Checking the range in which the given function is uniformly continuous

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Problem



For $p in mathbbR^1 $, let $f(p) = (2p+1,p^2)$.



a) Prove that $f:mathbbR^1 to mathbbR^2 $ is uniformly continuous on the closed interval $[0,2]$.



b) What is the largest interval in which the given function is uniformly continuous?



$Attempt$



a)
$4+(p_1+p_2)^2 leq 4+(|p_1|+|p_2|)^2 leq 4+(2+2)^2= 20$



$d(f(p_1),f(p_2))= (p_1-p_2)sqrt(4+(p_1+p_2)^2) leq delta *sqrt(20)= epsilon$.



Hence f is uniformly continuous on [0,2].



Doubt



How to approach part (b) of the question?







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    Problem



    For $p in mathbbR^1 $, let $f(p) = (2p+1,p^2)$.



    a) Prove that $f:mathbbR^1 to mathbbR^2 $ is uniformly continuous on the closed interval $[0,2]$.



    b) What is the largest interval in which the given function is uniformly continuous?



    $Attempt$



    a)
    $4+(p_1+p_2)^2 leq 4+(|p_1|+|p_2|)^2 leq 4+(2+2)^2= 20$



    $d(f(p_1),f(p_2))= (p_1-p_2)sqrt(4+(p_1+p_2)^2) leq delta *sqrt(20)= epsilon$.



    Hence f is uniformly continuous on [0,2].



    Doubt



    How to approach part (b) of the question?







    share|cite|improve this question






















      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      Problem



      For $p in mathbbR^1 $, let $f(p) = (2p+1,p^2)$.



      a) Prove that $f:mathbbR^1 to mathbbR^2 $ is uniformly continuous on the closed interval $[0,2]$.



      b) What is the largest interval in which the given function is uniformly continuous?



      $Attempt$



      a)
      $4+(p_1+p_2)^2 leq 4+(|p_1|+|p_2|)^2 leq 4+(2+2)^2= 20$



      $d(f(p_1),f(p_2))= (p_1-p_2)sqrt(4+(p_1+p_2)^2) leq delta *sqrt(20)= epsilon$.



      Hence f is uniformly continuous on [0,2].



      Doubt



      How to approach part (b) of the question?







      share|cite|improve this question












      Problem



      For $p in mathbbR^1 $, let $f(p) = (2p+1,p^2)$.



      a) Prove that $f:mathbbR^1 to mathbbR^2 $ is uniformly continuous on the closed interval $[0,2]$.



      b) What is the largest interval in which the given function is uniformly continuous?



      $Attempt$



      a)
      $4+(p_1+p_2)^2 leq 4+(|p_1|+|p_2|)^2 leq 4+(2+2)^2= 20$



      $d(f(p_1),f(p_2))= (p_1-p_2)sqrt(4+(p_1+p_2)^2) leq delta *sqrt(20)= epsilon$.



      Hence f is uniformly continuous on [0,2].



      Doubt



      How to approach part (b) of the question?









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 20 at 10:26









      blue boy

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          Any continuous function on a closed interval is uniformly continuous. So there is no largest interval on which $f$ is uniformly continuous. [It is not uniformly continuous on $mathbb R$].






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            up vote
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            Any continuous function on a closed interval is uniformly continuous. So there is no largest interval on which $f$ is uniformly continuous. [It is not uniformly continuous on $mathbb R$].






            share|cite|improve this answer
























              up vote
              2
              down vote













              Any continuous function on a closed interval is uniformly continuous. So there is no largest interval on which $f$ is uniformly continuous. [It is not uniformly continuous on $mathbb R$].






              share|cite|improve this answer






















                up vote
                2
                down vote










                up vote
                2
                down vote









                Any continuous function on a closed interval is uniformly continuous. So there is no largest interval on which $f$ is uniformly continuous. [It is not uniformly continuous on $mathbb R$].






                share|cite|improve this answer












                Any continuous function on a closed interval is uniformly continuous. So there is no largest interval on which $f$ is uniformly continuous. [It is not uniformly continuous on $mathbb R$].







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 20 at 10:32









                Kavi Rama Murthy

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