$f(a)-f(b)$ is rational iff $f(a-b) $ is rational












12















Prove that the continuous function $f:mathbb{R} to mathbb{R}$ satisfying $fleft(xright)-fleft(yright) inmathbb{Q} iff fleft(x-yright) in mathbb{Q}$ is of the form $
fleft(xright)=ax+b.$




My Attempt.



I tried considering the function $$gleft(xright)=frac{fleft(xright)-fleft(0right)}{fleft(1right)-fleft(0right)}$$ which also satisfies the property $gleft(aright)-gleft(bright) inmathbb{Q} iff gleft(a-bright) in mathbb{Q}$,



Now I am trying to prove that this function g is identity function and then I can prove that $fleft(xright)=left(fleft(1right)-fleft(0right)right)x+fleft(0right).$ And I am done.



Also This function has to be identity function because $gleft(0right)=0$ and $gleft(1right)=1$.



I tried assuming that the function g is such that $gleft(aright)neq a$ for some $ain mathbb{R}$. Then by continuity $gleft(xright)neq x$ for some $delta>0$ neighborhood of $a$.
But I cannot move further.



Also
Using a previously known result, I was able to prove that f must be monotonic. However I do not want to use any other result which is not known and not trivial.




If a function $f $ is continous in $left[a,bright]$ and $fleft(aright)=fleft(bright)$ then for any $epsilon >0$ there exists $m,n in left[a,bright] $such that $fleft(mright)=fleft(nright)$ and $m-n=epsilon$.




In this case choose $epsilon in mathbb{R}-mathbb{Q} $ and get $m,n in left[a,bright] $ such that $fleft(mright)-fleft(nright)=0 in mathbb{Q}$ but $m-n inmathbb{R}-mathbb{Q}$.






real-analysis continuity functional-equations rational-numbers fixed-point-theorems







share|cite|improve this question





























    12















    Prove that the continuous function $f:mathbb{R} to mathbb{R}$ satisfying $fleft(xright)-fleft(yright) inmathbb{Q} iff fleft(x-yright) in mathbb{Q}$ is of the form $
    fleft(xright)=ax+b.$




    My Attempt.



    I tried considering the function $$gleft(xright)=frac{fleft(xright)-fleft(0right)}{fleft(1right)-fleft(0right)}$$ which also satisfies the property $gleft(aright)-gleft(bright) inmathbb{Q} iff gleft(a-bright) in mathbb{Q}$,



    Now I am trying to prove that this function g is identity function and then I can prove that $fleft(xright)=left(fleft(1right)-fleft(0right)right)x+fleft(0right).$ And I am done.



    Also This function has to be identity function because $gleft(0right)=0$ and $gleft(1right)=1$.



    I tried assuming that the function g is such that $gleft(aright)neq a$ for some $ain mathbb{R}$. Then by continuity $gleft(xright)neq x$ for some $delta>0$ neighborhood of $a$.
    But I cannot move further.



    Also
    Using a previously known result, I was able to prove that f must be monotonic. However I do not want to use any other result which is not known and not trivial.




    If a function $f $ is continous in $left[a,bright]$ and $fleft(aright)=fleft(bright)$ then for any $epsilon >0$ there exists $m,n in left[a,bright] $such that $fleft(mright)=fleft(nright)$ and $m-n=epsilon$.




    In this case choose $epsilon in mathbb{R}-mathbb{Q} $ and get $m,n in left[a,bright] $ such that $fleft(mright)-fleft(nright)=0 in mathbb{Q}$ but $m-n inmathbb{R}-mathbb{Q}$.






    real-analysis continuity functional-equations rational-numbers fixed-point-theorems







    share|cite|improve this question



























      12












      12








      12


      5






      Prove that the continuous function $f:mathbb{R} to mathbb{R}$ satisfying $fleft(xright)-fleft(yright) inmathbb{Q} iff fleft(x-yright) in mathbb{Q}$ is of the form $
      fleft(xright)=ax+b.$




      My Attempt.



      I tried considering the function $$gleft(xright)=frac{fleft(xright)-fleft(0right)}{fleft(1right)-fleft(0right)}$$ which also satisfies the property $gleft(aright)-gleft(bright) inmathbb{Q} iff gleft(a-bright) in mathbb{Q}$,



      Now I am trying to prove that this function g is identity function and then I can prove that $fleft(xright)=left(fleft(1right)-fleft(0right)right)x+fleft(0right).$ And I am done.



      Also This function has to be identity function because $gleft(0right)=0$ and $gleft(1right)=1$.



      I tried assuming that the function g is such that $gleft(aright)neq a$ for some $ain mathbb{R}$. Then by continuity $gleft(xright)neq x$ for some $delta>0$ neighborhood of $a$.
      But I cannot move further.



      Also
      Using a previously known result, I was able to prove that f must be monotonic. However I do not want to use any other result which is not known and not trivial.




      If a function $f $ is continous in $left[a,bright]$ and $fleft(aright)=fleft(bright)$ then for any $epsilon >0$ there exists $m,n in left[a,bright] $such that $fleft(mright)=fleft(nright)$ and $m-n=epsilon$.




