$f(f(f(x)))=x,$ but $f(x)neq x$












4












$begingroup$


Let $f: mathbb{R}to mathbb{R}$ be strictly increasing function such that $f(f(f(x)))=x$ then by this $f(x)=x.$ Does the result fail if we drop the hypothesis that f is strictly increasing?
I find such an example if I change the domain and codomain to $mathbb{R}^2$ and $f=rotation by 120^o,$ but could not think of a map from $mathbb{R}to mathbb{R}.$
Any help is appreciated.






real-analysis functions







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    Let $f: mathbb{R}to mathbb{R}$ be strictly increasing function such that $f(f(f(x)))=x$ then by this $f(x)=x.$ Does the result fail if we drop the hypothesis that f is strictly increasing?
    I find such an example if I change the domain and codomain to $mathbb{R}^2$ and $f=rotation by 120^o,$ but could not think of a map from $mathbb{R}to mathbb{R}.$
    Any help is appreciated.






    real-analysis functions







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      Let $f: mathbb{R}to mathbb{R}$ be strictly increasing function such that $f(f(f(x)))=x$ then by this $f(x)=x.$ Does the result fail if we drop the hypothesis that f is strictly increasing?
      I find such an example if I change the domain and codomain to $mathbb{R}^2$ and $f=rotation by 120^o,$ but could not think of a map from $mathbb{R}to mathbb{R}.$
      Any help is appreciated.






      real-analysis functions







      share|cite|improve this question









      $endgroup$




      Let $f: mathbb{R}to mathbb{R}$ be strictly increasing function such that $f(f(f(x)))=x$ then by this $f(x)=x.$ Does the result fail if we drop the hypothesis that f is strictly increasing?
      I find such an example if I change the domain and codomain to $mathbb{R}^2$ and $f=rotation by 120^o,$ but could not think of a map from $mathbb{R}to mathbb{R}.$
      Any help is appreciated.






      real-analysis functions




      real-analysis functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 30 at 20:10









      user345777user345777

      432312




      432312






















          4 Answers
          4






          active

          oldest

          votes


















          9












          $begingroup$

          Let $g:mathbb{R^1}tomathbb{R^2}$ be a one-to-one mapping (which exists, because both sets have the same cardinality). Now set $f=g^{-1}circ Fcirc g:mathbb{R}^1tomathbb{R}^1$, where $F$ is the rotation by $120^circ$. So the result is not true: this is the desired counterexample.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Do you mean one-to-one and onto? Otherwise $g^{-1}$ doesn't make sense
            $endgroup$
            – leibnewtz
            Jan 30 at 20:27










          • $begingroup$
            @leibnewtz Yes, sure.
            $endgroup$
            – Vladimir
            Jan 30 at 20:30



















          3












          $begingroup$

          Define $f(0)=1, f(1)=2, f(2)=0$ and $f(x)=x$ for $x notin {0,1,2}$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            You can make $f$ act as a 3-cyclic permutation



            $ f(x) = begin{cases}
            0 & text{if $x = -1$ } \
            1 &text{if $x = 0$} \
            -1 &text{if $x = 1$} \
            x &text{otherwise}
            end{cases}$



            I suppose it's also possible to make a continuous version of this.






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              Continuous is not possible, because then f would be increasing.
              $endgroup$
              – user345777
              Jan 30 at 20:29












            • $begingroup$
              Sorry, it was a stupid guess. Can I ask you why it would be increasing?
              $endgroup$
              – Marco All-in Nervo
              Jan 30 at 20:33






            • 1




              $begingroup$
              Agree with @user345777 $f$ is obviously a bijection. A continuous bijection from $Bbb{R}$ to itself is either increasing or decreasing. The 3-fold iteration of a decreasing function is itself decreasing which is absurd. So it must increasing and that case has been handled.
              $endgroup$
              – Jyrki Lahtonen
              Jan 30 at 20:34



















            2












            $begingroup$

            If you allow yourself to consider the projective line (i.e., the reals with a point at infinity), then the function
            $$
            f(x) = frac{1}{1-x}
            $$
            is a nice continuous solution to the problem (although you have to make sense of "continuous" for this extended line). Pierre Samuel's book on Projective Geometry has a nice exposition of this around page 58.



