Question about continuous function.

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Question. Which of the following statements are true:



  1. If $f in C[0,2]$ is such that $f(0)=f(2)$, then there exist $x_1$ and $x_2$ in $[0,2]$ such that $x_1-x_2=1$ and $f(x_1)=f(x_2)$.


  2. Let $f$ and $g$ be continuous real valued functions on $mathbbR$ such that for all $xin mathbbR$, we have $f(g(x))=g(f(x))$. If there exists $x_0 in mathbbR$ such that $f(f(x_0))=g(g(x_0))$, then there exists $x_1in mathbbR$ such that $f(x_1)=g(x_1)$.




My Attempts.



  1. Here $f(0)=f(2)$. If $f(1) ne f(0)$, say, $f(1)>f(0)$ then for any $k$ such that $f(0)<k<f(1)$ there exists a $x_1,x_2$ with $0<x_1<1$ and $1<x_2<2$ such that $f(x_1)=f(x_2)=k$ (by Intermediate Value Property of $f$). But how can I prove $x_2-x_1=1$? Please help.


  2. I don't have any guess here to start...










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    up vote
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    Question. Which of the following statements are true:



    1. If $f in C[0,2]$ is such that $f(0)=f(2)$, then there exist $x_1$ and $x_2$ in $[0,2]$ such that $x_1-x_2=1$ and $f(x_1)=f(x_2)$.


    2. Let $f$ and $g$ be continuous real valued functions on $mathbbR$ such that for all $xin mathbbR$, we have $f(g(x))=g(f(x))$. If there exists $x_0 in mathbbR$ such that $f(f(x_0))=g(g(x_0))$, then there exists $x_1in mathbbR$ such that $f(x_1)=g(x_1)$.




    My Attempts.



    1. Here $f(0)=f(2)$. If $f(1) ne f(0)$, say, $f(1)>f(0)$ then for any $k$ such that $f(0)<k<f(1)$ there exists a $x_1,x_2$ with $0<x_1<1$ and $1<x_2<2$ such that $f(x_1)=f(x_2)=k$ (by Intermediate Value Property of $f$). But how can I prove $x_2-x_1=1$? Please help.


    2. I don't have any guess here to start...










    share|cite|improve this question























      up vote
      5
      down vote

      favorite
      2









      up vote
      5
      down vote

      favorite
      2






      2






      Question. Which of the following statements are true:



      1. If $f in C[0,2]$ is such that $f(0)=f(2)$, then there exist $x_1$ and $x_2$ in $[0,2]$ such that $x_1-x_2=1$ and $f(x_1)=f(x_2)$.


      2. Let $f$ and $g$ be continuous real valued functions on $mathbbR$ such that for all $xin mathbbR$, we have $f(g(x))=g(f(x))$. If there exists $x_0 in mathbbR$ such that $f(f(x_0))=g(g(x_0))$, then there exists $x_1in mathbbR$ such that $f(x_1)=g(x_1)$.




      My Attempts.



      1. Here $f(0)=f(2)$. If $f(1) ne f(0)$, say, $f(1)>f(0)$ then for any $k$ such that $f(0)<k<f(1)$ there exists a $x_1,x_2$ with $0<x_1<1$ and $1<x_2<2$ such that $f(x_1)=f(x_2)=k$ (by Intermediate Value Property of $f$). But how can I prove $x_2-x_1=1$? Please help.


      2. I don't have any guess here to start...










      share|cite|improve this question














      Question. Which of the following statements are true:



      1. If $f in C[0,2]$ is such that $f(0)=f(2)$, then there exist $x_1$ and $x_2$ in $[0,2]$ such that $x_1-x_2=1$ and $f(x_1)=f(x_2)$.


      2. Let $f$ and $g$ be continuous real valued functions on $mathbbR$ such that for all $xin mathbbR$, we have $f(g(x))=g(f(x))$. If there exists $x_0 in mathbbR$ such that $f(f(x_0))=g(g(x_0))$, then there exists $x_1in mathbbR$ such that $f(x_1)=g(x_1)$.




