Lipschitz continuity implies continuity












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Say $vert f(x)-f(y)vert le Lvert x-yvert$. How to prove the following: $forall lim_{n to infty} x_n = x_0 wedge x_0in Bbb{R}$: $lim_{n to infty}f(x_n) = f(x_0)$?



In other words: how to prove that Lipschitz-continuity implies regular continuity(without using differentiation or similar methods)?






real-analysis continuity lipschitz-functions







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    0












    $begingroup$


    Say $vert f(x)-f(y)vert le Lvert x-yvert$. How to prove the following: $forall lim_{n to infty} x_n = x_0 wedge x_0in Bbb{R}$: $lim_{n to infty}f(x_n) = f(x_0)$?



    In other words: how to prove that Lipschitz-continuity implies regular continuity(without using differentiation or similar methods)?






    real-analysis continuity lipschitz-functions







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Say $vert f(x)-f(y)vert le Lvert x-yvert$. How to prove the following: $forall lim_{n to infty} x_n = x_0 wedge x_0in Bbb{R}$: $lim_{n to infty}f(x_n) = f(x_0)$?



      In other words: how to prove that Lipschitz-continuity implies regular continuity(without using differentiation or similar methods)?






      real-analysis continuity lipschitz-functions







      share|cite|improve this question









      $endgroup$




      Say $vert f(x)-f(y)vert le Lvert x-yvert$. How to prove the following: $forall lim_{n to infty} x_n = x_0 wedge x_0in Bbb{R}$: $lim_{n to infty}f(x_n) = f(x_0)$?



      In other words: how to prove that Lipschitz-continuity implies regular continuity(without using differentiation or similar methods)?






      real-analysis continuity lipschitz-functions




      real-analysis continuity lipschitz-functions






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 6 '18 at 19:37









      Conny DagoConny Dago

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          3 Answers
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          1












          $begingroup$

          $x_nrightarrow x Leftrightarrow forall varepsilon >0 ,exists N_0: nge N_0 Rightarrow |x-x_n|le varepsilon$



          Now if $f$ is Lipshitz and $x_nrightarrow x$ this implies that



          $$|f(x)-f(x_n)|le L|x-x_n|le Lvarepsilon$$
          for such a sequence.



          Can you finish the proof with this information?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            No, unfortunately. I'm thinking of $lim_{n to infty} vert f(x_0)-f(x_n)vert le lim_{n to infty} vert x_0-x_n vert $ But I don't think it's gonna lead me further...
            $endgroup$
            – Conny Dago
            Dec 6 '18 at 20:50










          • $begingroup$
            Note: hamam_Abdallah's answer is less formal than Thomas' which is not to disrespect what anyone has written here. Hopefully @ConnyDago is able to recognize that as a matter of formality, there are different ways to write an argument.
            $endgroup$
            – Matt A Pelto
            Dec 7 '18 at 6:50



















          1












          $begingroup$

          hint



          $$|f(x_n)-f(x_0)|le L|x_n-x_0|$$



          $$implies$$



          $$f(x_0)-L|x_n-x_0|le f(x_n)le f(x_0)+L|x_n-x_0|$$



          now squeeze.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Assuming $L>0$ (otherwise the function $f$ is constant which is trivial):



            Let ${x_n}_{n=1}^infty$ be a sequence of real numbers such that $lim_{n to infty} x_n=x_0$.



            Let $varepsilon>0$ be given, and define $varepsilon':=min{varepsilon, frac{varepsilon}L}$. Since $x_n to x_0$ as $n to infty$ and $varepsilon'>0$, we may find a positive integer $N$ so that $|x_n-x_0|<varepsilon'$ whenever $n geq N$. Thus we have
            $$ |, f(x_n)-f(x_0)|leq L|x_n-x_0| < Lvarepsilon' leq varepsilon text{ whenever } n geq N.$$



            Therefore, $f$ is sequentially continuous on $mathbb R$.






            share|cite|improve this answer











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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              $x_nrightarrow x Leftrightarrow forall varepsilon >0 ,exists N_0: nge N_0 Rightarrow |x-x_n|le varepsilon$



              Now if $f$ is Lipshitz and $x_nrightarrow x$ this implies that



              $$|f(x)-f(x_n)|le L|x-x_n|le Lvarepsilon$$
              for such a sequence.



