$f$ is a $k$-Lipschitz continuous function & continuously differentiable $forall x in mathbb{R}^n$....












0












$begingroup$


EDIT - generalization for vector analysis



for $f: mathbb{R}^n to mathbb{R}^m$, $f$ is continuously differentiable $forall x in mathbb{R}^n$, and is
$k$-Lipschitz continuous function.



ProveDisprove: $$forall xin mathbb{R}^n: ||D_{f}(x)||_{op} leq k$$



$ $



previous Version:
I was asking the same question but for $f: mathbb{R} to mathbb{R}$



Let $f: mathbb{R} to mathbb{R}$ such that $f$ is continuously differentiable $forall x in mathbb{R}$.



Given that $f$ is a $k$-Lipschitz continuous, is it possible to prove that the derivative of $f$ is bounded by $k$?



Prove/disprove: $f'$ is bounded by $k$, as $f$ is a $k$-Lipschitz continuous function ,and continuously differentiable $forall x in mathbb{R}$



The only intuition I have in mind is that:



$$|f(y) - f(x)| = |f'(x)(x-y) + small{o}(x-y)| leq k|x-y|$$



Therefore
$$ left|f'(x) + frac{small{o}(|x-y)|}{x-y} right| leq k$$



But that isn't a sufficent term in order to prove that the derivative of $f$ is bounded by $k$.






vector-analysis lipschitz-functions







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    EDIT - generalization for vector analysis



    for $f: mathbb{R}^n to mathbb{R}^m$, $f$ is continuously differentiable $forall x in mathbb{R}^n$, and is
    $k$-Lipschitz continuous function.



    ProveDisprove: $$forall xin mathbb{R}^n: ||D_{f}(x)||_{op} leq k$$



    $ $



    previous Version:
    I was asking the same question but for $f: mathbb{R} to mathbb{R}$



    Let $f: mathbb{R} to mathbb{R}$ such that $f$ is continuously differentiable $forall x in mathbb{R}$.



    Given that $f$ is a $k$-Lipschitz continuous, is it possible to prove that the derivative of $f$ is bounded by $k$?



    Prove/disprove: $f'$ is bounded by $k$, as $f$ is a $k$-Lipschitz continuous function ,and continuously differentiable $forall x in mathbb{R}$



    The only intuition I have in mind is that:



    $$|f(y) - f(x)| = |f'(x)(x-y) + small{o}(x-y)| leq k|x-y|$$



    Therefore
    $$ left|f'(x) + frac{small{o}(|x-y)|}{x-y} right| leq k$$



    But that isn't a sufficent term in order to prove that the derivative of $f$ is bounded by $k$.






    vector-analysis lipschitz-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      EDIT - generalization for vector analysis



      for $f: mathbb{R}^n to mathbb{R}^m$, $f$ is continuously differentiable $forall x in mathbb{R}^n$, and is
      $k$-Lipschitz continuous function.



      ProveDisprove: $$forall xin mathbb{R}^n: ||D_{f}(x)||_{op} leq k$$



      $ $



      previous Version:
      I was asking the same question but for $f: mathbb{R} to mathbb{R}$



      Let $f: mathbb{R} to mathbb{R}$ such that $f$ is continuously differentiable $forall x in mathbb{R}$.



      Given that $f$ is a $k$-Lipschitz continuous, is it possible to prove that the derivative of $f$ is bounded by $k$?



      Prove/disprove: $f'$ is bounded by $k$, as $f$ is a $k$-Lipschitz continuous function ,and continuously differentiable $forall x in mathbb{R}$



      The only intuition I have in mind is that:



      $$|f(y) - f(x)| = |f'(x)(x-y) + small{o}(x-y)| leq k|x-y|$$



      Therefore
      $$ left|f'(x) + frac{small{o}(|x-y)|}{x-y} right| leq k$$



      But that isn't a sufficent term in order to prove that the derivative of $f$ is bounded by $k$.






      vector-analysis lipschitz-functions







      share|cite|improve this question











      $endgroup$




      EDIT - generalization for vector analysis



      for $f: mathbb{R}^n to mathbb{R}^m$, $f$ is continuously differentiable $forall x in mathbb{R}^n$, and is
      $k$-Lipschitz continuous function.