      My Attempts.



      1. Here $f(0)=f(2)$. If $f(1) ne f(0)$, say, $f(1)>f(0)$ then for any $k$ such that $f(0)<k<f(1)$ there exists a $x_1,x_2$ with $0<x_1<1$ and $1<x_2<2$ such that $f(x_1)=f(x_2)=k$ (by Intermediate Value Property of $f$). But how can I prove $x_2-x_1=1$? Please help.


      2. I don't have any guess here to start...







      real-analysis continuity






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      asked Aug 21 at 11:25









      Indrajit Ghosh

      879517




      879517




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          9
          down vote



          accepted










          For 1) you can consider :



          $$ g(x) = f(x+1) - f(x)$$



          $g$ is continuous and verify :



          $$ g(1) = f(2) - f(1) = f(0) - f(1) = -(f(1) - f(0)) =-g(0)$$



          Then $g$ sign is changing and by the intermediate value theorem you can such $x_1$ and $x_2$. Using the same idea for 2) you'll have the result.



          For 2), pose $h(x) = f(x) - g(x)$. Using the hypothesis :



          beginalign*
          h(f(x_0)) &= f(f(x_0)) - g(f(x_0)) \
          &=g(g(x_0)) - f(g(x_0)) \
          &= -h(g(x_0))
          endalign*



          So it means that $h$ sign is changing and since $h$ is continuous, the intermediate value theorem gives you that you can find $x_1$ such as $h(x_1) = f(x_1) - g(x_1) =0$.






          share|cite|improve this answer


















          • 4




            Anyway, it's a nice exercise, and a great answer. OP should accept it as the best answer.
            – Nicolas FRANCOIS
            Aug 21 at 12:34










          • @NicolasFRANCOIS It is very nice of you to say that!
            – amsmath
            Aug 21 at 12:38










          • Q1 is the case n=2 of the Horizontal Chord Theorwm: If $a<b$ and $f:[a,b]to Bbb R$ is continuous with $f(a)=f(b)$ then for any $nin Bbb N$ there exists $u,vin [a,b]$ with $|u-v|=(b-a)/n$ and $f(u)=f(v)$. Proof: For $xin [a, b-(b-a)/n]$ let $g(x)=f(x)-f(x+(b-a)/n).$ If $g$ were always positive or always negative then $0ne$ $ sum_j=0^n-1 g(a+j(b-a)/n)=$ $f(a)-f(b)=0.$
            – DanielWainfleet
            Aug 21 at 18:58

















          up vote
          5
          down vote













          As always in this kind of problems, to use intermediate values theorem, you have to convert your "equation" in the form $g(x)=0$. Here you have $f(x+1)=f(x)$, so let $g:xmapsto f(x+1)-f(x)$ be defined on interval $[0,1]$.



          You have $g(0)=f(1)-f(0)$, $g(1)=f(2)-f(1)$, so $g(0)+g(1)=f(2)-f(0)=0$.



          Either $g(0)$ and $g(1)$ are both $0$ (in which case you have found TWO solutions to your problem), or they are non null opposite. As $g$ is continuous, you can apply IVT to $g$ to prove existence of a solution.



          A note : with a little bit of work, you can generalize the result : for every fraction $frac2n$ of the complete interval, you can find two points distant of this fraction where $f$ takes the same value.



