A functional equation (another)












1












$begingroup$


I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $xin [1/2,1]$



$$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right).$$



I am actually not sure about existence of a solution.






convex-analysis functional-equations







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $xin [1/2,1]$



    $$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right).$$



    I am actually not sure about existence of a solution.






    convex-analysis functional-equations







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $xin [1/2,1]$



      $$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right).$$



      I am actually not sure about existence of a solution.






      convex-analysis functional-equations







      share|cite|improve this question











      $endgroup$




      I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $xin [1/2,1]$



      $$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right).$$



      I am actually not sure about existence of a solution.






      convex-analysis functional-equations




      convex-analysis functional-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 12:51









      Ethan Bolker

      42.6k549113




      42.6k549113










      asked Dec 13 '18 at 12:50









      S. CurryS. Curry

      92




      92






















          2 Answers
          2






          active

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          2












          $begingroup$

          First of all if we plug in $x=1$ we get



          $$f(1) = frac 12 + frac 14 f(1). $$



          This implies



          $$f(1) = frac 23$$



          so it appears that you cannot have $f(1)=1$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            $$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right) tag 1$$
            As already pointed out this implies $f(1)=frac23$ which is contradictory to the specified condition $f(1)=1$.



            Thus the answer is : There is no solution to the problem with the original wording.



            Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.



            We will check the solution :
            $$f(x)=frac23+cleft(frac{x}{x-1}right)^2 tag 2$$
            $c=$any constant.



            $$fleft( frac{2x}{1+x} right)=frac23+cleft(frac{ frac{2x}{1+x}}{ frac{2x}{1+x}-1}right)^2 tag 3$$
            After simplification :
            $$fleft( frac{2x}{1+x} right)= frac23+cleft(frac{2x}{x-1}right)^2$$
            We put $(3)$ in $(1)$ :
            $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac{1}{2} + frac{1}{4}left( frac23+cleft(frac{2x}{x-1}right)^2right)$$
            After simplification :
            $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac23+cleft(frac{x}{x-1}right)^2$$
            Comparing to Eq.$(2)$ we see that :
            $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=f(x)$$
            This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
            $$f(x)=frac23+cleft(frac{x}{x-1}right)^2 qquad (xneq 1)$$
            The particular case $c=0$ corresponds to the trivial solution $f(x)=frac23$.



            This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              active

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              2 Answers
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              2












              $begingroup$

              First of all if we plug in $x=1$ we get



              $$f(1) = frac 12 + frac 14 f(1). $$



              This implies



              $$f(1) = frac 23$$



              so it appears that you cannot have $f(1)=1$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                First of all if we plug in $x=1$ we get



                $$f(1) = frac 12 + frac 14 f(1). $$



                This implies



                $$f(1) = frac 23$$



                so it appears that you cannot have $f(1)=1$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  First of all if we plug in $x=1$ we get



                  $$f(1) = frac 12 + frac 14 f(1). $$



                  This implies



                  $$f(1) = frac 23$$



                  so it appears that you cannot have $f(1)=1$.






                  share|cite|improve this answer









                  $endgroup$



                  First of all if we plug in $x=1$ we get



                  $$f(1) = frac 12 + frac 14 f(1). $$



                  This implies



                  $$f(1) = frac 23$$



                  so it appears that you cannot have $f(1)=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 12:56









                  flawrflawr

                  11.5k32446




                  11.5k32446























                      0












                      $begingroup$

                      $$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right) tag 1$$
                      As already pointed out this implies $f(1)=frac23$ which is contradictory to the specified condition $f(1)=1$.



                      Thus the answer is : There is no solution to the problem with the original wording.



                      Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.



                      We will check the solution :
                      $$f(x)=frac23+cleft(frac{x}{x-1}right)^2 tag 2$$
                      $c=$any constant.



                      $$fleft( frac{2x}{1+x} right)=frac23+cleft(frac{ frac{2x}{1+x}}{ frac{2x}{1+x}-1}right)^2 tag 3$$
                      After simplification :
                      $$fleft( frac{2x}{1+x} right)= frac23+cleft(frac{2x}{x-1}right)^2$$
                      We put $(3)$ in $(1)$ :
                      $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac{1}{2} + frac{1}{4}left( frac23+cleft(frac{2x}{x-1}right)^2right)$$
                      After simplification :
                      $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac23+cleft(frac{x}{x-1}right)^2$$
                      Comparing to Eq.$(2)$ we see that :
                      $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=f(x)$$
                      This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
                      $$f(x)=frac23+cleft(frac{x}{x-1}right)^2 qquad (xneq 1)$$
                      The particular case $c=0$ corresponds to the trivial solution $f(x)=frac23$.