      ProveDisprove: $$forall xin mathbb{R}^n: ||D_{f}(x)||_{op} leq k$$



      $ $



      previous Version:
      I was asking the same question but for $f: mathbb{R} to mathbb{R}$



      Let $f: mathbb{R} to mathbb{R}$ such that $f$ is continuously differentiable $forall x in mathbb{R}$.



      Given that $f$ is a $k$-Lipschitz continuous, is it possible to prove that the derivative of $f$ is bounded by $k$?



      Prove/disprove: $f'$ is bounded by $k$, as $f$ is a $k$-Lipschitz continuous function ,and continuously differentiable $forall x in mathbb{R}$



      The only intuition I have in mind is that:



      $$|f(y) - f(x)| = |f'(x)(x-y) + small{o}(x-y)| leq k|x-y|$$



      Therefore
      $$ left|f'(x) + frac{small{o}(|x-y)|}{x-y} right| leq k$$



      But that isn't a sufficent term in order to prove that the derivative of $f$ is bounded by $k$.






      vector-analysis lipschitz-functions




      vector-analysis lipschitz-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 23 at 13:14







      Jneven

















      asked Jan 23 at 11:40









      JnevenJneven

      906322




      906322






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          We have
          $$ D_f(x) xi = lim_{trightarrow 0} frac{f(x+ tcdot xi) - f(x)}{t}.$$
          But by the lipschitz condition we get
          $$ left Vert frac{f(x+ tcdot xi) - f(x)}{t} rightVert leq k Vert xi Vert $$
          And thus, we have
          $$ Vert D_f(x) xi Vert leq k Vert xi Vert $$
          implying that
          $$ Vert D_f(x) Vert_{op} leq k. $$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            If $f$ is k -Lipschitz continuous, you have
            $$
            left| frac{f(y)-f(x)}{y-x} right| le k
            $$
            for all $y neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| le k.$






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              You should isolate instead $f'(x)$, in that case:



              $f'(x)=dfrac{f(x)-f(y)}{x-y}+ dfrac{o(vert x-yvert)}{x-y}= dfrac{f(x)-f(y)}{x-y}+o(1)$



              Then:



              $vert f'(x)vert leq Bigg vert dfrac{f(x)-f(y)}{x-y} Bigg vert+ vert o(1)vert leq k+ o(1)overset{y rightarrow x}{rightarrow}k$






              share|cite|improve this answer









              $endgroup$













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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                We have
                $$ D_f(x) xi = lim_{trightarrow 0} frac{f(x+ tcdot xi) - f(x)}{t}.$$
                But by the lipschitz condition we get
                $$ left Vert frac{f(x+ tcdot xi) - f(x)}{t} rightVert leq k Vert xi Vert $$
                And thus, we have
                $$ Vert D_f(x) xi Vert leq k Vert xi Vert $$
                implying that
                $$ Vert D_f(x) Vert_{op} leq k. $$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  We have
                  $$ D_f(x) xi = lim_{trightarrow 0} frac{f(x+ tcdot xi) - f(x)}{t}.$$
                  But by the lipschitz condition we get
                  $$ left Vert frac{f(x+ tcdot xi) - f(x)}{t} rightVert leq k Vert xi Vert $$
                  And thus, we have
                  $$ Vert D_f(x) xi Vert leq k Vert xi Vert $$
                  implying that
                  $$ Vert D_f(x) Vert_{op} leq k. $$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    We have
                    $$ D_f(x) xi = lim_{trightarrow 0} frac{f(x+ tcdot xi) - f(x)}{t}.$$
                    But by the lipschitz condition we get
                    $$ left Vert frac{f(x+ tcdot xi) - f(x)}{t} rightVert leq k Vert xi Vert $$
                    And thus, we have
                    $$ Vert D_f(x) xi Vert leq k Vert xi Vert $$
                    implying that
                    $$ Vert D_f(x) Vert_{op} leq k. $$






