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Show a continuous function with $f(x)=y$ and $f(y)=x$ has a fixed point.


Show that any continuous $f:[0,1] rightarrow [0,1]$ has a fixed point $zeta$Fixed point and period of continuous functionContinuous decreasing function has a fixed pointIVT and fixed point theorem$f$ is continuous, $f : X to X$, $X$ compact, and $f$ has an $epsilon$-fixed point for each $epsilon > 0$. Show $f$ has a fixed point.Showing that $f$ has a fixed point.Prove that a continuous function has a fixed pointContinuous function and fixed pointProve that $f:[0,1] to [0,1]$ has a fixed pointFor a continuous function $f$ satisfying $f(f(x))=x$ has exactly one fixed point













3












$begingroup$


Suppose $a<b$ and $f:[a,b] to [a,b]$ be continous. Suppose that $x neq y$ in $[a,b]$ with $f(x)=y$ and $f(y)=x$. Prove that $f$ has a fixed point in $(x,y)$.



So I was thinking of considering the function $g(x)=f(x)-x$, which we know is continuous. Then we also know that because $f(a) geq a$ that $g(a)=f(a)-a geq 0$. Similarly, because $f(b) leq b$ then $g(b)=f(b)-b leq 0$.



Can we just use the fact that because $g(x)$ is continuous, $0 in [g(b),g(a)]$, the IVT says there exists $c in [a,b]$ such that $g(c)=f(c)-c=0$ so $f(c)=c$? Then we know $c$ is a fixed point.



How do we show that $c$ is in $(x,y)$??



We know that $g(x)=f(x)-x=y-x neq 0$
and $g(y)=f(y)-y=x-y neq 0$ but we don't know that those are in $(a,b)$?










share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Suppose $a<b$ and $f:[a,b] to [a,b]$ be continous. Suppose that $x neq y$ in $[a,b]$ with $f(x)=y$ and $f(y)=x$. Prove that $f$ has a fixed point in $(x,y)$.



    So I was thinking of considering the function $g(x)=f(x)-x$, which we know is continuous. Then we also know that because $f(a) geq a$ that $g(a)=f(a)-a geq 0$. Similarly, because $f(b) leq b$ then $g(b)=f(b)-b leq 0$.



    Can we just use the fact that because $g(x)$ is continuous, $0 in [g(b),g(a)]$, the IVT says there exists $c in [a,b]$ such that $g(c)=f(c)-c=0$ so $f(c)=c$? Then we know $c$ is a fixed point.



    How do we show that $c$ is in $(x,y)$??



    We know that $g(x)=f(x)-x=y-x neq 0$
    and $g(y)=f(y)-y=x-y neq 0$ but we don't know that those are in $(a,b)$?










    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Suppose $a<b$ and $f:[a,b] to [a,b]$ be continous. Suppose that $x neq y$ in $[a,b]$ with $f(x)=y$ and $f(y)=x$. Prove that $f$ has a fixed point in $(x,y)$.



      So I was thinking of considering the function $g(x)=f(x)-x$, which we know is continuous. Then we also know that because $f(a) geq a$ that $g(a)=f(a)-a geq 0$. Similarly, because $f(b) leq b$ then $g(b)=f(b)-b leq 0$.



      Can we just use the fact that because $g(x)$ is continuous, $0 in [g(b),g(a)]$, the IVT says there exists $c in [a,b]$ such that $g(c)=f(c)-c=0$ so $f(c)=c$? Then we know $c$ is a fixed point.



      How do we show that $c$ is in $(x,y)$??



      We know that $g(x)=f(x)-x=y-x neq 0$
      and $g(y)=f(y)-y=x-y neq 0$ but we don't know that those are in $(a,b)$?










      share|cite|improve this question











      $endgroup$




      Suppose $a<b$ and $f:[a,b] to [a,b]$ be continous. Suppose that $x neq y$ in $[a,b]$ with $f(x)=y$ and $f(y)=x$. Prove that $f$ has a fixed point in $(x,y)$.



      So I was thinking of considering the function $g(x)=f(x)-x$, which we know is continuous. Then we also know that because $f(a) geq a$ that $g(a)=f(a)-a geq 0$. Similarly, because $f(b) leq b$ then $g(b)=f(b)-b leq 0$.



      Can we just use the fact that because $g(x)$ is continuous, $0 in [g(b),g(a)]$, the IVT says there exists $c in [a,b]$ such that $g(c)=f(c)-c=0$ so $f(c)=c$? Then we know $c$ is a fixed point.



      How do we show that $c$ is in $(x,y)$??



      We know that $g(x)=f(x)-x=y-x neq 0$
      and $g(y)=f(y)-y=x-y neq 0$ but we don't know that those are in $(a,b)$?







      real-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 12 hours ago









      YuiTo Cheng

      2,3084937




      2,3084937










      asked 13 hours ago









      big_math_boybig_math_boy

      303




      303






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          You've essentially stated the argument. But, rather restrict $f$ to $[x,y]$ (where w.l.o.g $x<y$). Then, define $g:[x,y]tomathbb R$ by



          $$g(t)=f(t)-t$$



          for any $tin [x,y]$. As $f$ is continuous on $[a,b]$ and $[x,y]subseteq [a,b]$, $g$ is continuous on $[x,y]$. Also, you have $g(x)=f(x)-x=y-x>0$ and $g(y)=f(y)-y=x-y<0$ as $x<y$.



