Prove a function is constant function with the given conditions.












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If $f(4-x)=f (4+x)$ and $f(2-x)=f(2+x)$, prove that $f$ is constant function.




I tried to solve by assuming x and y are two variable then if we get $f (x)=f (y)$ then $f$ is constant function, but can't complete the problem. Someone please help me.






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    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
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    – José Carlos Santos
    Oct 27 '18 at 9:21
















0












$begingroup$



If $f(4-x)=f (4+x)$ and $f(2-x)=f(2+x)$, prove that $f$ is constant function.




I tried to solve by assuming x and y are two variable then if we get $f (x)=f (y)$ then $f$ is constant function, but can't complete the problem. Someone please help me.






functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    Oct 27 '18 at 9:21














0












0








0





$begingroup$



If $f(4-x)=f (4+x)$ and $f(2-x)=f(2+x)$, prove that $f$ is constant function.




I tried to solve by assuming x and y are two variable then if we get $f (x)=f (y)$ then $f$ is constant function, but can't complete the problem. Someone please help me.






functional-equations







share|cite|improve this question











$endgroup$





If $f(4-x)=f (4+x)$ and $f(2-x)=f(2+x)$, prove that $f$ is constant function.




I tried to solve by assuming x and y are two variable then if we get $f (x)=f (y)$ then $f$ is constant function, but can't complete the problem. Someone please help me.






functional-equations




functional-equations






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edited Jan 6 at 2:38









amWhy

1




1










asked Oct 27 '18 at 9:13









Pramit BanerjeePramit Banerjee

11




11












  • $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    Oct 27 '18 at 9:21


















  • $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    Oct 27 '18 at 9:21
















$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Oct 27 '18 at 9:21




$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Oct 27 '18 at 9:21










3 Answers
3






active

oldest

votes


















2












$begingroup$

$f$ is not necessarily constant. Here is a counter-example:
$$
f(x) =
begin{cases}
0, & text{if $xinBbb Q$} \
1, & text{if $xnotinBbb Q$}
end{cases}$$



And another one, continuous this time:



$$f(x)=cospi x$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The two equations $,f(4-x)=f(4+x),$ and $,f(2-x)=f(2+x),$ require that the function is invariant with respect to two reflections and hence a translation by $4$. The simple calculation is $, x mapsto 8-x ,$ in the first reflection and then $ 8-x mapsto x-4 ,$ after the second reflection yielding a translation by $4.$ Assuming that the function is defined for all real numbers, and that no other conditions apply, then you can let the function be arbitrarily defined on the interval $,[0,2].,$
    The function must be an even function with a period of $4$. That is, $,f(x)=f(-x),$ and $,f(x)=f(x+4),$ for all real $,x.,$ This determines the function for all real numbers and that it satisfies $,f(2 n-x)=f(2 n+x)=f(x),$ and $, f(4 n + x) = f(x) ,$ for all integer $,n,$ and all real $,x.,$



    For a simple example, consider $, f(4n+x) := |x|,$ for all $,|x|le 2,$ and integer $,n.,$ This is a sawtooth function with fundamental period $,4.,$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      That should read, "and hence a translation by 4." For instance, $f(x)=cosfrac{pi}{2}x$ satisfies the conditions.
      $endgroup$
      – TonyK
      Oct 27 '18 at 10:37










    • $begingroup$
      A reflection about $x=2$, followed by a reflection about $x=4$, amounts to a translation by $4$, not $2$.
      $endgroup$
      – TonyK
      Oct 27 '18 at 10:53










    • $begingroup$
      @TonyK Thanks for that correction. I should be more careful.
      $endgroup$
      – Somos
      Oct 27 '18 at 11:01



















    0












    $begingroup$

    Without tellings what are the domain and the codomain of $f$, nobody can answer that. However, if the domain is, say, $mathbb Q$, then the statement is false. Consider, for instance,$$begin{array}{rccc}fcolon&mathbb Q&longrightarrow&mathbb Z\&q&mapsto&begin{cases}1&text{ if }qinmathbb Z\0&text{ otherwise.}end{cases}end{array}$$






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      $f$ is not necessarily constant. Here is a counter-example:
      $$
      f(x) =
      begin{cases}
      0, & text{if $xinBbb Q$} \
      1, & text{if $xnotinBbb Q$}
      end{cases}$$



      And another one, continuous this time:



      $$f(x)=cospi x$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $f$ is not necessarily constant. Here is a counter-example:
        $$
        f(x) =
        begin{cases}
        0, & text{if $xinBbb Q$} \
        1, & text{if $xnotinBbb Q$}
        end{cases}$$



        And another one, continuous this time:



        $$f(x)=cospi x$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $f$ is not necessarily constant. Here is a counter-example:
          $$
          f(x) =
          begin{cases}
          0, & text{if $xinBbb Q$} \
          1, & text{if $xnotinBbb Q$}
          end{cases}$$



