supremum of a function with integral












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$begingroup$


Let $mathcal F$ be the set of continuous functions $f:[0,1]to mathbb R$ and $max_{0le xle1} |f(x)|=1$. Now let $mathcal I:mathcal Fto R,mathcal I(f)=int_0^1f(x)dx-f(0)+f(1)$. Firstly I had to show $mathcal I(f)<3,forall f in mathcal F$ and this is pretty easy. Now I have to determine $sup({mathcal I(f)}|fin mathcal F)$. Doesn't result that the supremum is $3$ from the statement before? I know that I am surely wrong. Can somebody give me some tips on how can I get the supremum, please?






calculus supremum-and-infimum







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$endgroup$

















    1












    $begingroup$


    Let $mathcal F$ be the set of continuous functions $f:[0,1]to mathbb R$ and $max_{0le xle1} |f(x)|=1$. Now let $mathcal I:mathcal Fto R,mathcal I(f)=int_0^1f(x)dx-f(0)+f(1)$. Firstly I had to show $mathcal I(f)<3,forall f in mathcal F$ and this is pretty easy. Now I have to determine $sup({mathcal I(f)}|fin mathcal F)$. Doesn't result that the supremum is $3$ from the statement before? I know that I am surely wrong. Can somebody give me some tips on how can I get the supremum, please?






    calculus supremum-and-infimum







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $mathcal F$ be the set of continuous functions $f:[0,1]to mathbb R$ and $max_{0le xle1} |f(x)|=1$. Now let $mathcal I:mathcal Fto R,mathcal I(f)=int_0^1f(x)dx-f(0)+f(1)$. Firstly I had to show $mathcal I(f)<3,forall f in mathcal F$ and this is pretty easy. Now I have to determine $sup({mathcal I(f)}|fin mathcal F)$. Doesn't result that the supremum is $3$ from the statement before? I know that I am surely wrong. Can somebody give me some tips on how can I get the supremum, please?






      calculus supremum-and-infimum







      share|cite|improve this question











      $endgroup$




      Let $mathcal F$ be the set of continuous functions $f:[0,1]to mathbb R$ and $max_{0le xle1} |f(x)|=1$. Now let $mathcal I:mathcal Fto R,mathcal I(f)=int_0^1f(x)dx-f(0)+f(1)$. Firstly I had to show $mathcal I(f)<3,forall f in mathcal F$ and this is pretty easy. Now I have to determine $sup({mathcal I(f)}|fin mathcal F)$. Doesn't result that the supremum is $3$ from the statement before? I know that I am surely wrong. Can somebody give me some tips on how can I get the supremum, please?






      calculus supremum-and-infimum




      calculus supremum-and-infimum






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 31 at 18:45









      Julián Aguirre

      69.5k24297




      69.5k24297










      asked Jan 31 at 18:01









      GaboruGaboru

      5078




      5078






















          2 Answers
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          active

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          3












          $begingroup$

          The fact that $mathcal I(f)<3$ implies that $sup{mathcal I(f)mid fin mathcal F}le3$. To show that the supremum is equal to $3$, you have to find for every $epsilon>0$ a function $finmathcal F$ such that $mathcal I(f)ge3-epsilon$. This can be done for instance with functions like
          $$
          f_delta(x)=begin{cases}
          dfrac{2}{delta},x-1 & 0le xledelta,\
          1 & delta<xle1.
          end{cases}
          $$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            obvious part is that $forall fin mathcal F $



            $$|mathcal I(f) | leq |int_0^1f(x)dx|+|f(0)|+|f(1)| leq 1+1+1leq 3$$
            inequality here is not strict ( $leq$ instead of $<$)



            Thus, we have shown that $sup({mathcal I(f)}|fin mathcal F) leq 3$
            To prove that it is equal to 3 we need to find a function $g$ or a sequence $g_n$ of functions in $mathcal F $, such that



            $$mathcal I(g) = 3; quad or quad lim_{n rightarrow infty} mathcal I(g_n) = 3$$



            (this would prove that $sup({mathcal I(f)}|fin mathcal F) geq 3$, which would solve the problem)



            First thing that comes to mind is the function $g$ defined as $g(t) = 1, ; forall t in (0, 1]$ and $g(0)=-1$. Then $mathcal I(g)$ would be equal to 3, but function $g$ is not contionious.
            Only one step is left to complete the solution: approximate $g$ with a sequence $g_n$ of functions from $mathcal F$ such that $lim_{n rightarrow infty} mathcal I(g_n) = 3$






            share|cite|improve this answer









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              2 Answers
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              2 Answers
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              3












              $begingroup$

              The fact that $mathcal I(f)<3$ implies that $sup{mathcal I(f)mid fin mathcal F}le3$. To show that the supremum is equal to $3$, you have to find for every $epsilon>0$ a function $finmathcal F$ such that $mathcal I(f)ge3-epsilon$. This can be done for instance with functions like
              $$
              f_delta(x)=begin{cases}
              dfrac{2}{delta},x-1 & 0le xledelta,\
              1 & delta<xle1.
              end{cases}
              $$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                The fact that $mathcal I(f)<3$ implies that $sup{mathcal I(f)mid fin mathcal F}le3$. To show that the supremum is equal to $3$, you have to find for every $epsilon>0$ a function $finmathcal F$ such that $mathcal I(f)ge3-epsilon$. This can be done for instance with functions like
                $$
                f_delta(x)=begin{cases}
                dfrac{2}{delta},x-1 & 0le xledelta,\
                1 & delta<xle1.
                end{cases}
                $$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The fact that $mathcal I(f)<3$ implies that $sup{mathcal I(f)mid fin mathcal F}le3$. To show that the supremum is equal to $3$, you have to find for every $epsilon>0$ a function $finmathcal F$ such that $mathcal I(f)ge3-epsilon$. This can be done for instance with functions like
                  $$
                  f_delta(x)=begin{cases}
                  dfrac{2}{delta},x-1 & 0le xledelta,\
                  1 & delta<xle1.
                  end{cases}
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  The fact that $mathcal I(f)<3$ implies that $sup{mathcal I(f)mid fin mathcal F}le3$. To show that the supremum is equal to $3$, you have to find for every $epsilon>0$ a function $finmathcal F$ such that $mathcal I(f)ge3-epsilon$. This can be done for instance with functions like
                  $$
                  f_delta(x)=begin{cases}
                  dfrac{2}{delta},x-1 & 0le xledelta,\
                  1 & delta<xle1.
                  end{cases}
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 31 at 18:52









