nov 29 '18 at 21:49














1












1








1








Say $f : mathbb{R} to mathbb{R}$ is differentiable.



Show if $f'$ is bounded, then $f$ is uniformly continuous.





Try



For any $[a, b] subset mathbb{R}$$(a<b)$, I can show $f$ is uniformly continuous, as follows.




Say $x_0 in [a,b]$ is given. Fix $epsilon >0$.



Since $exists M >0$ s.t. $|f'(x)| le M$, we show $|y - x_0| < epsilon/M Rightarrow |f(y) - f(x_0)| < epsilon$.



Assume $exists y_0 in [a,b]$ s.t. $|y_0 - x_0| < epsilon/M$ and $|f(y_0) - f(x_0)| ge epsilon$.



Since $frac{f(y_0) - f(x_0)}{y_0 - x_0} > epsilon times M/epsilon = M$, by MVT $exists c in left[min{x_0, y_0} max{x_0, y_0}right]$ s.t. $f'(c) = frac{f(y_0) - f(x_0)}{y_0 - x_0}$. Observe $|f'(c)| > M$, which is a contradiction to $|f'(x)| le M$.




But I'm not sure I can generalize this fact on $[a,b]$ to $mathbb{R}$.






real-analysis







share|cite|improve this question














Say $f : mathbb{R} to mathbb{R}$ is differentiable.



Show if $f'$ is bounded, then $f$ is uniformly continuous.





Try



For any $[a, b] subset mathbb{R}$$(a<b)$, I can show $f$ is uniformly continuous, as follows.




Say $x_0 in [a,b]$ is given. Fix $epsilon >0$.



Since $exists M >0$ s.t. $|f'(x)| le M$, we show $|y - x_0| < epsilon/M Rightarrow |f(y) - f(x_0)| < epsilon$.



Assume $exists y_0 in [a,b]$ s.t. $|y_0 - x_0| < epsilon/M$ and $|f(y_0) - f(x_0)| ge epsilon$.



Since $frac{f(y_0) - f(x_0)}{y_0 - x_0} > epsilon times M/epsilon = M$, by MVT $exists c in left[min{x_0, y_0} max{x_0, y_0}right]$ s.t. $f'(c) = frac{f(y_0) - f(x_0)}{y_0 - x_0}$. Observe $|f'(c)| > M$, which is a contradiction to $|f'(x)| le M$.




But I'm not sure I can generalize this fact on $[a,b]$ to $mathbb{R}$.






real-analysis




real-analysis






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share|cite|improve this question











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asked Nov 29 '18 at 21:17









MoreblueMoreblue

8671216




8671216












  • Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
    – Peter Szilas
    Nov 29 '18 at 21:49


















  • Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
    – Peter Szilas
    Nov 29 '18 at 21:49
















Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
– Peter Szilas
Nov 29 '18 at 21:49




Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
– Peter Szilas
Nov 29 '18 at 21:49










2 Answers
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4














hint



$f$ is differentiable at $Bbb R$ thus by MVT, for all $x,y$, there exists $c$ in $Bbb R$ such that



$$f(x)-f(y)=(x-y)f'(c)$$
but $f'$ is bounded, so



there exist $K> 0$ :
$$|f(x)-f(y)|le K|x-y|.$$



you can take $eta=frac{epsilon}{K}$.






share|cite|improve this answer





























    1














    I don't think there is any trouble to generalize.



    If $x,y in mathbb R$ the mean-value theorem gives you a point $z$ in between $x$ and $y$ satisfying $$f(x) - f(y) = f'(z)(x-y)$$ so that
    $$|f(x) - f(y)| = |f'(z)| |x-y| le M |x-y|.$$
    This holds for all $x$ and $y$.



