Finding the Cardinality of a set.












1












$begingroup$


Let $A$ be the set of all continuous functions $f :[0,1] → [0,∞)$ satisfying the following condition:
$int_{0}^{x} f(t) dt geq f(x) $ for all $ x in [0,1]$.



Find the cardinality of the set $A$ .



I have no idea how to procede . Please help.






real-analysis







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $A$ be the set of all continuous functions $f :[0,1] → [0,∞)$ satisfying the following condition:
    $int_{0}^{x} f(t) dt geq f(x) $ for all $ x in [0,1]$.



    Find the cardinality of the set $A$ .



    I have no idea how to procede . Please help.






    real-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $A$ be the set of all continuous functions $f :[0,1] → [0,∞)$ satisfying the following condition:
      $int_{0}^{x} f(t) dt geq f(x) $ for all $ x in [0,1]$.



      Find the cardinality of the set $A$ .



      I have no idea how to procede . Please help.






      real-analysis







      share|cite|improve this question









      $endgroup$




      Let $A$ be the set of all continuous functions $f :[0,1] → [0,∞)$ satisfying the following condition:
      $int_{0}^{x} f(t) dt geq f(x) $ for all $ x in [0,1]$.



      Find the cardinality of the set $A$ .



      I have no idea how to procede . Please help.






      real-analysis




      real-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 6 '18 at 14:29









      blue boyblue boy

      1,236613




      1,236613






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Clearly the identically zero function ($f(x)=0$ for all $x$) belongs to $A$.



          I will prove that $A$ has no other element.



          Let $f in A$. Then, for all $x in [0,1]$
          $$int_0^x f(t) mathrm d t ge f(x)$$
          integrating on $x$ you have
          $$int_0^1 int_0^x f(t) mathrm d t mathrm d x ge int_0^1 f(x) mathrm d x$$
          The first integral can be manipulated into
          $$int_0^1 int_0^x f(t) mathrm d t mathrm d x =
          int_0^1 int_t^1 f(t) mathrm d x mathrm d t = int_0^1 (1-t) f(t) mathrm d t = int_0^1 f(t) mathrm d t - int_0^1 tf(t) mathrm d t$$
          Thus you have
          $$int_0^1 f(t) mathrm d t - int_0^1 tf(t) mathrm d t ge int_0^1 f(x) mathrm d x$$
          i.e.
          $$int_0^1 t f(t) mathrm d t le 0$$



          Since $tf(t)$ is continuous and non-negative, its integral must be non-negative. Thus you have
          $$0 le int_0^1 t f(t) mathrm d t le 0$$
          which implies that $f(t)$ is identically zero.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How did you get the idea to do this ? :)
            $endgroup$
            – blue boy
            Dec 8 '18 at 13:11






          • 1




            $begingroup$
            I saw a lot of tricks here at Math Stack Exchange :P. Anyway, when you are in doubt, integrate and the inequality must be preserved. Other times differentiating is a good idea: it depends on the problem.
            $endgroup$
            – Crostul
            Dec 8 '18 at 14:10












          • $begingroup$
            Ok. Thanks. I think it takes time to get used to pure mathematics.
            $endgroup$
            – blue boy
            Dec 8 '18 at 14:55










          • $begingroup$
            I began studying pure mathematics last year. Sometimes i feel depressed not able to solve questions. But i think that is the part of learning. :)
            $endgroup$
            – blue boy
            Dec 8 '18 at 15:02



















          1












          $begingroup$

          Hint: how many continuous functions $f:[0,,1]mapsto [0,,infty)$ exist, satisfying the last condition or otherwise? (You can get an upper bound from the fact $f$ is specified by its values on $Bbb Qcap [0,,1]$, and this upper bound is also an obvious lower bound by constructing a specific family of solutions.) And can you show this cardinality also lower-bounds $|A|$?






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Clearly the identically zero function ($f(x)=0$ for all $x$) belongs to $A$.



            I will prove that $A$ has no other element.



            Let $f in A$. Then, for all $x in [0,1]$
            $$int_0^x f(t) mathrm d t ge f(x)$$
            integrating on $x$ you have
            $$int_0^1 int_0^x f(t) mathrm d t mathrm d x ge int_0^1 f(x) mathrm d x$$
            The first integral can be manipulated into
            $$int_0^1 int_0^x f(t) mathrm d t mathrm d x =
            int_0^1 int_t^1 f(t) mathrm d x mathrm d t = int_0^1 (1-t) f(t) mathrm d t = int_0^1 f(t) mathrm d t - int_0^1 tf(t) mathrm d t$$
            Thus you have
            $$int_0^1 f(t) mathrm d t - int_0^1 tf(t) mathrm d t ge int_0^1 f(x) mathrm d x$$
            i.e.
            $$int_0^1 t f(t) mathrm d t le 0$$



            Since $tf(t)$ is continuous and non-negative, its integral must be non-negative. Thus you have
            $$0 le int_0^1 t f(t) mathrm d t le 0$$
            which implies that $f(t)$ is identically zero.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How did you get the idea to do this ? :)
              $endgroup$
              – blue boy
              Dec 8 '18 at 13:11






