Bounded function in $[0,1]$ without max and min.












4












$begingroup$


Do we have a function $f$ defined in $[0, 1]$, which is bounded but has no maximum and minimum ?



I do know that $arctan x$ can give me a hint, which is bounded without max and min but that's in $mathbb R$.
Thanks!




EDITED:

Firstly thank all the answers that I received.

I would like to ask a little additional question: could we find some edited or mixed trigonometric functions to satisfy this problem?

If yes, a example please!

The reason is, that I got stuck with the idea of $arctan x$ and other possible trigonometric functions.



Thanks!






real-analysis calculus trigonometry







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Not if your function is continuous. Check for instance en.m.wikipedia.org/wiki/Extreme_value_theorem
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:36










  • $begingroup$
    Thanks for the re-edit!
    $endgroup$
    – Jigao
    Dec 16 '18 at 11:52
















4












$begingroup$


Do we have a function $f$ defined in $[0, 1]$, which is bounded but has no maximum and minimum ?



I do know that $arctan x$ can give me a hint, which is bounded without max and min but that's in $mathbb R$.
Thanks!




EDITED:

Firstly thank all the answers that I received.

I would like to ask a little additional question: could we find some edited or mixed trigonometric functions to satisfy this problem?

If yes, a example please!

The reason is, that I got stuck with the idea of $arctan x$ and other possible trigonometric functions.



Thanks!






real-analysis calculus trigonometry







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Not if your function is continuous. Check for instance en.m.wikipedia.org/wiki/Extreme_value_theorem
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:36










  • $begingroup$
    Thanks for the re-edit!
    $endgroup$
    – Jigao
    Dec 16 '18 at 11:52














4












4








4





$begingroup$


Do we have a function $f$ defined in $[0, 1]$, which is bounded but has no maximum and minimum ?



I do know that $arctan x$ can give me a hint, which is bounded without max and min but that's in $mathbb R$.
Thanks!




EDITED:

Firstly thank all the answers that I received.

I would like to ask a little additional question: could we find some edited or mixed trigonometric functions to satisfy this problem?

If yes, a example please!

The reason is, that I got stuck with the idea of $arctan x$ and other possible trigonometric functions.



Thanks!






real-analysis calculus trigonometry







share|cite|improve this question











$endgroup$




Do we have a function $f$ defined in $[0, 1]$, which is bounded but has no maximum and minimum ?



I do know that $arctan x$ can give me a hint, which is bounded without max and min but that's in $mathbb R$.
Thanks!




EDITED:

Firstly thank all the answers that I received.

I would like to ask a little additional question: could we find some edited or mixed trigonometric functions to satisfy this problem?

If yes, a example please!

The reason is, that I got stuck with the idea of $arctan x$ and other possible trigonometric functions.



Thanks!






real-analysis calculus trigonometry




real-analysis calculus trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 12:22







Jigao

















asked Dec 16 '18 at 11:34









JigaoJigao

236




236








  • 5




    $begingroup$
    Not if your function is continuous. Check for instance en.m.wikipedia.org/wiki/Extreme_value_theorem
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:36










  • $begingroup$
    Thanks for the re-edit!
    $endgroup$
    – Jigao
    Dec 16 '18 at 11:52














  • 5




    $begingroup$
    Not if your function is continuous. Check for instance en.m.wikipedia.org/wiki/Extreme_value_theorem
    $endgroup$
    – Mindlack
    Dec 16 '18 at 11:36










  • $begingroup$
    Thanks for the re-edit!
    $endgroup$
    – Jigao
    Dec 16 '18 at 11:52








5




5




$begingroup$
Not if your function is continuous. Check for instance en.m.wikipedia.org/wiki/Extreme_value_theorem
$endgroup$
– Mindlack
Dec 16 '18 at 11:36




$begingroup$
Not if your function is continuous. Check for instance en.m.wikipedia.org/wiki/Extreme_value_theorem
$endgroup$
– Mindlack
Dec 16 '18 at 11:36












$begingroup$
Thanks for the re-edit!
$endgroup$
– Jigao
Dec 16 '18 at 11:52




$begingroup$
Thanks for the re-edit!
$endgroup$
– Jigao
Dec 16 '18 at 11:52










4 Answers
4






active

oldest

votes


















2












$begingroup$

You cannot have a purely trigonometric function defined on $[0,1]$ that does not attain its max/min because it would be continuous and the extreme value theorem would apply.



$$
f(x)=begin{cases}(1-x)sin(frac 1x) &; xin(0,1]\
0 &; x=0
end{cases}
$$
This function is almost a trigonometric function as you specified. It has $sup_{xin[0,1]} f(x)=1$ and $inf_{xin[0,1]} f(x)=-1$ but those values are not attained.






share|cite|improve this answer









$endgroup$





















    7












    $begingroup$

    $f(x)=x$ if $.25<x<.75$, and $f(x)=0.5$ otherwise






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      The function $f$ defined below has $sup_{xin[0,1]} f(x)=1$ and $inf_{xin[0,1]} f(x)=-1$ but doesn't attain those values.
      $$
      f(x)=begin{cases}frac {(-1)^nn}{n+1} &; x=frac 1n text{ for $ninBbb N$}\
      0 &; text{otherwise}
      end{cases}
      $$






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        The key point is continuity, without that we can construct counterexample as




        • $f(x)=x quad xin [0,1/2)$


        • $f(x)=0 quad x=1/2$


        • $f(x)=1-x quad xin (1/2,1]$



        Refer also to Extreme value theorem.