      In this case choose $epsilon in mathbb{R}-mathbb{Q} $ and get $m,n in left[a,bright] $ such that $fleft(mright)-fleft(nright)=0 in mathbb{Q}$ but $m-n inmathbb{R}-mathbb{Q}$.






      real-analysis continuity functional-equations rational-numbers fixed-point-theorems







      share|cite|improve this question
















      Prove that the continuous function $f:mathbb{R} to mathbb{R}$ satisfying $fleft(xright)-fleft(yright) inmathbb{Q} iff fleft(x-yright) in mathbb{Q}$ is of the form $
      fleft(xright)=ax+b.$




      My Attempt.



      I tried considering the function $$gleft(xright)=frac{fleft(xright)-fleft(0right)}{fleft(1right)-fleft(0right)}$$ which also satisfies the property $gleft(aright)-gleft(bright) inmathbb{Q} iff gleft(a-bright) in mathbb{Q}$,



      Now I am trying to prove that this function g is identity function and then I can prove that $fleft(xright)=left(fleft(1right)-fleft(0right)right)x+fleft(0right).$ And I am done.



      Also This function has to be identity function because $gleft(0right)=0$ and $gleft(1right)=1$.



      I tried assuming that the function g is such that $gleft(aright)neq a$ for some $ain mathbb{R}$. Then by continuity $gleft(xright)neq x$ for some $delta>0$ neighborhood of $a$.
      But I cannot move further.



      Also
      Using a previously known result, I was able to prove that f must be monotonic. However I do not want to use any other result which is not known and not trivial.




      If a function $f $ is continous in $left[a,bright]$ and $fleft(aright)=fleft(bright)$ then for any $epsilon >0$ there exists $m,n in left[a,bright] $such that $fleft(mright)=fleft(nright)$ and $m-n=epsilon$.




      In this case choose $epsilon in mathbb{R}-mathbb{Q} $ and get $m,n in left[a,bright] $ such that $fleft(mright)-fleft(nright)=0 in mathbb{Q}$ but $m-n inmathbb{R}-mathbb{Q}$.






      real-analysis continuity functional-equations rational-numbers fixed-point-theorems




      real-analysis continuity functional-equations rational-numbers fixed-point-theorems






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 3:14







      Rakesh Bhatt

















      asked Dec 5 '18 at 17:03









      Rakesh BhattRakesh Bhatt

      1,015114




      1,015114






















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          For any $y$, we have either $f(x+y) - f(x) in mathbb Q$ for all $x$, or $f(x+y) - f(x) notin mathbb Q$ for all $x$, depending on whether or not $f(y) in mathbb Q$. But $f(x+y)-f(x)$ is continuous, so by the Intermediate Value Theorem we conclude $f(x+y) - f(x)$ is constant. Thus
          $$ f(x+y) - f(x) = f(y) - f(0) $$
          This says $f(x) - f(0)$ is an additive function. And it's not hard to prove that continuous additive functions are linear.






          share|cite|improve this answer





















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            For any $y$, we have either $f(x+y) - f(x) in mathbb Q$ for all $x$, or $f(x+y) - f(x) notin mathbb Q$ for all $x$, depending on whether or not $f(y) in mathbb Q$. But $f(x+y)-f(x)$ is continuous, so by the Intermediate Value Theorem we conclude $f(x+y) - f(x)$ is constant. Thus
            $$ f(x+y) - f(x) = f(y) - f(0) $$
            This says $f(x) - f(0)$ is an additive function. And it's not hard to prove that continuous additive functions are linear.






            share|cite|improve this answer


























              9














              For any $y$, we have either $f(x+y) - f(x) in mathbb Q$ for all $x$, or $f(x+y) - f(x) notin mathbb Q$ for all $x$, depending on whether or not $f(y) in mathbb Q$. But $f(x+y)-f(x)$ is continuous, so by the Intermediate Value Theorem we conclude $f(x+y) - f(x)$ is constant. Thus
              $$ f(x+y) - f(x) = f(y) - f(0) $$
              This says $f(x) - f(0)$ is an additive function. And it's not hard to prove that continuous additive functions are linear.






              share|cite|improve this answer
























                9












                9








                9






                For any $y$, we have either $f(x+y) - f(x) in mathbb Q$ for all $x$, or $f(x+y) - f(x) notin mathbb Q$ for all $x$, depending on whether or not $f(y) in mathbb Q$. But $f(x+y)-f(x)$ is continuous, so by the Intermediate Value Theorem we conclude $f(x+y) - f(x)$ is constant. Thus
                $$ f(x+y) - f(x) = f(y) - f(0) $$
                This says $f(x) - f(0)$ is an additive function. And it's not hard to prove that continuous additive functions are linear.






                share|cite|improve this answer












                For any $y$, we have either $f(x+y) - f(x) in mathbb Q$ for all $x$, or $f(x+y) - f(x) notin mathbb Q$ for all $x$, depending on whether or not $f(y) in mathbb Q$. But $f(x+y)-f(x)$ is continuous, so by the Intermediate Value Theorem we conclude $f(x+y) - f(x)$ is constant. Thus
                $$ f(x+y) - f(x) = f(y) - f(0) $$
                This says $f(x) - f(0)$ is an additive function. And it's not hard to prove that continuous additive functions are linear.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 5 '18 at 17:22









                Robert IsraelRobert Israel

                319k23208458




                319k23208458






























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