            (By the way, this answer was inspired by @DonaldSplutterwit's deleted answer, although I might have stumbled on it myself, since I've been thinking about projective geometry and looking at Samuel's book today anyhow. Thanks, Donald!)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Then you could "cram $infty$ into $mathbb{R}$" using a variant of the infinite hotel paradox if you really wanted to, and then apply the same sort of construction as in Vladimir's solution. Of course, that won't result in a continuous function $mathbb{R} to mathbb{R}$.
              $endgroup$
              – Daniel Schepler
              Jan 30 at 20:41












            Your Answer





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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            9












            $begingroup$

            Let $g:mathbb{R^1}tomathbb{R^2}$ be a one-to-one mapping (which exists, because both sets have the same cardinality). Now set $f=g^{-1}circ Fcirc g:mathbb{R}^1tomathbb{R}^1$, where $F$ is the rotation by $120^circ$. So the result is not true: this is the desired counterexample.






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              Do you mean one-to-one and onto? Otherwise $g^{-1}$ doesn't make sense
              $endgroup$
              – leibnewtz
              Jan 30 at 20:27










            • $begingroup$
              @leibnewtz Yes, sure.
              $endgroup$
              – Vladimir
              Jan 30 at 20:30
















            9












            $begingroup$

            Let $g:mathbb{R^1}tomathbb{R^2}$ be a one-to-one mapping (which exists, because both sets have the same cardinality). Now set $f=g^{-1}circ Fcirc g:mathbb{R}^1tomathbb{R}^1$, where $F$ is the rotation by $120^circ$. So the result is not true: this is the desired counterexample.






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              Do you mean one-to-one and onto? Otherwise $g^{-1}$ doesn't make sense
              $endgroup$
              – leibnewtz
              Jan 30 at 20:27










            • $begingroup$
              @leibnewtz Yes, sure.
              $endgroup$
              – Vladimir
              Jan 30 at 20:30














            9












            9








            9





            $begingroup$

            Let $g:mathbb{R^1}tomathbb{R^2}$ be a one-to-one mapping (which exists, because both sets have the same cardinality). Now set $f=g^{-1}circ Fcirc g:mathbb{R}^1tomathbb{R}^1$, where $F$ is the rotation by $120^circ$. So the result is not true: this is the desired counterexample.






            share|cite|improve this answer









            $endgroup$



            Let $g:mathbb{R^1}tomathbb{R^2}$ be a one-to-one mapping (which exists, because both sets have the same cardinality). Now set $f=g^{-1}circ Fcirc g:mathbb{R}^1tomathbb{R}^1$, where $F$ is the rotation by $120^circ$. So the result is not true: this is the desired counterexample.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 30 at 20:19









            VladimirVladimir

            5,413618




            5,413618








            • 2




              $begingroup$
              Do you mean one-to-one and onto? Otherwise $g^{-1}$ doesn't make sense
              $endgroup$
              – leibnewtz
              Jan 30 at 20:27










            • $begingroup$
              @leibnewtz Yes, sure.
              $endgroup$
              – Vladimir
              Jan 30 at 20:30














            • 2




              $begingroup$
              Do you mean one-to-one and onto? Otherwise $g^{-1}$ doesn't make sense
              $endgroup$
              – leibnewtz
              Jan 30 at 20:27










            • $begingroup$
              @leibnewtz Yes, sure.
              $endgroup$
              – Vladimir
              Jan 30 at 20:30








            2




            2




            $begingroup$
            Do you mean one-to-one and onto? Otherwise $g^{-1}$ doesn't make sense
            $endgroup$
            – leibnewtz
            Jan 30 at 20:27




            $begingroup$
            Do you mean one-to-one and onto? Otherwise $g^{-1}$ doesn't make sense
            $endgroup$
            – leibnewtz
            Jan 30 at 20:27