    Thus if $epsilon > 0$ and $|x-y| < epsilon /M$, then $|f(x) - f(y)| < epsilon$.






    share|cite|improve this answer





















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      2 Answers
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      active

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      2 Answers
      2






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      4














      hint



      $f$ is differentiable at $Bbb R$ thus by MVT, for all $x,y$, there exists $c$ in $Bbb R$ such that



      $$f(x)-f(y)=(x-y)f'(c)$$
      but $f'$ is bounded, so



      there exist $K> 0$ :
      $$|f(x)-f(y)|le K|x-y|.$$



      you can take $eta=frac{epsilon}{K}$.






      share|cite|improve this answer


























        4














        hint



        $f$ is differentiable at $Bbb R$ thus by MVT, for all $x,y$, there exists $c$ in $Bbb R$ such that



        $$f(x)-f(y)=(x-y)f'(c)$$
        but $f'$ is bounded, so



        there exist $K> 0$ :
        $$|f(x)-f(y)|le K|x-y|.$$



        you can take $eta=frac{epsilon}{K}$.






        share|cite|improve this answer
























          4












          4








          4






          hint



          $f$ is differentiable at $Bbb R$ thus by MVT, for all $x,y$, there exists $c$ in $Bbb R$ such that



          $$f(x)-f(y)=(x-y)f'(c)$$
          but $f'$ is bounded, so



          there exist $K> 0$ :
          $$|f(x)-f(y)|le K|x-y|.$$



          you can take $eta=frac{epsilon}{K}$.






          share|cite|improve this answer












          hint



          $f$ is differentiable at $Bbb R$ thus by MVT, for all $x,y$, there exists $c$ in $Bbb R$ such that



          $$f(x)-f(y)=(x-y)f'(c)$$
          but $f'$ is bounded, so



          there exist $K> 0$ :
          $$|f(x)-f(y)|le K|x-y|.$$



          you can take $eta=frac{epsilon}{K}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 21:24









          hamam_Abdallahhamam_Abdallah

          38.1k21634




          38.1k21634























              1














              I don't think there is any trouble to generalize.



              If $x,y in mathbb R$ the mean-value theorem gives you a point $z$ in between $x$ and $y$ satisfying $$f(x) - f(y) = f'(z)(x-y)$$ so that
              $$|f(x) - f(y)| = |f'(z)| |x-y| le M |x-y|.$$
              This holds for all $x$ and $y$.



              Thus if $epsilon > 0$ and $|x-y| < epsilon /M$, then $|f(x) - f(y)| < epsilon$.






              share|cite|improve this answer


























                1














                I don't think there is any trouble to generalize.



                If $x,y in mathbb R$ the mean-value theorem gives you a point $z$ in between $x$ and $y$ satisfying $$f(x) - f(y) = f'(z)(x-y)$$ so that
                $$|f(x) - f(y)| = |f'(z)| |x-y| le M |x-y|.$$
                This holds for all $x$ and $y$.



                Thus if $epsilon > 0$ and $|x-y| < epsilon /M$, then $|f(x) - f(y)| < epsilon$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  I don't think there is any trouble to generalize.



                  If $x,y in mathbb R$ the mean-value theorem gives you a point $z$ in between $x$ and $y$ satisfying $$f(x) - f(y) = f'(z)(x-y)$$ so that
                  $$|f(x) - f(y)| = |f'(z)| |x-y| le M |x-y|.$$
                  This holds for all $x$ and $y$.



                  Thus if $epsilon > 0$ and $|x-y| < epsilon /M$, then $|f(x) - f(y)| < epsilon$.






                  share|cite|improve this answer












                  I don't think there is any trouble to generalize.



                  If $x,y in mathbb R$ the mean-value theorem gives you a point $z$ in between $x$ and $y$ satisfying $$f(x) - f(y) = f'(z)(x-y)$$ so that
                  $$|f(x) - f(y)| = |f'(z)| |x-y| le M |x-y|.$$
                  This holds for all $x$ and $y$.



                  Thus if $epsilon > 0$ and $|x-y| < epsilon /M$, then $|f(x) - f(y)| < epsilon$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 '18 at 21:22









                  Umberto P.Umberto P.

                  38.6k13064




                  38.6k13064






























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