            • 1




              $begingroup$
              I saw a lot of tricks here at Math Stack Exchange :P. Anyway, when you are in doubt, integrate and the inequality must be preserved. Other times differentiating is a good idea: it depends on the problem.
              $endgroup$
              – Crostul
              Dec 8 '18 at 14:10












            • $begingroup$
              Ok. Thanks. I think it takes time to get used to pure mathematics.
              $endgroup$
              – blue boy
              Dec 8 '18 at 14:55










            • $begingroup$
              I began studying pure mathematics last year. Sometimes i feel depressed not able to solve questions. But i think that is the part of learning. :)
              $endgroup$
              – blue boy
              Dec 8 '18 at 15:02
















            1












            $begingroup$

            Clearly the identically zero function ($f(x)=0$ for all $x$) belongs to $A$.



            I will prove that $A$ has no other element.



            Let $f in A$. Then, for all $x in [0,1]$
            $$int_0^x f(t) mathrm d t ge f(x)$$
            integrating on $x$ you have
            $$int_0^1 int_0^x f(t) mathrm d t mathrm d x ge int_0^1 f(x) mathrm d x$$
            The first integral can be manipulated into
            $$int_0^1 int_0^x f(t) mathrm d t mathrm d x =
            int_0^1 int_t^1 f(t) mathrm d x mathrm d t = int_0^1 (1-t) f(t) mathrm d t = int_0^1 f(t) mathrm d t - int_0^1 tf(t) mathrm d t$$
            Thus you have
            $$int_0^1 f(t) mathrm d t - int_0^1 tf(t) mathrm d t ge int_0^1 f(x) mathrm d x$$
            i.e.
            $$int_0^1 t f(t) mathrm d t le 0$$



            Since $tf(t)$ is continuous and non-negative, its integral must be non-negative. Thus you have
            $$0 le int_0^1 t f(t) mathrm d t le 0$$
            which implies that $f(t)$ is identically zero.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How did you get the idea to do this ? :)
              $endgroup$
              – blue boy
              Dec 8 '18 at 13:11






            • 1




              $begingroup$
              I saw a lot of tricks here at Math Stack Exchange :P. Anyway, when you are in doubt, integrate and the inequality must be preserved. Other times differentiating is a good idea: it depends on the problem.
              $endgroup$
              – Crostul
              Dec 8 '18 at 14:10












            • $begingroup$
              Ok. Thanks. I think it takes time to get used to pure mathematics.
              $endgroup$
              – blue boy
              Dec 8 '18 at 14:55










            • $begingroup$
              I began studying pure mathematics last year. Sometimes i feel depressed not able to solve questions. But i think that is the part of learning. :)
              $endgroup$
              – blue boy
              Dec 8 '18 at 15:02














            1












            1








            1





            $begingroup$

            Clearly the identically zero function ($f(x)=0$ for all $x$) belongs to $A$.



            I will prove that $A$ has no other element.



            Let $f in A$. Then, for all $x in [0,1]$
            $$int_0^x f(t) mathrm d t ge f(x)$$
            integrating on $x$ you have
            $$int_0^1 int_0^x f(t) mathrm d t mathrm d x ge int_0^1 f(x) mathrm d x$$
            The first integral can be manipulated into
            $$int_0^1 int_0^x f(t) mathrm d t mathrm d x =
            int_0^1 int_t^1 f(t) mathrm d x mathrm d t = int_0^1 (1-t) f(t) mathrm d t = int_0^1 f(t) mathrm d t - int_0^1 tf(t) mathrm d t$$
            Thus you have
            $$int_0^1 f(t) mathrm d t - int_0^1 tf(t) mathrm d t ge int_0^1 f(x) mathrm d x$$
            i.e.
            $$int_0^1 t f(t) mathrm d t le 0$$



            Since $tf(t)$ is continuous and non-negative, its integral must be non-negative. Thus you have
            $$0 le int_0^1 t f(t) mathrm d t le 0$$
            which implies that $f(t)$ is identically zero.






            share|cite|improve this answer









            $endgroup$



            Clearly the identically zero function ($f(x)=0$ for all $x$) belongs to $A$.



            I will prove that $A$ has no other element.



            Let $f in A$. Then, for all $x in [0,1]$
            $$int_0^x f(t) mathrm d t ge f(x)$$
            integrating on $x$ you have
            $$int_0^1 int_0^x f(t) mathrm d t mathrm d x ge int_0^1 f(x) mathrm d x$$
            The first integral can be manipulated into
            $$int_0^1 int_0^x f(t) mathrm d t mathrm d x =
            int_0^1 int_t^1 f(t) mathrm d x mathrm d t = int_0^1 (1-t) f(t) mathrm d t = int_0^1 f(t) mathrm d t - int_0^1 tf(t) mathrm d t$$
            Thus you have
            $$int_0^1 f(t) mathrm d t - int_0^1 tf(t) mathrm d t ge int_0^1 f(x) mathrm d x$$
            i.e.
            $$int_0^1 t f(t) mathrm d t le 0$$