          And another one, continuous this time:



          $$f(x)=cospi x$$






          share|cite|improve this answer









          $endgroup$



          $f$ is not necessarily constant. Here is a counter-example:
          $$
          f(x) =
          begin{cases}
          0, & text{if $xinBbb Q$} \
          1, & text{if $xnotinBbb Q$}
          end{cases}$$



          And another one, continuous this time:



          $$f(x)=cospi x$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 27 '18 at 9:23









          TonyKTonyK

          42.2k355134




          42.2k355134























              2












              $begingroup$

              The two equations $,f(4-x)=f(4+x),$ and $,f(2-x)=f(2+x),$ require that the function is invariant with respect to two reflections and hence a translation by $4$. The simple calculation is $, x mapsto 8-x ,$ in the first reflection and then $ 8-x mapsto x-4 ,$ after the second reflection yielding a translation by $4.$ Assuming that the function is defined for all real numbers, and that no other conditions apply, then you can let the function be arbitrarily defined on the interval $,[0,2].,$
              The function must be an even function with a period of $4$. That is, $,f(x)=f(-x),$ and $,f(x)=f(x+4),$ for all real $,x.,$ This determines the function for all real numbers and that it satisfies $,f(2 n-x)=f(2 n+x)=f(x),$ and $, f(4 n + x) = f(x) ,$ for all integer $,n,$ and all real $,x.,$



              For a simple example, consider $, f(4n+x) := |x|,$ for all $,|x|le 2,$ and integer $,n.,$ This is a sawtooth function with fundamental period $,4.,$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                That should read, "and hence a translation by 4." For instance, $f(x)=cosfrac{pi}{2}x$ satisfies the conditions.
                $endgroup$
                – TonyK
                Oct 27 '18 at 10:37










              • $begingroup$
                A reflection about $x=2$, followed by a reflection about $x=4$, amounts to a translation by $4$, not $2$.
                $endgroup$
                – TonyK
                Oct 27 '18 at 10:53










              • $begingroup$
                @TonyK Thanks for that correction. I should be more careful.
                $endgroup$
                – Somos
                Oct 27 '18 at 11:01
















              2












              $begingroup$

              The two equations $,f(4-x)=f(4+x),$ and $,f(2-x)=f(2+x),$ require that the function is invariant with respect to two reflections and hence a translation by $4$. The simple calculation is $, x mapsto 8-x ,$ in the first reflection and then $ 8-x mapsto x-4 ,$ after the second reflection yielding a translation by $4.$ Assuming that the function is defined for all real numbers, and that no other conditions apply, then you can let the function be arbitrarily defined on the interval $,[0,2].,$
              The function must be an even function with a period of $4$. That is, $,f(x)=f(-x),$ and $,f(x)=f(x+4),$ for all real $,x.,$ This determines the function for all real numbers and that it satisfies $,f(2 n-x)=f(2 n+x)=f(x),$ and $, f(4 n + x) = f(x) ,$ for all integer $,n,$ and all real $,x.,$



              For a simple example, consider $, f(4n+x) := |x|,$ for all $,|x|le 2,$ and integer $,n.,$ This is a sawtooth function with fundamental period $,4.,$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                That should read, "and hence a translation by 4." For instance, $f(x)=cosfrac{pi}{2}x$ satisfies the conditions.
                $endgroup$
                – TonyK
                Oct 27 '18 at 10:37










              • $begingroup$
                A reflection about $x=2$, followed by a reflection about $x=4$, amounts to a translation by $4$, not $2$.
                $endgroup$
                – TonyK
                Oct 27 '18 at 10:53










              • $begingroup$
                @TonyK Thanks for that correction. I should be more careful.
                $endgroup$
                – Somos
                Oct 27 '18 at 11:01














              2












              2








              2





              $begingroup$

              The two equations $,f(4-x)=f(4+x),$ and $,f(2-x)=f(2+x),$ require that the function is invariant with respect to two reflections and hence a translation by $4$. The simple calculation is $, x mapsto 8-x ,$ in the first reflection and then $ 8-x mapsto x-4 ,$ after the second reflection yielding a translation by $4.$ Assuming that the function is defined for all real numbers, and that no other conditions apply, then you can let the function be arbitrarily defined on the interval $,[0,2].,$
              The function must be an even function with a period of $4$. That is, $,f(x)=f(-x),$ and $,f(x)=f(x+4),$ for all real $,x.,$ This determines the function for all real numbers and that it satisfies $,f(2 n-x)=f(2 n+x)=f(x),$ and $, f(4 n + x) = f(x) ,$ for all integer $,n,$ and all real $,x.,$