          For example, if you run $10$ miles in an hour, there is a quarter of an hour where you actually ran $2.5$ miles.






          share|cite|improve this answer




















          • @Nicola...can you please tell me the general statement more precisely in your note...I don't get that general one...
            – Indrajit Ghosh
            Aug 21 at 11:54







          • 1




            This is what I mean : for any positive integer $n$, there exists two points $x_1^(n)$ and $x_2^(n)$ such that $x_2^(n)-x_1^(n)=frac1n$ and $fleft(x_1^(n)right)=fleft(x_2^(n)right)$ (it's a result from Paul Levy).
            – Nicolas FRANCOIS
            Aug 21 at 11:57











          • @Nicolas..What is $f$..? Is $fin C[0,2]$?
            – Indrajit Ghosh
            Aug 21 at 12:00










          • Yes, as in your initial post.
            – Nicolas FRANCOIS
            Aug 21 at 12:00










          • Does $[0,2]$ is necessary...I mean what about any $[a,b]$ ?
            – Indrajit Ghosh
            Aug 21 at 12:01










          Your Answer




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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          9
          down vote



          accepted










          For 1) you can consider :



          $$ g(x) = f(x+1) - f(x)$$



          $g$ is continuous and verify :



          $$ g(1) = f(2) - f(1) = f(0) - f(1) = -(f(1) - f(0)) =-g(0)$$



          Then $g$ sign is changing and by the intermediate value theorem you can such $x_1$ and $x_2$. Using the same idea for 2) you'll have the result.



          For 2), pose $h(x) = f(x) - g(x)$. Using the hypothesis :



          beginalign*
          h(f(x_0)) &= f(f(x_0)) - g(f(x_0)) \
          &=g(g(x_0)) - f(g(x_0)) \
          &= -h(g(x_0))
          endalign*



          So it means that $h$ sign is changing and since $h$ is continuous, the intermediate value theorem gives you that you can find $x_1$ such as $h(x_1) = f(x_1) - g(x_1) =0$.






          share|cite|improve this answer


















          • 4




            Anyway, it's a nice exercise, and a great answer. OP should accept it as the best answer.
            – Nicolas FRANCOIS
            Aug 21 at 12:34










          • @NicolasFRANCOIS It is very nice of you to say that!
            – amsmath
            Aug 21 at 12:38










          • Q1 is the case n=2 of the Horizontal Chord Theorwm: If $a<b$ and $f:[a,b]to Bbb R$ is continuous with $f(a)=f(b)$ then for any $nin Bbb N$ there exists $u,vin [a,b]$ with $|u-v|=(b-a)/n$ and $f(u)=f(v)$. Proof: For $xin [a, b-(b-a)/n]$ let $g(x)=f(x)-f(x+(b-a)/n).$ If $g$ were always positive or always negative then $0ne$ $ sum_j=0^n-1 g(a+j(b-a)/n)=$ $f(a)-f(b)=0.$
            – DanielWainfleet
            Aug 21 at 18:58














          up vote
          9
          down vote



          accepted










          For 1) you can consider :



          $$ g(x) = f(x+1) - f(x)$$



          $g$ is continuous and verify :



          $$ g(1) = f(2) - f(1) = f(0) - f(1) = -(f(1) - f(0)) =-g(0)$$



          Then $g$ sign is changing and by the intermediate value theorem you can such $x_1$ and $x_2$. Using the same idea for 2) you'll have the result.



          For 2), pose $h(x) = f(x) - g(x)$. Using the hypothesis :



          beginalign*
          h(f(x_0)) &= f(f(x_0)) - g(f(x_0)) \
          &=g(g(x_0)) - f(g(x_0)) \
          &= -h(g(x_0))
          endalign*



          So it means that $h$ sign is changing and since $h$ is continuous, the intermediate value theorem gives you that you can find $x_1$ such as $h(x_1) = f(x_1) - g(x_1) =0$.






          share|cite|improve this answer


















          • 4




            Anyway, it's a nice exercise, and a great answer. OP should accept it as the best answer.
            – Nicolas FRANCOIS
            Aug 21 at 12:34










          • @NicolasFRANCOIS It is very nice of you to say that!
            – amsmath
            Aug 21 at 12:38