            $begingroup$
            @leibnewtz Yes, sure.
            $endgroup$
            – Vladimir
            Jan 30 at 20:30




            $begingroup$
            @leibnewtz Yes, sure.
            $endgroup$
            – Vladimir
            Jan 30 at 20:30











            3












            $begingroup$

            Define $f(0)=1, f(1)=2, f(2)=0$ and $f(x)=x$ for $x notin {0,1,2}$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Define $f(0)=1, f(1)=2, f(2)=0$ and $f(x)=x$ for $x notin {0,1,2}$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Define $f(0)=1, f(1)=2, f(2)=0$ and $f(x)=x$ for $x notin {0,1,2}$.






                share|cite|improve this answer









                $endgroup$



                Define $f(0)=1, f(1)=2, f(2)=0$ and $f(x)=x$ for $x notin {0,1,2}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 at 20:23









                VincentVincent

                1,067614




                1,067614























                    2












                    $begingroup$

                    You can make $f$ act as a 3-cyclic permutation



                    $ f(x) = begin{cases}
                    0 & text{if $x = -1$ } \
                    1 &text{if $x = 0$} \
                    -1 &text{if $x = 1$} \
                    x &text{otherwise}
                    end{cases}$



                    I suppose it's also possible to make a continuous version of this.






                    share|cite|improve this answer









                    $endgroup$









                    • 2




                      $begingroup$
                      Continuous is not possible, because then f would be increasing.
                      $endgroup$
                      – user345777
                      Jan 30 at 20:29












                    • $begingroup$
                      Sorry, it was a stupid guess. Can I ask you why it would be increasing?
                      $endgroup$
                      – Marco All-in Nervo
                      Jan 30 at 20:33






                    • 1




                      $begingroup$
                      Agree with @user345777 $f$ is obviously a bijection. A continuous bijection from $Bbb{R}$ to itself is either increasing or decreasing. The 3-fold iteration of a decreasing function is itself decreasing which is absurd. So it must increasing and that case has been handled.
                      $endgroup$
                      – Jyrki Lahtonen
                      Jan 30 at 20:34
















                    2












                    $begingroup$

                    You can make $f$ act as a 3-cyclic permutation



                    $ f(x) = begin{cases}
                    0 & text{if $x = -1$ } \
                    1 &text{if $x = 0$} \
                    -1 &text{if $x = 1$} \
                    x &text{otherwise}
                    end{cases}$



                    I suppose it's also possible to make a continuous version of this.






                    share|cite|improve this answer









                    $endgroup$









                    • 2




                      $begingroup$
                      Continuous is not possible, because then f would be increasing.
                      $endgroup$
                      – user345777
                      Jan 30 at 20:29












                    • $begingroup$
                      Sorry, it was a stupid guess. Can I ask you why it would be increasing?
                      $endgroup$
                      – Marco All-in Nervo
                      Jan 30 at 20:33






                    • 1




                      $begingroup$
                      Agree with @user345777 $f$ is obviously a bijection. A continuous bijection from $Bbb{R}$ to itself is either increasing or decreasing. The 3-fold iteration of a decreasing function is itself decreasing which is absurd. So it must increasing and that case has been handled.
                      $endgroup$
                      – Jyrki Lahtonen
                      Jan 30 at 20:34














                    2












                    2








                    2





                    $begingroup$

                    You can make $f$ act as a 3-cyclic permutation



                    $ f(x) = begin{cases}
                    0 & text{if $x = -1$ } \
                    1 &text{if $x = 0$} \
                    -1 &text{if $x = 1$} \
                    x &text{otherwise}
                    end{cases}$



                    I suppose it's also possible to make a continuous version of this.






                    share|cite|improve this answer









                    $endgroup$



                    You can make $f$ act as a 3-cyclic permutation



                    $ f(x) = begin{cases}
                    0 & text{if $x = -1$ } \
                    1 &text{if $x = 0$} \
                    -1 &text{if $x = 1$} \
                    x &text{otherwise}
                    end{cases}$