                  Julián AguirreJulián Aguirre

                  69.5k24297




                  69.5k24297























                      1












                      $begingroup$

                      obvious part is that $forall fin mathcal F $



                      $$|mathcal I(f) | leq |int_0^1f(x)dx|+|f(0)|+|f(1)| leq 1+1+1leq 3$$
                      inequality here is not strict ( $leq$ instead of $<$)



                      Thus, we have shown that $sup({mathcal I(f)}|fin mathcal F) leq 3$
                      To prove that it is equal to 3 we need to find a function $g$ or a sequence $g_n$ of functions in $mathcal F $, such that



                      $$mathcal I(g) = 3; quad or quad lim_{n rightarrow infty} mathcal I(g_n) = 3$$



                      (this would prove that $sup({mathcal I(f)}|fin mathcal F) geq 3$, which would solve the problem)



                      First thing that comes to mind is the function $g$ defined as $g(t) = 1, ; forall t in (0, 1]$ and $g(0)=-1$. Then $mathcal I(g)$ would be equal to 3, but function $g$ is not contionious.
                      Only one step is left to complete the solution: approximate $g$ with a sequence $g_n$ of functions from $mathcal F$ such that $lim_{n rightarrow infty} mathcal I(g_n) = 3$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        obvious part is that $forall fin mathcal F $



                        $$|mathcal I(f) | leq |int_0^1f(x)dx|+|f(0)|+|f(1)| leq 1+1+1leq 3$$
                        inequality here is not strict ( $leq$ instead of $<$)



                        Thus, we have shown that $sup({mathcal I(f)}|fin mathcal F) leq 3$
                        To prove that it is equal to 3 we need to find a function $g$ or a sequence $g_n$ of functions in $mathcal F $, such that



                        $$mathcal I(g) = 3; quad or quad lim_{n rightarrow infty} mathcal I(g_n) = 3$$



                        (this would prove that $sup({mathcal I(f)}|fin mathcal F) geq 3$, which would solve the problem)



                        First thing that comes to mind is the function $g$ defined as $g(t) = 1, ; forall t in (0, 1]$ and $g(0)=-1$. Then $mathcal I(g)$ would be equal to 3, but function $g$ is not contionious.
                        Only one step is left to complete the solution: approximate $g$ with a sequence $g_n$ of functions from $mathcal F$ such that $lim_{n rightarrow infty} mathcal I(g_n) = 3$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          obvious part is that $forall fin mathcal F $



                          $$|mathcal I(f) | leq |int_0^1f(x)dx|+|f(0)|+|f(1)| leq 1+1+1leq 3$$
                          inequality here is not strict ( $leq$ instead of $<$)



                          Thus, we have shown that $sup({mathcal I(f)}|fin mathcal F) leq 3$
                          To prove that it is equal to 3 we need to find a function $g$ or a sequence $g_n$ of functions in $mathcal F $, such that



                          $$mathcal I(g) = 3; quad or quad lim_{n rightarrow infty} mathcal I(g_n) = 3$$



                          (this would prove that $sup({mathcal I(f)}|fin mathcal F) geq 3$, which would solve the problem)



                          First thing that comes to mind is the function $g$ defined as $g(t) = 1, ; forall t in (0, 1]$ and $g(0)=-1$. Then $mathcal I(g)$ would be equal to 3, but function $g$ is not contionious.
                          Only one step is left to complete the solution: approximate $g$ with a sequence $g_n$ of functions from $mathcal F$ such that $lim_{n rightarrow infty} mathcal I(g_n) = 3$






                          share|cite|improve this answer









                          $endgroup$



                          obvious part is that $forall fin mathcal F $



                          $$|mathcal I(f) | leq |int_0^1f(x)dx|+|f(0)|+|f(1)| leq 1+1+1leq 3$$
                          inequality here is not strict ( $leq$ instead of $<$)



                          Thus, we have shown that $sup({mathcal I(f)}|fin mathcal F) leq 3$
                          To prove that it is equal to 3 we need to find a function $g$ or a sequence $g_n$ of functions in $mathcal F $, such that



                          $$mathcal I(g) = 3; quad or quad lim_{n rightarrow infty} mathcal I(g_n) = 3$$



                          (this would prove that $sup({mathcal I(f)}|fin mathcal F) geq 3$, which would solve the problem)



                          First thing that comes to mind is the function $g$ defined as $g(t) = 1, ; forall t in (0, 1]$ and $g(0)=-1$. Then $mathcal I(g)$ would be equal to 3, but function $g$ is not contionious.
                          Only one step is left to complete the solution: approximate $g$ with a sequence $g_n$ of functions from $mathcal F$ such that $lim_{n rightarrow infty} mathcal I(g_n) = 3$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 31 at 19:02









                          asdfasdf

                          561




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