          • Q1 is the case n=2 of the Horizontal Chord Theorwm: If $a<b$ and $f:[a,b]to Bbb R$ is continuous with $f(a)=f(b)$ then for any $nin Bbb N$ there exists $u,vin [a,b]$ with $|u-v|=(b-a)/n$ and $f(u)=f(v)$. Proof: For $xin [a, b-(b-a)/n]$ let $g(x)=f(x)-f(x+(b-a)/n).$ If $g$ were always positive or always negative then $0ne$ $ sum_j=0^n-1 g(a+j(b-a)/n)=$ $f(a)-f(b)=0.$
            – DanielWainfleet
            Aug 21 at 18:58












          up vote
          9
          down vote



          accepted







          up vote
          9
          down vote



          accepted






          For 1) you can consider :



          $$ g(x) = f(x+1) - f(x)$$



          $g$ is continuous and verify :



          $$ g(1) = f(2) - f(1) = f(0) - f(1) = -(f(1) - f(0)) =-g(0)$$



          Then $g$ sign is changing and by the intermediate value theorem you can such $x_1$ and $x_2$. Using the same idea for 2) you'll have the result.



          For 2), pose $h(x) = f(x) - g(x)$. Using the hypothesis :



          beginalign*
          h(f(x_0)) &= f(f(x_0)) - g(f(x_0)) \
          &=g(g(x_0)) - f(g(x_0)) \
          &= -h(g(x_0))
          endalign*



          So it means that $h$ sign is changing and since $h$ is continuous, the intermediate value theorem gives you that you can find $x_1$ such as $h(x_1) = f(x_1) - g(x_1) =0$.






          share|cite|improve this answer














          For 1) you can consider :



          $$ g(x) = f(x+1) - f(x)$$



          $g$ is continuous and verify :



          $$ g(1) = f(2) - f(1) = f(0) - f(1) = -(f(1) - f(0)) =-g(0)$$



          Then $g$ sign is changing and by the intermediate value theorem you can such $x_1$ and $x_2$. Using the same idea for 2) you'll have the result.



          For 2), pose $h(x) = f(x) - g(x)$. Using the hypothesis :



          beginalign*
          h(f(x_0)) &= f(f(x_0)) - g(f(x_0)) \
          &=g(g(x_0)) - f(g(x_0)) \
          &= -h(g(x_0))
          endalign*



          So it means that $h$ sign is changing and since $h$ is continuous, the intermediate value theorem gives you that you can find $x_1$ such as $h(x_1) = f(x_1) - g(x_1) =0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 21 at 12:04

























          answered Aug 21 at 11:33









          tmaths

          1,334113




          1,334113







          • 4




            Anyway, it's a nice exercise, and a great answer. OP should accept it as the best answer.
            – Nicolas FRANCOIS
            Aug 21 at 12:34










          • @NicolasFRANCOIS It is very nice of you to say that!
            – amsmath
            Aug 21 at 12:38










          • Q1 is the case n=2 of the Horizontal Chord Theorwm: If $a<b$ and $f:[a,b]to Bbb R$ is continuous with $f(a)=f(b)$ then for any $nin Bbb N$ there exists $u,vin [a,b]$ with $|u-v|=(b-a)/n$ and $f(u)=f(v)$. Proof: For $xin [a, b-(b-a)/n]$ let $g(x)=f(x)-f(x+(b-a)/n).$ If $g$ were always positive or always negative then $0ne$ $ sum_j=0^n-1 g(a+j(b-a)/n)=$ $f(a)-f(b)=0.$
            – DanielWainfleet
            Aug 21 at 18:58












          • 4




            Anyway, it's a nice exercise, and a great answer. OP should accept it as the best answer.
            – Nicolas FRANCOIS
            Aug 21 at 12:34










          • @NicolasFRANCOIS It is very nice of you to say that!
            – amsmath
            Aug 21 at 12:38