              For a simple example, consider $, f(4n+x) := |x|,$ for all $,|x|le 2,$ and integer $,n.,$ This is a sawtooth function with fundamental period $,4.,$






              share|cite|improve this answer











              $endgroup$



              The two equations $,f(4-x)=f(4+x),$ and $,f(2-x)=f(2+x),$ require that the function is invariant with respect to two reflections and hence a translation by $4$. The simple calculation is $, x mapsto 8-x ,$ in the first reflection and then $ 8-x mapsto x-4 ,$ after the second reflection yielding a translation by $4.$ Assuming that the function is defined for all real numbers, and that no other conditions apply, then you can let the function be arbitrarily defined on the interval $,[0,2].,$
              The function must be an even function with a period of $4$. That is, $,f(x)=f(-x),$ and $,f(x)=f(x+4),$ for all real $,x.,$ This determines the function for all real numbers and that it satisfies $,f(2 n-x)=f(2 n+x)=f(x),$ and $, f(4 n + x) = f(x) ,$ for all integer $,n,$ and all real $,x.,$



              For a simple example, consider $, f(4n+x) := |x|,$ for all $,|x|le 2,$ and integer $,n.,$ This is a sawtooth function with fundamental period $,4.,$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Oct 27 '18 at 11:05

























              answered Oct 27 '18 at 10:08









              SomosSomos

              13.1k11034




              13.1k11034












              • $begingroup$
                That should read, "and hence a translation by 4." For instance, $f(x)=cosfrac{pi}{2}x$ satisfies the conditions.
                $endgroup$
                – TonyK
                Oct 27 '18 at 10:37










              • $begingroup$
                A reflection about $x=2$, followed by a reflection about $x=4$, amounts to a translation by $4$, not $2$.
                $endgroup$
                – TonyK
                Oct 27 '18 at 10:53










              • $begingroup$
                @TonyK Thanks for that correction. I should be more careful.
                $endgroup$
                – Somos
                Oct 27 '18 at 11:01


















              • $begingroup$
                That should read, "and hence a translation by 4." For instance, $f(x)=cosfrac{pi}{2}x$ satisfies the conditions.
                $endgroup$
                – TonyK
                Oct 27 '18 at 10:37










              • $begingroup$
                A reflection about $x=2$, followed by a reflection about $x=4$, amounts to a translation by $4$, not $2$.
                $endgroup$
                – TonyK
                Oct 27 '18 at 10:53










              • $begingroup$
                @TonyK Thanks for that correction. I should be more careful.
                $endgroup$
                – Somos
                Oct 27 '18 at 11:01
















              $begingroup$
              That should read, "and hence a translation by 4." For instance, $f(x)=cosfrac{pi}{2}x$ satisfies the conditions.
              $endgroup$
              – TonyK
              Oct 27 '18 at 10:37




              $begingroup$
              That should read, "and hence a translation by 4." For instance, $f(x)=cosfrac{pi}{2}x$ satisfies the conditions.
              $endgroup$
              – TonyK
              Oct 27 '18 at 10:37












              $begingroup$
              A reflection about $x=2$, followed by a reflection about $x=4$, amounts to a translation by $4$, not $2$.
              $endgroup$
              – TonyK
              Oct 27 '18 at 10:53




              $begingroup$
              A reflection about $x=2$, followed by a reflection about $x=4$, amounts to a translation by $4$, not $2$.
              $endgroup$
              – TonyK
              Oct 27 '18 at 10:53












              $begingroup$
              @TonyK Thanks for that correction. I should be more careful.
              $endgroup$
              – Somos
              Oct 27 '18 at 11:01




              $begingroup$
              @TonyK Thanks for that correction. I should be more careful.
              $endgroup$
              – Somos
              Oct 27 '18 at 11:01











              0












              $begingroup$

              Without tellings what are the domain and the codomain of $f$, nobody can answer that. However, if the domain is, say, $mathbb Q$, then the statement is false. Consider, for instance,$$begin{array}{rccc}fcolon&mathbb Q&longrightarrow&mathbb Z\&q&mapsto&begin{cases}1&text{ if }qinmathbb Z\0&text{ otherwise.}end{cases}end{array}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Without tellings what are the domain and the codomain of $f$, nobody can answer that. However, if the domain is, say, $mathbb Q$, then the statement is false. Consider, for instance,$$begin{array}{rccc}fcolon&mathbb Q&longrightarrow&mathbb Z\&q&mapsto&begin{cases}1&text{ if }qinmathbb Z\0&text{ otherwise.}end{cases}end{array}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Without tellings what are the domain and the codomain of $f$, nobody can answer that. However, if the domain is, say, $mathbb Q$, then the statement is false. Consider, for instance,$$begin{array}{rccc}fcolon&mathbb Q&longrightarrow&mathbb Z\&q&mapsto&begin{cases}1&text{ if }qinmathbb Z\0&text{ otherwise.}end{cases}end{array}$$






                  share|cite|improve this answer









                  $endgroup$



                  Without tellings what are the domain and the codomain of $f$, nobody can answer that. However, if the domain is, say, $mathbb Q$, then the statement is false. Consider, for instance,$$begin{array}{rccc}fcolon&mathbb Q&longrightarrow&mathbb Z\&q&mapsto&begin{cases}1&text{ if }qinmathbb Z\0&text{ otherwise.}end{cases}end{array}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 27 '18 at 9:25









                  José Carlos SantosJosé Carlos Santos

                  154k22123226




                  154k22123226






























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