                    I suppose it's also possible to make a continuous version of this.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 30 at 20:27









                    Marco All-in NervoMarco All-in Nervo

                    254210




                    254210








                    • 2




                      $begingroup$
                      Continuous is not possible, because then f would be increasing.
                      $endgroup$
                      – user345777
                      Jan 30 at 20:29












                    • $begingroup$
                      Sorry, it was a stupid guess. Can I ask you why it would be increasing?
                      $endgroup$
                      – Marco All-in Nervo
                      Jan 30 at 20:33






                    • 1




                      $begingroup$
                      Agree with @user345777 $f$ is obviously a bijection. A continuous bijection from $Bbb{R}$ to itself is either increasing or decreasing. The 3-fold iteration of a decreasing function is itself decreasing which is absurd. So it must increasing and that case has been handled.
                      $endgroup$
                      – Jyrki Lahtonen
                      Jan 30 at 20:34














                    • 2




                      $begingroup$
                      Continuous is not possible, because then f would be increasing.
                      $endgroup$
                      – user345777
                      Jan 30 at 20:29












                    • $begingroup$
                      Sorry, it was a stupid guess. Can I ask you why it would be increasing?
                      $endgroup$
                      – Marco All-in Nervo
                      Jan 30 at 20:33






                    • 1




                      $begingroup$
                      Agree with @user345777 $f$ is obviously a bijection. A continuous bijection from $Bbb{R}$ to itself is either increasing or decreasing. The 3-fold iteration of a decreasing function is itself decreasing which is absurd. So it must increasing and that case has been handled.
                      $endgroup$
                      – Jyrki Lahtonen
                      Jan 30 at 20:34








                    2




                    2




                    $begingroup$
                    Continuous is not possible, because then f would be increasing.
                    $endgroup$
                    – user345777
                    Jan 30 at 20:29






                    $begingroup$
                    Continuous is not possible, because then f would be increasing.
                    $endgroup$
                    – user345777
                    Jan 30 at 20:29














                    $begingroup$
                    Sorry, it was a stupid guess. Can I ask you why it would be increasing?
                    $endgroup$
                    – Marco All-in Nervo
                    Jan 30 at 20:33




                    $begingroup$
                    Sorry, it was a stupid guess. Can I ask you why it would be increasing?
                    $endgroup$
                    – Marco All-in Nervo
                    Jan 30 at 20:33




                    1




                    1




                    $begingroup$
                    Agree with @user345777 $f$ is obviously a bijection. A continuous bijection from $Bbb{R}$ to itself is either increasing or decreasing. The 3-fold iteration of a decreasing function is itself decreasing which is absurd. So it must increasing and that case has been handled.
                    $endgroup$
                    – Jyrki Lahtonen
                    Jan 30 at 20:34




                    $begingroup$
                    Agree with @user345777 $f$ is obviously a bijection. A continuous bijection from $Bbb{R}$ to itself is either increasing or decreasing. The 3-fold iteration of a decreasing function is itself decreasing which is absurd. So it must increasing and that case has been handled.
                    $endgroup$
                    – Jyrki Lahtonen
                    Jan 30 at 20:34











                    2












                    $begingroup$

                    If you allow yourself to consider the projective line (i.e., the reals with a point at infinity), then the function
                    $$
                    f(x) = frac{1}{1-x}
                    $$
                    is a nice continuous solution to the problem (although you have to make sense of "continuous" for this extended line). Pierre Samuel's book on Projective Geometry has a nice exposition of this around page 58.



                    (By the way, this answer was inspired by @DonaldSplutterwit's deleted answer, although I might have stumbled on it myself, since I've been thinking about projective geometry and looking at Samuel's book today anyhow. Thanks, Donald!)