                    share|cite|improve this answer









                    $endgroup$



                    We have
                    $$ D_f(x) xi = lim_{trightarrow 0} frac{f(x+ tcdot xi) - f(x)}{t}.$$
                    But by the lipschitz condition we get
                    $$ left Vert frac{f(x+ tcdot xi) - f(x)}{t} rightVert leq k Vert xi Vert $$
                    And thus, we have
                    $$ Vert D_f(x) xi Vert leq k Vert xi Vert $$
                    implying that
                    $$ Vert D_f(x) Vert_{op} leq k. $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 23 at 13:30









                    Severin SchravenSeverin Schraven

                    6,3381934




                    6,3381934























                        2












                        $begingroup$

                        If $f$ is k -Lipschitz continuous, you have
                        $$
                        left| frac{f(y)-f(x)}{y-x} right| le k
                        $$
                        for all $y neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| le k.$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          If $f$ is k -Lipschitz continuous, you have
                          $$
                          left| frac{f(y)-f(x)}{y-x} right| le k
                          $$
                          for all $y neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| le k.$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            If $f$ is k -Lipschitz continuous, you have
                            $$
                            left| frac{f(y)-f(x)}{y-x} right| le k
                            $$
                            for all $y neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| le k.$






                            share|cite|improve this answer









                            $endgroup$



                            If $f$ is k -Lipschitz continuous, you have
                            $$
                            left| frac{f(y)-f(x)}{y-x} right| le k
                            $$
                            for all $y neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| le k.$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 23 at 11:46









                            Reiner MartinReiner Martin

                            3,509414




                            3,509414























                                1












                                $begingroup$

                                You should isolate instead $f'(x)$, in that case:



                                $f'(x)=dfrac{f(x)-f(y)}{x-y}+ dfrac{o(vert x-yvert)}{x-y}= dfrac{f(x)-f(y)}{x-y}+o(1)$



                                Then:



                                $vert f'(x)vert leq Bigg vert dfrac{f(x)-f(y)}{x-y} Bigg vert+ vert o(1)vert leq k+ o(1)overset{y rightarrow x}{rightarrow}k$






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  You should isolate instead $f'(x)$, in that case:



                                  $f'(x)=dfrac{f(x)-f(y)}{x-y}+ dfrac{o(vert x-yvert)}{x-y}= dfrac{f(x)-f(y)}{x-y}+o(1)$



                                  Then:



                                  $vert f'(x)vert leq Bigg vert dfrac{f(x)-f(y)}{x-y} Bigg vert+ vert o(1)vert leq k+ o(1)overset{y rightarrow x}{rightarrow}k$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    You should isolate instead $f'(x)$, in that case:



                                    $f'(x)=dfrac{f(x)-f(y)}{x-y}+ dfrac{o(vert x-yvert)}{x-y}= dfrac{f(x)-f(y)}{x-y}+o(1)$



                                    Then:



                                    $vert f'(x)vert leq Bigg vert dfrac{f(x)-f(y)}{x-y} Bigg vert+ vert o(1)vert leq k+ o(1)overset{y rightarrow x}{rightarrow}k$






                                    share|cite|improve this answer









                                    $endgroup$



                                    You should isolate instead $f'(x)$, in that case:



                                    $f'(x)=dfrac{f(x)-f(y)}{x-y}+ dfrac{o(vert x-yvert)}{x-y}= dfrac{f(x)-f(y)}{x-y}+o(1)$



                                    Then:



                                    $vert f'(x)vert leq Bigg vert dfrac{f(x)-f(y)}{x-y} Bigg vert+ vert o(1)vert leq k+ o(1)overset{y rightarrow x}{rightarrow}k$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 23 at 11:49









                                    Keen-ameteurKeen-ameteur

                                    1,477416




                                    1,477416






























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