Uniform continuity implies existence of limit of integral












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Let $f: (0,1) rightarrow mathbb{R} $ be uniformly continuous. Prove that $$ lim_{epsilon to 0} int^{1-epsilon}_{epsilon}!!f(t)dt in mathbb{R}$$ Any ideas?? $f$ can be extended to a continuous function $hat{f}$ in $[0,1]$, so an integrable one. My best guess is the integral will equal $int^{1}_{0}hat{f}(t),dt$. Am I correct?? A rigorous proof will be very much appreciated. Thanks in advance.






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    $begingroup$


    Let $f: (0,1) rightarrow mathbb{R} $ be uniformly continuous. Prove that $$ lim_{epsilon to 0} int^{1-epsilon}_{epsilon}!!f(t)dt in mathbb{R}$$ Any ideas?? $f$ can be extended to a continuous function $hat{f}$ in $[0,1]$, so an integrable one. My best guess is the integral will equal $int^{1}_{0}hat{f}(t),dt$. Am I correct?? A rigorous proof will be very much appreciated. Thanks in advance.






    real-analysis integration uniform-continuity







    share|cite|improve this question









    New contributor




    nikos steb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      $begingroup$


      Let $f: (0,1) rightarrow mathbb{R} $ be uniformly continuous. Prove that $$ lim_{epsilon to 0} int^{1-epsilon}_{epsilon}!!f(t)dt in mathbb{R}$$ Any ideas?? $f$ can be extended to a continuous function $hat{f}$ in $[0,1]$, so an integrable one. My best guess is the integral will equal $int^{1}_{0}hat{f}(t),dt$. Am I correct?? A rigorous proof will be very much appreciated. Thanks in advance.






      real-analysis integration uniform-continuity







      share|cite|improve this question









      New contributor




      nikos steb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Let $f: (0,1) rightarrow mathbb{R} $ be uniformly continuous. Prove that $$ lim_{epsilon to 0} int^{1-epsilon}_{epsilon}!!f(t)dt in mathbb{R}$$ Any ideas?? $f$ can be extended to a continuous function $hat{f}$ in $[0,1]$, so an integrable one. My best guess is the integral will equal $int^{1}_{0}hat{f}(t),dt$. Am I correct?? A rigorous proof will be very much appreciated. Thanks in advance.






      real-analysis integration uniform-continuity




      real-analysis integration uniform-continuity






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      nikos steb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      edited 4 hours ago









      Bernard

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      asked 5 hours ago









      nikos stebnikos steb

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          3 Answers
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          $begingroup$

          Correct.



          As the extension is bounded, where $|hat{f}(x)| leqslant M$ on $[0,1]$, we have as $epsilon to 0$,



          $$left|int_0^1hat{f} - int_epsilon^{1-epsilon} f right| = left|int_0^1hat{f} - int_epsilon^{1-epsilon} hat{f} right|leqslant left|int_0^epsilon hat{f}right| + left|int_{1-epsilon}^1 hat{f} right| leqslant 2Mepsilon to 0$$






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            $begingroup$

            Hint: $int_epsilon^{1-epsilon} f = int_epsilon^{1-epsilon} hat f.$






            share|cite|improve this answer









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              1












              $begingroup$

              Indeed, $f$ can be extended to $hat f$ on $[0,1]$. Let $epsilon > 0$. Then begin{align} bigg| int_{0}^1 hat fdx - int_{epsilon}^{1- epsilon} hat f dxbigg| &= bigg| int_{0}^{epsilon} hat f dx + int_{1-epsilon}^{epsilon} hat f dx bigg|
              \
              &leq 2epsilon ||hat f||_{infty}end{align}
              where $||hat f||_{infty}$ is the supremum of $hat f$ over $[0,1]$ and is finite by uniform continuity on a compact set.






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                3 Answers
                3






                active

                oldest

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                3 Answers
                3






                active

                oldest

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                active

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                active

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                3












                $begingroup$

                Correct.



                As the extension is bounded, where $|hat{f}(x)| leqslant M$ on $[0,1]$, we have as $epsilon to 0$,



                $$left|int_0^1hat{f} - int_epsilon^{1-epsilon} f right| = left|int_0^1hat{f} - int_epsilon^{1-epsilon} hat{f} right|leqslant left|int_0^epsilon hat{f}right| + left|int_{1-epsilon}^1 hat{f} right| leqslant 2Mepsilon to 0$$






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Correct.