              Can you finish the proof with this information?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                No, unfortunately. I'm thinking of $lim_{n to infty} vert f(x_0)-f(x_n)vert le lim_{n to infty} vert x_0-x_n vert $ But I don't think it's gonna lead me further...
                $endgroup$
                – Conny Dago
                Dec 6 '18 at 20:50










              • $begingroup$
                Note: hamam_Abdallah's answer is less formal than Thomas' which is not to disrespect what anyone has written here. Hopefully @ConnyDago is able to recognize that as a matter of formality, there are different ways to write an argument.
                $endgroup$
                – Matt A Pelto
                Dec 7 '18 at 6:50
















              1












              $begingroup$

              $x_nrightarrow x Leftrightarrow forall varepsilon >0 ,exists N_0: nge N_0 Rightarrow |x-x_n|le varepsilon$



              Now if $f$ is Lipshitz and $x_nrightarrow x$ this implies that



              $$|f(x)-f(x_n)|le L|x-x_n|le Lvarepsilon$$
              for such a sequence.



              Can you finish the proof with this information?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                No, unfortunately. I'm thinking of $lim_{n to infty} vert f(x_0)-f(x_n)vert le lim_{n to infty} vert x_0-x_n vert $ But I don't think it's gonna lead me further...
                $endgroup$
                – Conny Dago
                Dec 6 '18 at 20:50










              • $begingroup$
                Note: hamam_Abdallah's answer is less formal than Thomas' which is not to disrespect what anyone has written here. Hopefully @ConnyDago is able to recognize that as a matter of formality, there are different ways to write an argument.
                $endgroup$
                – Matt A Pelto
                Dec 7 '18 at 6:50














              1












              1








              1





              $begingroup$

              $x_nrightarrow x Leftrightarrow forall varepsilon >0 ,exists N_0: nge N_0 Rightarrow |x-x_n|le varepsilon$



              Now if $f$ is Lipshitz and $x_nrightarrow x$ this implies that



              $$|f(x)-f(x_n)|le L|x-x_n|le Lvarepsilon$$
              for such a sequence.



              Can you finish the proof with this information?






              share|cite|improve this answer









              $endgroup$



              $x_nrightarrow x Leftrightarrow forall varepsilon >0 ,exists N_0: nge N_0 Rightarrow |x-x_n|le varepsilon$



              Now if $f$ is Lipshitz and $x_nrightarrow x$ this implies that



              $$|f(x)-f(x_n)|le L|x-x_n|le Lvarepsilon$$
              for such a sequence.



              Can you finish the proof with this information?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 6 '18 at 19:53









              ThomasThomas

              16.8k21631




              16.8k21631












              • $begingroup$
                No, unfortunately. I'm thinking of $lim_{n to infty} vert f(x_0)-f(x_n)vert le lim_{n to infty} vert x_0-x_n vert $ But I don't think it's gonna lead me further...
                $endgroup$
                – Conny Dago
                Dec 6 '18 at 20:50










              • $begingroup$
                Note: hamam_Abdallah's answer is less formal than Thomas' which is not to disrespect what anyone has written here. Hopefully @ConnyDago is able to recognize that as a matter of formality, there are different ways to write an argument.
                $endgroup$
                – Matt A Pelto
                Dec 7 '18 at 6:50


















              • $begingroup$
                No, unfortunately. I'm thinking of $lim_{n to infty} vert f(x_0)-f(x_n)vert le lim_{n to infty} vert x_0-x_n vert $ But I don't think it's gonna lead me further...
                $endgroup$
                – Conny Dago
                Dec 6 '18 at 20:50










              • $begingroup$
                Note: hamam_Abdallah's answer is less formal than Thomas' which is not to disrespect what anyone has written here. Hopefully @ConnyDago is able to recognize that as a matter of formality, there are different ways to write an argument.
                $endgroup$
                – Matt A Pelto
                Dec 7 '18 at 6:50
















              $begingroup$
              No, unfortunately. I'm thinking of $lim_{n to infty} vert f(x_0)-f(x_n)vert le lim_{n to infty} vert x_0-x_n vert $ But I don't think it's gonna lead me further...
              $endgroup$
              – Conny Dago
              Dec 6 '18 at 20:50




              $begingroup$
              No, unfortunately. I'm thinking of $lim_{n to infty} vert f(x_0)-f(x_n)vert le lim_{n to infty} vert x_0-x_n vert $ But I don't think it's gonna lead me further...
              $endgroup$
              – Conny Dago
              Dec 6 '18 at 20:50












              $begingroup$
              Note: hamam_Abdallah's answer is less formal than Thomas' which is not to disrespect what anyone has written here. Hopefully @ConnyDago is able to recognize that as a matter of formality, there are different ways to write an argument.
              $endgroup$
              – Matt A Pelto
              Dec 7 '18 at 6:50




              $begingroup$
              Note: hamam_Abdallah's answer is less formal than Thomas' which is not to disrespect what anyone has written here. Hopefully @ConnyDago is able to recognize that as a matter of formality, there are different ways to write an argument.
              $endgroup$
              – Matt A Pelto
              Dec 7 '18 at 6:50