          • Q1 is the case n=2 of the Horizontal Chord Theorwm: If $a<b$ and $f:[a,b]to Bbb R$ is continuous with $f(a)=f(b)$ then for any $nin Bbb N$ there exists $u,vin [a,b]$ with $|u-v|=(b-a)/n$ and $f(u)=f(v)$. Proof: For $xin [a, b-(b-a)/n]$ let $g(x)=f(x)-f(x+(b-a)/n).$ If $g$ were always positive or always negative then $0ne$ $ sum_j=0^n-1 g(a+j(b-a)/n)=$ $f(a)-f(b)=0.$
            – DanielWainfleet
            Aug 21 at 18:58







          4




          4




          Anyway, it's a nice exercise, and a great answer. OP should accept it as the best answer.
          – Nicolas FRANCOIS
          Aug 21 at 12:34




          Anyway, it's a nice exercise, and a great answer. OP should accept it as the best answer.
          – Nicolas FRANCOIS
          Aug 21 at 12:34












          @NicolasFRANCOIS It is very nice of you to say that!
          – amsmath
          Aug 21 at 12:38




          @NicolasFRANCOIS It is very nice of you to say that!
          – amsmath
          Aug 21 at 12:38












          Q1 is the case n=2 of the Horizontal Chord Theorwm: If $a<b$ and $f:[a,b]to Bbb R$ is continuous with $f(a)=f(b)$ then for any $nin Bbb N$ there exists $u,vin [a,b]$ with $|u-v|=(b-a)/n$ and $f(u)=f(v)$. Proof: For $xin [a, b-(b-a)/n]$ let $g(x)=f(x)-f(x+(b-a)/n).$ If $g$ were always positive or always negative then $0ne$ $ sum_j=0^n-1 g(a+j(b-a)/n)=$ $f(a)-f(b)=0.$
          – DanielWainfleet
          Aug 21 at 18:58




          Q1 is the case n=2 of the Horizontal Chord Theorwm: If $a<b$ and $f:[a,b]to Bbb R$ is continuous with $f(a)=f(b)$ then for any $nin Bbb N$ there exists $u,vin [a,b]$ with $|u-v|=(b-a)/n$ and $f(u)=f(v)$. Proof: For $xin [a, b-(b-a)/n]$ let $g(x)=f(x)-f(x+(b-a)/n).$ If $g$ were always positive or always negative then $0ne$ $ sum_j=0^n-1 g(a+j(b-a)/n)=$ $f(a)-f(b)=0.$
          – DanielWainfleet
          Aug 21 at 18:58










          up vote
          5
          down vote













          As always in this kind of problems, to use intermediate values theorem, you have to convert your "equation" in the form $g(x)=0$. Here you have $f(x+1)=f(x)$, so let $g:xmapsto f(x+1)-f(x)$ be defined on interval $[0,1]$.



          You have $g(0)=f(1)-f(0)$, $g(1)=f(2)-f(1)$, so $g(0)+g(1)=f(2)-f(0)=0$.



          Either $g(0)$ and $g(1)$ are both $0$ (in which case you have found TWO solutions to your problem), or they are non null opposite. As $g$ is continuous, you can apply IVT to $g$ to prove existence of a solution.



          A note : with a little bit of work, you can generalize the result : for every fraction $frac2n$ of the complete interval, you can find two points distant of this fraction where $f$ takes the same value.



          For example, if you run $10$ miles in an hour, there is a quarter of an hour where you actually ran $2.5$ miles.






          share|cite|improve this answer




















          • @Nicola...can you please tell me the general statement more precisely in your note...I don't get that general one...
            – Indrajit Ghosh
            Aug 21 at 11:54







          • 1




            This is what I mean : for any positive integer $n$, there exists two points $x_1^(n)$ and $x_2^(n)$ such that $x_2^(n)-x_1^(n)=frac1n$ and $fleft(x_1^(n)right)=fleft(x_2^(n)right)$ (it's a result from Paul Levy).
            – Nicolas FRANCOIS
            Aug 21 at 11:57











          • @Nicolas..What is $f$..? Is $fin C[0,2]$?
            – Indrajit Ghosh
            Aug 21 at 12:00










          • Yes, as in your initial post.
            – Nicolas FRANCOIS
            Aug 21 at 12:00










          • Does $[0,2]$ is necessary...I mean what about any $[a,b]$ ?
            – Indrajit Ghosh
            Aug 21 at 12:01














          up vote
          5
          down vote













          As always in this kind of problems, to use intermediate values theorem, you have to convert your "equation" in the form $g(x)=0$. Here you have $f(x+1)=f(x)$, so let $g:xmapsto f(x+1)-f(x)$ be defined on interval $[0,1]$.