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          You cannot have a purely trigonometric function defined on $[0,1]$ that does not attain its max/min because it would be continuous and the extreme value theorem would apply.



          $$
          f(x)=begin{cases}(1-x)sin(frac 1x) &; xin(0,1]\
          0 &; x=0
          end{cases}
          $$
          This function is almost a trigonometric function as you specified. It has $sup_{xin[0,1]} f(x)=1$ and $inf_{xin[0,1]} f(x)=-1$ but those values are not attained.






          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            You cannot have a purely trigonometric function defined on $[0,1]$ that does not attain its max/min because it would be continuous and the extreme value theorem would apply.



            $$
            f(x)=begin{cases}(1-x)sin(frac 1x) &; xin(0,1]\
            0 &; x=0
            end{cases}
            $$
            This function is almost a trigonometric function as you specified. It has $sup_{xin[0,1]} f(x)=1$ and $inf_{xin[0,1]} f(x)=-1$ but those values are not attained.






            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              You cannot have a purely trigonometric function defined on $[0,1]$ that does not attain its max/min because it would be continuous and the extreme value theorem would apply.



              $$
              f(x)=begin{cases}(1-x)sin(frac 1x) &; xin(0,1]\
              0 &; x=0
              end{cases}
              $$
              This function is almost a trigonometric function as you specified. It has $sup_{xin[0,1]} f(x)=1$ and $inf_{xin[0,1]} f(x)=-1$ but those values are not attained.






              share|cite|improve this answer









              $endgroup$



              You cannot have a purely trigonometric function defined on $[0,1]$ that does not attain its max/min because it would be continuous and the extreme value theorem would apply.



              $$
              f(x)=begin{cases}(1-x)sin(frac 1x) &; xin(0,1]\
              0 &; x=0
              end{cases}
              $$
              This function is almost a trigonometric function as you specified. It has $sup_{xin[0,1]} f(x)=1$ and $inf_{xin[0,1]} f(x)=-1$ but those values are not attained.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 16 '18 at 12:20









              BigbearZzzBigbearZzz

              8,49221652




              8,49221652























                  7












                  $begingroup$

                  $f(x)=x$ if $.25<x<.75$, and $f(x)=0.5$ otherwise






                  share|cite|improve this answer









                  $endgroup$


















                    7












                    $begingroup$

                    $f(x)=x$ if $.25<x<.75$, and $f(x)=0.5$ otherwise






                    share|cite|improve this answer









                    $endgroup$
















                      7












                      7








                      7





                      $begingroup$

                      $f(x)=x$ if $.25<x<.75$, and $f(x)=0.5$ otherwise






                      share|cite|improve this answer









                      $endgroup$



                      $f(x)=x$ if $.25<x<.75$, and $f(x)=0.5$ otherwise







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 16 '18 at 11:38









                      David PetersonDavid Peterson

                      8,75121935




                      8,75121935























                          2












                          $begingroup$

                          The function $f$ defined below has $sup_{xin[0,1]} f(x)=1$ and $inf_{xin[0,1]} f(x)=-1$ but doesn't attain those values.
                          $$
                          f(x)=begin{cases}frac {(-1)^nn}{n+1} &; x=frac 1n text{ for $ninBbb N$}\
                          0 &; text{otherwise}
                          end{cases}
                          $$






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            The function $f$ defined below has $sup_{xin[0,1]} f(x)=1$ and $inf_{xin[0,1]} f(x)=-1$ but doesn't attain those values.
                            $$
                            f(x)=begin{cases}frac {(-1)^nn}{n+1} &; x=frac 1n text{ for $ninBbb N$}\
                            0 &; text{otherwise}
                            end{cases}
                            $$






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              The function $f$ defined below has $sup_{xin[0,1]} f(x)=1$ and $inf_{xin[0,1]} f(x)=-1$ but doesn't attain those values.
                              $$
                              f(x)=begin{cases}frac {(-1)^nn}{n+1} &; x=frac 1n text{ for $ninBbb N$}\
                              0 &; text{otherwise}
                              end{cases}
                              $$






                              share|cite|improve this answer









                              $endgroup$



                              The function $f$ defined below has $sup_{xin[0,1]} f(x)=1$ and $inf_{xin[0,1]} f(x)=-1$ but doesn't attain those values.
                              $$
                              f(x)=begin{cases}frac {(-1)^nn}{n+1} &; x=frac 1n text{ for $ninBbb N$}\
                              0 &; text{otherwise}
                              end{cases}
                              $$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 16 '18 at 12:01









                              BigbearZzzBigbearZzz

                              8,49221652




                              8,49221652























                                  1












                                  $begingroup$

                                  The key point is continuity, without that we can construct counterexample as




                                  • $f(x)=x quad xin [0,1/2)$


                                  • $f(x)=0 quad x=1/2$


                                  • $f(x)=1-x quad xin (1/2,1]$



                                  Refer also to Extreme value theorem.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    The key point is continuity, without that we can construct counterexample as




                                    • $f(x)=x quad xin [0,1/2)$


                                    • $f(x)=0 quad x=1/2$


                                    • $f(x)=1-x quad xin (1/2,1]$



                                    Refer also to Extreme value theorem.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      The key point is continuity, without that we can construct counterexample as




                                      • $f(x)=x quad xin [0,1/2)$


                                      • $f(x)=0 quad x=1/2$


                                      • $f(x)=1-x quad xin (1/2,1]$



                                      Refer also to Extreme value theorem.






                                      share|cite|improve this answer









                                      $endgroup$



                                      The key point is continuity, without that we can construct counterexample as




                                      • $f(x)=x quad xin [0,1/2)$


                                      • $f(x)=0 quad x=1/2$


                                      • $f(x)=1-x quad xin (1/2,1]$



                                      Refer also to Extreme value theorem.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 16 '18 at 12:08









                                      gimusigimusi

                                      92.9k94494




                                      92.9k94494






























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