                  As the extension is bounded, where $|hat{f}(x)| leqslant M$ on $[0,1]$, we have as $epsilon to 0$,



                  $$left|int_0^1hat{f} - int_epsilon^{1-epsilon} f right| = left|int_0^1hat{f} - int_epsilon^{1-epsilon} hat{f} right|leqslant left|int_0^epsilon hat{f}right| + left|int_{1-epsilon}^1 hat{f} right| leqslant 2Mepsilon to 0$$






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Correct.



                    As the extension is bounded, where $|hat{f}(x)| leqslant M$ on $[0,1]$, we have as $epsilon to 0$,



                    $$left|int_0^1hat{f} - int_epsilon^{1-epsilon} f right| = left|int_0^1hat{f} - int_epsilon^{1-epsilon} hat{f} right|leqslant left|int_0^epsilon hat{f}right| + left|int_{1-epsilon}^1 hat{f} right| leqslant 2Mepsilon to 0$$






                    share|cite|improve this answer









                    $endgroup$



                    Correct.



                    As the extension is bounded, where $|hat{f}(x)| leqslant M$ on $[0,1]$, we have as $epsilon to 0$,



                    $$left|int_0^1hat{f} - int_epsilon^{1-epsilon} f right| = left|int_0^1hat{f} - int_epsilon^{1-epsilon} hat{f} right|leqslant left|int_0^epsilon hat{f}right| + left|int_{1-epsilon}^1 hat{f} right| leqslant 2Mepsilon to 0$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 4 hours ago









                    RRLRRL

                    50.5k42573




                    50.5k42573























                        2












                        $begingroup$

                        Hint: $int_epsilon^{1-epsilon} f = int_epsilon^{1-epsilon} hat f.$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Hint: $int_epsilon^{1-epsilon} f = int_epsilon^{1-epsilon} hat f.$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Hint: $int_epsilon^{1-epsilon} f = int_epsilon^{1-epsilon} hat f.$






                            share|cite|improve this answer









                            $endgroup$



                            Hint: $int_epsilon^{1-epsilon} f = int_epsilon^{1-epsilon} hat f.$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 4 hours ago









                            zhw.zhw.

                            72.4k43175




                            72.4k43175























                                1












                                $begingroup$

                                Indeed, $f$ can be extended to $hat f$ on $[0,1]$. Let $epsilon > 0$. Then begin{align} bigg| int_{0}^1 hat fdx - int_{epsilon}^{1- epsilon} hat f dxbigg| &= bigg| int_{0}^{epsilon} hat f dx + int_{1-epsilon}^{epsilon} hat f dx bigg|
                                \
                                &leq 2epsilon ||hat f||_{infty}end{align}
                                where $||hat f||_{infty}$ is the supremum of $hat f$ over $[0,1]$ and is finite by uniform continuity on a compact set.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  Indeed, $f$ can be extended to $hat f$ on $[0,1]$. Let $epsilon > 0$. Then begin{align} bigg| int_{0}^1 hat fdx - int_{epsilon}^{1- epsilon} hat f dxbigg| &= bigg| int_{0}^{epsilon} hat f dx + int_{1-epsilon}^{epsilon} hat f dx bigg|
                                  \
                                  &leq 2epsilon ||hat f||_{infty}end{align}
                                  where $||hat f||_{infty}$ is the supremum of $hat f$ over $[0,1]$ and is finite by uniform continuity on a compact set.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Indeed, $f$ can be extended to $hat f$ on $[0,1]$. Let $epsilon > 0$. Then begin{align} bigg| int_{0}^1 hat fdx - int_{epsilon}^{1- epsilon} hat f dxbigg| &= bigg| int_{0}^{epsilon} hat f dx + int_{1-epsilon}^{epsilon} hat f dx bigg|
                                    \
                                    &leq 2epsilon ||hat f||_{infty}end{align}
                                    where $||hat f||_{infty}$ is the supremum of $hat f$ over $[0,1]$ and is finite by uniform continuity on a compact set.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Indeed, $f$ can be extended to $hat f$ on $[0,1]$. Let $epsilon > 0$. Then begin{align} bigg| int_{0}^1 hat fdx - int_{epsilon}^{1- epsilon} hat f dxbigg| &= bigg| int_{0}^{epsilon} hat f dx + int_{1-epsilon}^{epsilon} hat f dx bigg|
                                    \
                                    &leq 2epsilon ||hat f||_{infty}end{align}
                                    where $||hat f||_{infty}$ is the supremum of $hat f$ over $[0,1]$ and is finite by uniform continuity on a compact set.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 4 hours ago









                                    OldGodzillaOldGodzilla

                                    57425




                                    57425






















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