          You have $g(0)=f(1)-f(0)$, $g(1)=f(2)-f(1)$, so $g(0)+g(1)=f(2)-f(0)=0$.



          Either $g(0)$ and $g(1)$ are both $0$ (in which case you have found TWO solutions to your problem), or they are non null opposite. As $g$ is continuous, you can apply IVT to $g$ to prove existence of a solution.



          A note : with a little bit of work, you can generalize the result : for every fraction $frac2n$ of the complete interval, you can find two points distant of this fraction where $f$ takes the same value.



          For example, if you run $10$ miles in an hour, there is a quarter of an hour where you actually ran $2.5$ miles.






          share|cite|improve this answer




















          • @Nicola...can you please tell me the general statement more precisely in your note...I don't get that general one...
            – Indrajit Ghosh
            Aug 21 at 11:54







          • 1




            This is what I mean : for any positive integer $n$, there exists two points $x_1^(n)$ and $x_2^(n)$ such that $x_2^(n)-x_1^(n)=frac1n$ and $fleft(x_1^(n)right)=fleft(x_2^(n)right)$ (it's a result from Paul Levy).
            – Nicolas FRANCOIS
            Aug 21 at 11:57











          • @Nicolas..What is $f$..? Is $fin C[0,2]$?
            – Indrajit Ghosh
            Aug 21 at 12:00










          • Yes, as in your initial post.
            – Nicolas FRANCOIS
            Aug 21 at 12:00










          • Does $[0,2]$ is necessary...I mean what about any $[a,b]$ ?
            – Indrajit Ghosh
            Aug 21 at 12:01












          up vote
          5
          down vote










          up vote
          5
          down vote









          As always in this kind of problems, to use intermediate values theorem, you have to convert your "equation" in the form $g(x)=0$. Here you have $f(x+1)=f(x)$, so let $g:xmapsto f(x+1)-f(x)$ be defined on interval $[0,1]$.



          You have $g(0)=f(1)-f(0)$, $g(1)=f(2)-f(1)$, so $g(0)+g(1)=f(2)-f(0)=0$.



          Either $g(0)$ and $g(1)$ are both $0$ (in which case you have found TWO solutions to your problem), or they are non null opposite. As $g$ is continuous, you can apply IVT to $g$ to prove existence of a solution.



          A note : with a little bit of work, you can generalize the result : for every fraction $frac2n$ of the complete interval, you can find two points distant of this fraction where $f$ takes the same value.



          For example, if you run $10$ miles in an hour, there is a quarter of an hour where you actually ran $2.5$ miles.






          share|cite|improve this answer












          As always in this kind of problems, to use intermediate values theorem, you have to convert your "equation" in the form $g(x)=0$. Here you have $f(x+1)=f(x)$, so let $g:xmapsto f(x+1)-f(x)$ be defined on interval $[0,1]$.



          You have $g(0)=f(1)-f(0)$, $g(1)=f(2)-f(1)$, so $g(0)+g(1)=f(2)-f(0)=0$.



          Either $g(0)$ and $g(1)$ are both $0$ (in which case you have found TWO solutions to your problem), or they are non null opposite. As $g$ is continuous, you can apply IVT to $g$ to prove existence of a solution.



          A note : with a little bit of work, you can generalize the result : for every fraction $frac2n$ of the complete interval, you can find two points distant of this fraction where $f$ takes the same value.