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Then you could "cram $infty$ into $mathbb{R}$" using a variant of the infinite hotel paradox if you really wanted to, and then apply the same sort of construction as in Vladimir's solution. Of course, that won't result in a continuous function $mathbb{R} to mathbb{R}$.
                      $endgroup$
                      – Daniel Schepler
                      Jan 30 at 20:41
















                    2












                    $begingroup$

                    If you allow yourself to consider the projective line (i.e., the reals with a point at infinity), then the function
                    $$
                    f(x) = frac{1}{1-x}
                    $$
                    is a nice continuous solution to the problem (although you have to make sense of "continuous" for this extended line). Pierre Samuel's book on Projective Geometry has a nice exposition of this around page 58.



                    (By the way, this answer was inspired by @DonaldSplutterwit's deleted answer, although I might have stumbled on it myself, since I've been thinking about projective geometry and looking at Samuel's book today anyhow. Thanks, Donald!)






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Then you could "cram $infty$ into $mathbb{R}$" using a variant of the infinite hotel paradox if you really wanted to, and then apply the same sort of construction as in Vladimir's solution. Of course, that won't result in a continuous function $mathbb{R} to mathbb{R}$.
                      $endgroup$
                      – Daniel Schepler
                      Jan 30 at 20:41














                    2












                    2








                    2





                    $begingroup$

                    If you allow yourself to consider the projective line (i.e., the reals with a point at infinity), then the function
                    $$
                    f(x) = frac{1}{1-x}
                    $$
                    is a nice continuous solution to the problem (although you have to make sense of "continuous" for this extended line). Pierre Samuel's book on Projective Geometry has a nice exposition of this around page 58.



                    (By the way, this answer was inspired by @DonaldSplutterwit's deleted answer, although I might have stumbled on it myself, since I've been thinking about projective geometry and looking at Samuel's book today anyhow. Thanks, Donald!)






                    share|cite|improve this answer











                    $endgroup$



                    If you allow yourself to consider the projective line (i.e., the reals with a point at infinity), then the function
                    $$
                    f(x) = frac{1}{1-x}
                    $$
                    is a nice continuous solution to the problem (although you have to make sense of "continuous" for this extended line). Pierre Samuel's book on Projective Geometry has a nice exposition of this around page 58.



                    (By the way, this answer was inspired by @DonaldSplutterwit's deleted answer, although I might have stumbled on it myself, since I've been thinking about projective geometry and looking at Samuel's book today anyhow. Thanks, Donald!)







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 30 at 22:30

























                    answered Jan 30 at 20:34









                    John HughesJohn Hughes

                    65.2k24293




                    65.2k24293












                    • $begingroup$
                      Then you could "cram $infty$ into $mathbb{R}$" using a variant of the infinite hotel paradox if you really wanted to, and then apply the same sort of construction as in Vladimir's solution. Of course, that won't result in a continuous function $mathbb{R} to mathbb{R}$.
                      $endgroup$
                      – Daniel Schepler
                      Jan 30 at 20:41


















                    • $begingroup$
                      Then you could "cram $infty$ into $mathbb{R}$" using a variant of the infinite hotel paradox if you really wanted to, and then apply the same sort of construction as in Vladimir's solution. Of course, that won't result in a continuous function $mathbb{R} to mathbb{R}$.
                      $endgroup$
                      – Daniel Schepler
                      Jan 30 at 20:41
















                    $begingroup$
                    Then you could "cram $infty$ into $mathbb{R}$" using a variant of the infinite hotel paradox if you really wanted to, and then apply the same sort of construction as in Vladimir's solution. Of course, that won't result in a continuous function $mathbb{R} to mathbb{R}$.
                    $endgroup$
                    – Daniel Schepler
                    Jan 30 at 20:41




                    $begingroup$
                    Then you could "cram $infty$ into $mathbb{R}$" using a variant of the infinite hotel paradox if you really wanted to, and then apply the same sort of construction as in Vladimir's solution. Of course, that won't result in a continuous function $mathbb{R} to mathbb{R}$.
                    $endgroup$
                    – Daniel Schepler
                    Jan 30 at 20:41


















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                    in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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                    1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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