              1












              $begingroup$

              hint



              $$|f(x_n)-f(x_0)|le L|x_n-x_0|$$



              $$implies$$



              $$f(x_0)-L|x_n-x_0|le f(x_n)le f(x_0)+L|x_n-x_0|$$



              now squeeze.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                hint



                $$|f(x_n)-f(x_0)|le L|x_n-x_0|$$



                $$implies$$



                $$f(x_0)-L|x_n-x_0|le f(x_n)le f(x_0)+L|x_n-x_0|$$



                now squeeze.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  hint



                  $$|f(x_n)-f(x_0)|le L|x_n-x_0|$$



                  $$implies$$



                  $$f(x_0)-L|x_n-x_0|le f(x_n)le f(x_0)+L|x_n-x_0|$$



                  now squeeze.






                  share|cite|improve this answer









                  $endgroup$



                  hint



                  $$|f(x_n)-f(x_0)|le L|x_n-x_0|$$



                  $$implies$$



                  $$f(x_0)-L|x_n-x_0|le f(x_n)le f(x_0)+L|x_n-x_0|$$



                  now squeeze.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 19:55









                  hamam_Abdallahhamam_Abdallah

                  38k21634




                  38k21634























                      0












                      $begingroup$

                      Assuming $L>0$ (otherwise the function $f$ is constant which is trivial):



                      Let ${x_n}_{n=1}^infty$ be a sequence of real numbers such that $lim_{n to infty} x_n=x_0$.



                      Let $varepsilon>0$ be given, and define $varepsilon':=min{varepsilon, frac{varepsilon}L}$. Since $x_n to x_0$ as $n to infty$ and $varepsilon'>0$, we may find a positive integer $N$ so that $|x_n-x_0|<varepsilon'$ whenever $n geq N$. Thus we have
                      $$ |, f(x_n)-f(x_0)|leq L|x_n-x_0| < Lvarepsilon' leq varepsilon text{ whenever } n geq N.$$



                      Therefore, $f$ is sequentially continuous on $mathbb R$.






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Assuming $L>0$ (otherwise the function $f$ is constant which is trivial):



                        Let ${x_n}_{n=1}^infty$ be a sequence of real numbers such that $lim_{n to infty} x_n=x_0$.



                        Let $varepsilon>0$ be given, and define $varepsilon':=min{varepsilon, frac{varepsilon}L}$. Since $x_n to x_0$ as $n to infty$ and $varepsilon'>0$, we may find a positive integer $N$ so that $|x_n-x_0|<varepsilon'$ whenever $n geq N$. Thus we have
                        $$ |, f(x_n)-f(x_0)|leq L|x_n-x_0| < Lvarepsilon' leq varepsilon text{ whenever } n geq N.$$



                        Therefore, $f$ is sequentially continuous on $mathbb R$.






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Assuming $L>0$ (otherwise the function $f$ is constant which is trivial):



                          Let ${x_n}_{n=1}^infty$ be a sequence of real numbers such that $lim_{n to infty} x_n=x_0$.



                          Let $varepsilon>0$ be given, and define $varepsilon':=min{varepsilon, frac{varepsilon}L}$. Since $x_n to x_0$ as $n to infty$ and $varepsilon'>0$, we may find a positive integer $N$ so that $|x_n-x_0|<varepsilon'$ whenever $n geq N$. Thus we have
                          $$ |, f(x_n)-f(x_0)|leq L|x_n-x_0| < Lvarepsilon' leq varepsilon text{ whenever } n geq N.$$



                          Therefore, $f$ is sequentially continuous on $mathbb R$.






                          share|cite|improve this answer











                          $endgroup$



                          Assuming $L>0$ (otherwise the function $f$ is constant which is trivial):



                          Let ${x_n}_{n=1}^infty$ be a sequence of real numbers such that $lim_{n to infty} x_n=x_0$.



                          Let $varepsilon>0$ be given, and define $varepsilon':=min{varepsilon, frac{varepsilon}L}$. Since $x_n to x_0$ as $n to infty$ and $varepsilon'>0$, we may find a positive integer $N$ so that $|x_n-x_0|<varepsilon'$ whenever $n geq N$. Thus we have
                          $$ |, f(x_n)-f(x_0)|leq L|x_n-x_0| < Lvarepsilon' leq varepsilon text{ whenever } n geq N.$$



                          Therefore, $f$ is sequentially continuous on $mathbb R$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 7 '18 at 3:15

























                          answered Dec 7 '18 at 0:00









                          Matt A PeltoMatt A Pelto

                          2,537620




                          2,537620






























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