          For example, if you run $10$ miles in an hour, there is a quarter of an hour where you actually ran $2.5$ miles.







          share|cite|improve this answer












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          share|cite|improve this answer










          answered Aug 21 at 11:32









          Nicolas FRANCOIS

          3,3221415




          3,3221415











          • @Nicola...can you please tell me the general statement more precisely in your note...I don't get that general one...
            – Indrajit Ghosh
            Aug 21 at 11:54







          • 1




            This is what I mean : for any positive integer $n$, there exists two points $x_1^(n)$ and $x_2^(n)$ such that $x_2^(n)-x_1^(n)=frac1n$ and $fleft(x_1^(n)right)=fleft(x_2^(n)right)$ (it's a result from Paul Levy).
            – Nicolas FRANCOIS
            Aug 21 at 11:57











          • @Nicolas..What is $f$..? Is $fin C[0,2]$?
            – Indrajit Ghosh
            Aug 21 at 12:00










          • Yes, as in your initial post.
            – Nicolas FRANCOIS
            Aug 21 at 12:00










          • Does $[0,2]$ is necessary...I mean what about any $[a,b]$ ?
            – Indrajit Ghosh
            Aug 21 at 12:01
















          • @Nicola...can you please tell me the general statement more precisely in your note...I don't get that general one...
            – Indrajit Ghosh
            Aug 21 at 11:54







          • 1




            This is what I mean : for any positive integer $n$, there exists two points $x_1^(n)$ and $x_2^(n)$ such that $x_2^(n)-x_1^(n)=frac1n$ and $fleft(x_1^(n)right)=fleft(x_2^(n)right)$ (it's a result from Paul Levy).
            – Nicolas FRANCOIS
            Aug 21 at 11:57











          • @Nicolas..What is $f$..? Is $fin C[0,2]$?
            – Indrajit Ghosh
            Aug 21 at 12:00










          • Yes, as in your initial post.
            – Nicolas FRANCOIS
            Aug 21 at 12:00










          • Does $[0,2]$ is necessary...I mean what about any $[a,b]$ ?
            – Indrajit Ghosh
            Aug 21 at 12:01















          @Nicola...can you please tell me the general statement more precisely in your note...I don't get that general one...
          – Indrajit Ghosh
          Aug 21 at 11:54





          @Nicola...can you please tell me the general statement more precisely in your note...I don't get that general one...
          – Indrajit Ghosh
          Aug 21 at 11:54





          1




          1




          This is what I mean : for any positive integer $n$, there exists two points $x_1^(n)$ and $x_2^(n)$ such that $x_2^(n)-x_1^(n)=frac1n$ and $fleft(x_1^(n)right)=fleft(x_2^(n)right)$ (it's a result from Paul Levy).
          – Nicolas FRANCOIS
          Aug 21 at 11:57





          This is what I mean : for any positive integer $n$, there exists two points $x_1^(n)$ and $x_2^(n)$ such that $x_2^(n)-x_1^(n)=frac1n$ and $fleft(x_1^(n)right)=fleft(x_2^(n)right)$ (it's a result from Paul Levy).
          – Nicolas FRANCOIS
          Aug 21 at 11:57













          @Nicolas..What is $f$..? Is $fin C[0,2]$?
          – Indrajit Ghosh
          Aug 21 at 12:00




          @Nicolas..What is $f$..? Is $fin C[0,2]$?
          – Indrajit Ghosh
          Aug 21 at 12:00












          Yes, as in your initial post.
          – Nicolas FRANCOIS
          Aug 21 at 12:00




          Yes, as in your initial post.
          – Nicolas FRANCOIS
          Aug 21 at 12:00












          Does $[0,2]$ is necessary...I mean what about any $[a,b]$ ?
          – Indrajit Ghosh
          Aug 21 at 12:01




          Does $[0,2]$ is necessary...I mean what about any $[a,b]$ ?
          – Indrajit Ghosh
          Aug 21 at 12:01

















           

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