          Thus, by the intermediate value theorem, there is a $sin (x,y)$ such that $g(s)=0$, i.e. $f(s)=s$.






          share|cite|improve this answer









          $endgroup$





















            7












            $begingroup$

            Without loss of generality you can assume that $x < y$. Now consider $g(t) = f(t) - t$ not on the entire interval $[a, b]$ but only on $[x, y]$.



            Then $ g(x) = y- x$ and $g(y) = x-y$ have opposite sign, so that you can apply the intermediate value theorem.



            Note also that I have chosen a different variable name ($t$ instead of $x$) for defining $g$, in order to avoid confusion between that variable and the given (fixed) value $x$.






            share|cite|improve this answer











            $endgroup$














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              2 Answers
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              2 Answers
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              active

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              active

              oldest

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              2












              $begingroup$

              You've essentially stated the argument. But, rather restrict $f$ to $[x,y]$ (where w.l.o.g $x<y$). Then, define $g:[x,y]tomathbb R$ by



              $$g(t)=f(t)-t$$



              for any $tin [x,y]$. As $f$ is continuous on $[a,b]$ and $[x,y]subseteq [a,b]$, $g$ is continuous on $[x,y]$. Also, you have $g(x)=f(x)-x=y-x>0$ and $g(y)=f(y)-y=x-y<0$ as $x<y$.



              Thus, by the intermediate value theorem, there is a $sin (x,y)$ such that $g(s)=0$, i.e. $f(s)=s$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                You've essentially stated the argument. But, rather restrict $f$ to $[x,y]$ (where w.l.o.g $x<y$). Then, define $g:[x,y]tomathbb R$ by



                $$g(t)=f(t)-t$$



                for any $tin [x,y]$. As $f$ is continuous on $[a,b]$ and $[x,y]subseteq [a,b]$, $g$ is continuous on $[x,y]$. Also, you have $g(x)=f(x)-x=y-x>0$ and $g(y)=f(y)-y=x-y<0$ as $x<y$.



                Thus, by the intermediate value theorem, there is a $sin (x,y)$ such that $g(s)=0$, i.e. $f(s)=s$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You've essentially stated the argument. But, rather restrict $f$ to $[x,y]$ (where w.l.o.g $x<y$). Then, define $g:[x,y]tomathbb R$ by



                  $$g(t)=f(t)-t$$



                  for any $tin [x,y]$. As $f$ is continuous on $[a,b]$ and $[x,y]subseteq [a,b]$, $g$ is continuous on $[x,y]$. Also, you have $g(x)=f(x)-x=y-x>0$ and $g(y)=f(y)-y=x-y<0$ as $x<y$.



                  Thus, by the intermediate value theorem, there is a $sin (x,y)$ such that $g(s)=0$, i.e. $f(s)=s$.






                  share|cite|improve this answer









                  $endgroup$



                  You've essentially stated the argument. But, rather restrict $f$ to $[x,y]$ (where w.l.o.g $x<y$). Then, define $g:[x,y]tomathbb R$ by



                  $$g(t)=f(t)-t$$



                  for any $tin [x,y]$. As $f$ is continuous on $[a,b]$ and $[x,y]subseteq [a,b]$, $g$ is continuous on $[x,y]$. Also, you have $g(x)=f(x)-x=y-x>0$ and $g(y)=f(y)-y=x-y<0$ as $x<y$.



                  Thus, by the intermediate value theorem, there is a $sin (x,y)$ such that $g(s)=0$, i.e. $f(s)=s$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 13 hours ago









                  blubblub

                  3,241829




                  3,241829























                      7












                      $begingroup$

                      Without loss of generality you can assume that $x < y$. Now consider $g(t) = f(t) - t$ not on the entire interval $[a, b]$ but only on $[x, y]$.



                      Then $ g(x) = y- x$ and $g(y) = x-y$ have opposite sign, so that you can apply the intermediate value theorem.



                      Note also that I have chosen a different variable name ($t$ instead of $x$) for defining $g$, in order to avoid confusion between that variable and the given (fixed) value $x$.






                      share|cite|improve this answer











                      $endgroup$


















                        7












                        $begingroup$

                        Without loss of generality you can assume that $x < y$. Now consider $g(t) = f(t) - t$ not on the entire interval $[a, b]$ but only on $[x, y]$.



                        Then $ g(x) = y- x$ and $g(y) = x-y$ have opposite sign, so that you can apply the intermediate value theorem.



                        Note also that I have chosen a different variable name ($t$ instead of $x$) for defining $g$, in order to avoid confusion between that variable and the given (fixed) value $x$.






                        share|cite|improve this answer











                        $endgroup$
















                          7












                          7








                          7





                          $begingroup$

                          Without loss of generality you can assume that $x < y$. Now consider $g(t) = f(t) - t$ not on the entire interval $[a, b]$ but only on $[x, y]$.



                          Then $ g(x) = y- x$ and $g(y) = x-y$ have opposite sign, so that you can apply the intermediate value theorem.



                          Note also that I have chosen a different variable name ($t$ instead of $x$) for defining $g$, in order to avoid confusion between that variable and the given (fixed) value $x$.






                          share|cite|improve this answer











                          $endgroup$



                          Without loss of generality you can assume that $x < y$. Now consider $g(t) = f(t) - t$ not on the entire interval $[a, b]$ but only on $[x, y]$.



                          Then $ g(x) = y- x$ and $g(y) = x-y$ have opposite sign, so that you can apply the intermediate value theorem.



                          Note also that I have chosen a different variable name ($t$ instead of $x$) for defining $g$, in order to avoid confusion between that variable and the given (fixed) value $x$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 13 hours ago

























                          answered 13 hours ago









                          Martin RMartin R

                          30.8k33560




                          30.8k33560






























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