            Since $tf(t)$ is continuous and non-negative, its integral must be non-negative. Thus you have
            $$0 le int_0^1 t f(t) mathrm d t le 0$$
            which implies that $f(t)$ is identically zero.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 6 '18 at 14:52









            CrostulCrostul

            27.8k22352




            27.8k22352












            • $begingroup$
              How did you get the idea to do this ? :)
              $endgroup$
              – blue boy
              Dec 8 '18 at 13:11






            • 1




              $begingroup$
              I saw a lot of tricks here at Math Stack Exchange :P. Anyway, when you are in doubt, integrate and the inequality must be preserved. Other times differentiating is a good idea: it depends on the problem.
              $endgroup$
              – Crostul
              Dec 8 '18 at 14:10












            • $begingroup$
              Ok. Thanks. I think it takes time to get used to pure mathematics.
              $endgroup$
              – blue boy
              Dec 8 '18 at 14:55










            • $begingroup$
              I began studying pure mathematics last year. Sometimes i feel depressed not able to solve questions. But i think that is the part of learning. :)
              $endgroup$
              – blue boy
              Dec 8 '18 at 15:02


















            • $begingroup$
              How did you get the idea to do this ? :)
              $endgroup$
              – blue boy
              Dec 8 '18 at 13:11






            • 1




              $begingroup$
              I saw a lot of tricks here at Math Stack Exchange :P. Anyway, when you are in doubt, integrate and the inequality must be preserved. Other times differentiating is a good idea: it depends on the problem.
              $endgroup$
              – Crostul
              Dec 8 '18 at 14:10












            • $begingroup$
              Ok. Thanks. I think it takes time to get used to pure mathematics.
              $endgroup$
              – blue boy
              Dec 8 '18 at 14:55










            • $begingroup$
              I began studying pure mathematics last year. Sometimes i feel depressed not able to solve questions. But i think that is the part of learning. :)
              $endgroup$
              – blue boy
              Dec 8 '18 at 15:02
















            $begingroup$
            How did you get the idea to do this ? :)
            $endgroup$
            – blue boy
            Dec 8 '18 at 13:11




            $begingroup$
            How did you get the idea to do this ? :)
            $endgroup$
            – blue boy
            Dec 8 '18 at 13:11




            1




            1




            $begingroup$
            I saw a lot of tricks here at Math Stack Exchange :P. Anyway, when you are in doubt, integrate and the inequality must be preserved. Other times differentiating is a good idea: it depends on the problem.
            $endgroup$
            – Crostul
            Dec 8 '18 at 14:10






            $begingroup$
            I saw a lot of tricks here at Math Stack Exchange :P. Anyway, when you are in doubt, integrate and the inequality must be preserved. Other times differentiating is a good idea: it depends on the problem.
            $endgroup$
            – Crostul
            Dec 8 '18 at 14:10














            $begingroup$
            Ok. Thanks. I think it takes time to get used to pure mathematics.
            $endgroup$
            – blue boy
            Dec 8 '18 at 14:55




            $begingroup$
            Ok. Thanks. I think it takes time to get used to pure mathematics.
            $endgroup$
            – blue boy
            Dec 8 '18 at 14:55












            $begingroup$
            I began studying pure mathematics last year. Sometimes i feel depressed not able to solve questions. But i think that is the part of learning. :)
            $endgroup$
            – blue boy
            Dec 8 '18 at 15:02




            $begingroup$
            I began studying pure mathematics last year. Sometimes i feel depressed not able to solve questions. But i think that is the part of learning. :)
            $endgroup$
            – blue boy
            Dec 8 '18 at 15:02











            1












            $begingroup$

            Hint: how many continuous functions $f:[0,,1]mapsto [0,,infty)$ exist, satisfying the last condition or otherwise? (You can get an upper bound from the fact $f$ is specified by its values on $Bbb Qcap [0,,1]$, and this upper bound is also an obvious lower bound by constructing a specific family of solutions.) And can you show this cardinality also lower-bounds $|A|$?






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Hint: how many continuous functions $f:[0,,1]mapsto [0,,infty)$ exist, satisfying the last condition or otherwise? (You can get an upper bound from the fact $f$ is specified by its values on $Bbb Qcap [0,,1]$, and this upper bound is also an obvious lower bound by constructing a specific family of solutions.) And can you show this cardinality also lower-bounds $|A|$?






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Hint: how many continuous functions $f:[0,,1]mapsto [0,,infty)$ exist, satisfying the last condition or otherwise? (You can get an upper bound from the fact $f$ is specified by its values on $Bbb Qcap [0,,1]$, and this upper bound is also an obvious lower bound by constructing a specific family of solutions.) And can you show this cardinality also lower-bounds $|A|$?






                share|cite|improve this answer









                $endgroup$



                Hint: how many continuous functions $f:[0,,1]mapsto [0,,infty)$ exist, satisfying the last condition or otherwise? (You can get an upper bound from the fact $f$ is specified by its values on $Bbb Qcap [0,,1]$, and this upper bound is also an obvious lower bound by constructing a specific family of solutions.) And can you show this cardinality also lower-bounds $|A|$?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 14:41









                J.G.J.G.

                24.9k22539




                24.9k22539






























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