                      This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        $$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right) tag 1$$
                        As already pointed out this implies $f(1)=frac23$ which is contradictory to the specified condition $f(1)=1$.



                        Thus the answer is : There is no solution to the problem with the original wording.



                        Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.



                        We will check the solution :
                        $$f(x)=frac23+cleft(frac{x}{x-1}right)^2 tag 2$$
                        $c=$any constant.



                        $$fleft( frac{2x}{1+x} right)=frac23+cleft(frac{ frac{2x}{1+x}}{ frac{2x}{1+x}-1}right)^2 tag 3$$
                        After simplification :
                        $$fleft( frac{2x}{1+x} right)= frac23+cleft(frac{2x}{x-1}right)^2$$
                        We put $(3)$ in $(1)$ :
                        $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac{1}{2} + frac{1}{4}left( frac23+cleft(frac{2x}{x-1}right)^2right)$$
                        After simplification :
                        $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac23+cleft(frac{x}{x-1}right)^2$$
                        Comparing to Eq.$(2)$ we see that :
                        $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=f(x)$$
                        This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
                        $$f(x)=frac23+cleft(frac{x}{x-1}right)^2 qquad (xneq 1)$$
                        The particular case $c=0$ corresponds to the trivial solution $f(x)=frac23$.



                        This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          $$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right) tag 1$$
                          As already pointed out this implies $f(1)=frac23$ which is contradictory to the specified condition $f(1)=1$.



                          Thus the answer is : There is no solution to the problem with the original wording.



                          Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.



                          We will check the solution :
                          $$f(x)=frac23+cleft(frac{x}{x-1}right)^2 tag 2$$
                          $c=$any constant.



                          $$fleft( frac{2x}{1+x} right)=frac23+cleft(frac{ frac{2x}{1+x}}{ frac{2x}{1+x}-1}right)^2 tag 3$$
                          After simplification :
                          $$fleft( frac{2x}{1+x} right)= frac23+cleft(frac{2x}{x-1}right)^2$$
                          We put $(3)$ in $(1)$ :
                          $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac{1}{2} + frac{1}{4}left( frac23+cleft(frac{2x}{x-1}right)^2right)$$
                          After simplification :
                          $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac23+cleft(frac{x}{x-1}right)^2$$
                          Comparing to Eq.$(2)$ we see that :
                          $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=f(x)$$
                          This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
                          $$f(x)=frac23+cleft(frac{x}{x-1}right)^2 qquad (xneq 1)$$
                          The particular case $c=0$ corresponds to the trivial solution $f(x)=frac23$.



                          This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.






                          share|cite|improve this answer









                          $endgroup$



                          $$f(x)= frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right) tag 1$$
                          As already pointed out this implies $f(1)=frac23$ which is contradictory to the specified condition $f(1)=1$.



                          Thus the answer is : There is no solution to the problem with the original wording.



                          Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.



                          We will check the solution :
                          $$f(x)=frac23+cleft(frac{x}{x-1}right)^2 tag 2$$
                          $c=$any constant.



                          $$fleft( frac{2x}{1+x} right)=frac23+cleft(frac{ frac{2x}{1+x}}{ frac{2x}{1+x}-1}right)^2 tag 3$$
                          After simplification :
                          $$fleft( frac{2x}{1+x} right)= frac23+cleft(frac{2x}{x-1}right)^2$$
                          We put $(3)$ in $(1)$ :
                          $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac{1}{2} + frac{1}{4}left( frac23+cleft(frac{2x}{x-1}right)^2right)$$
                          After simplification :
                          $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=frac23+cleft(frac{x}{x-1}right)^2$$
                          Comparing to Eq.$(2)$ we see that :
                          $$frac{1}{2} + frac{1}{4}fleft(frac{2x}{1+x}right)=f(x)$$
                          This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
                          $$f(x)=frac23+cleft(frac{x}{x-1}right)^2 qquad (xneq 1)$$
                          The particular case $c=0$ corresponds to the trivial solution $f(x)=frac23$.



                          This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 14 '18 at 11:55









                          JJacquelinJJacquelin

                          43.7k21853




                          43.7k21853






























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