is this set invariant under a operator?












0












$begingroup$


Let $E$ be a Banach space, and $T:Erightarrow E$ a continuous bounded mapping.



Let $x_0in E$ and $x_n=T(x_{n-1})$, $U=overline{conv}(x_0,x_1,...,x_n,...)$.



Is $U$ invariant under the operator $T$, i.e $T(U)subset U$?



Edit:
We donotes by $overline{conv}(M)$ the closure of the convex hull of $M$.






real-analysis banach-spaces fixed-point-theorems







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $U$? ${}{}$
    $endgroup$
    – Will M.
    Dec 18 '18 at 17:37










  • $begingroup$
    @Will M. I edited my post, thank you
    $endgroup$
    – Motaka
    Dec 19 '18 at 9:19






  • 2




    $begingroup$
    This is false in general. If you assume that $T$ is affine, then it's true.
    $endgroup$
    – Song
    Dec 19 '18 at 12:50
















0












$begingroup$


Let $E$ be a Banach space, and $T:Erightarrow E$ a continuous bounded mapping.



Let $x_0in E$ and $x_n=T(x_{n-1})$, $U=overline{conv}(x_0,x_1,...,x_n,...)$.



Is $U$ invariant under the operator $T$, i.e $T(U)subset U$?



Edit:
We donotes by $overline{conv}(M)$ the closure of the convex hull of $M$.






real-analysis banach-spaces fixed-point-theorems







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $U$? ${}{}$
    $endgroup$
    – Will M.
    Dec 18 '18 at 17:37










  • $begingroup$
    @Will M. I edited my post, thank you
    $endgroup$
    – Motaka
    Dec 19 '18 at 9:19






  • 2




    $begingroup$
    This is false in general. If you assume that $T$ is affine, then it's true.
    $endgroup$
    – Song
    Dec 19 '18 at 12:50














0












0








0





$begingroup$


Let $E$ be a Banach space, and $T:Erightarrow E$ a continuous bounded mapping.



Let $x_0in E$ and $x_n=T(x_{n-1})$, $U=overline{conv}(x_0,x_1,...,x_n,...)$.



Is $U$ invariant under the operator $T$, i.e $T(U)subset U$?



Edit:
We donotes by $overline{conv}(M)$ the closure of the convex hull of $M$.






real-analysis banach-spaces fixed-point-theorems







share|cite|improve this question











$endgroup$




Let $E$ be a Banach space, and $T:Erightarrow E$ a continuous bounded mapping.



Let $x_0in E$ and $x_n=T(x_{n-1})$, $U=overline{conv}(x_0,x_1,...,x_n,...)$.



Is $U$ invariant under the operator $T$, i.e $T(U)subset U$?



Edit:
We donotes by $overline{conv}(M)$ the closure of the convex hull of $M$.






real-analysis banach-spaces fixed-point-theorems




real-analysis banach-spaces 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 19 '18 at 12:29







Motaka

















asked Dec 17 '18 at 16:08









MotakaMotaka

238111




238111












  • $begingroup$
    What is $U$? ${}{}$
    $endgroup$
    – Will M.
    Dec 18 '18 at 17:37










  • $begingroup$
    @Will M. I edited my post, thank you
    $endgroup$
    – Motaka
    Dec 19 '18 at 9:19






  • 2




    $begingroup$
    This is false in general. If you assume that $T$ is affine, then it's true.
    $endgroup$
    – Song
    Dec 19 '18 at 12:50


















  • $begingroup$
    What is $U$? ${}{}$
    $endgroup$
    – Will M.
    Dec 18 '18 at 17:37










  • $begingroup$
    @Will M. I edited my post, thank you
    $endgroup$
    – Motaka
    Dec 19 '18 at 9:19






  • 2




    $begingroup$
    This is false in general. If you assume that $T$ is affine, then it's true.
    $endgroup$
    – Song
    Dec 19 '18 at 12:50
















$begingroup$
What is $U$? ${}{}$
$endgroup$
– Will M.
Dec 18 '18 at 17:37




$begingroup$
What is $U$? ${}{}$
$endgroup$
– Will M.
Dec 18 '18 at 17:37












$begingroup$
@Will M. I edited my post, thank you
$endgroup$
– Motaka
Dec 19 '18 at 9:19




$begingroup$
@Will M. I edited my post, thank you
$endgroup$
– Motaka
Dec 19 '18 at 9:19




2




2




$begingroup$
This is false in general. If you assume that $T$ is affine, then it's true.
$endgroup$
– Song
Dec 19 '18 at 12:50




$begingroup$
This is false in general. If you assume that $T$ is affine, then it's true.
$endgroup$
– Song
Dec 19 '18 at 12:50










1 Answer
1






active

oldest

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1












$begingroup$

Take $E = mathbb{R}$, $T(x) = 1 + sin(pi x)$ and $x_0=0$.



$x_1 = T(0) = 1$, and $forall n>0, x_n = T(x_{n-1}) = T(1) = 1$, so :



$$U=overline{conv}(x_0,x_1,...,x_n,...) = overline{conv}(0,1,...,1,...) = [0, 1]$$
But :
$$T(U) = T([0,1]) = [1,2]$$






share|cite|improve this answer











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    1












    $begingroup$

    Take $E = mathbb{R}$, $T(x) = 1 + sin(pi x)$ and $x_0=0$.



    $x_1 = T(0) = 1$, and $forall n>0, x_n = T(x_{n-1}) = T(1) = 1$, so :



    $$U=overline{conv}(x_0,x_1,...,x_n,...) = overline{conv}(0,1,...,1,...) = [0, 1]$$
    But :
    $$T(U) = T([0,1]) = [1,2]$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Take $E = mathbb{R}$, $T(x) = 1 + sin(pi x)$ and $x_0=0$.



      $x_1 = T(0) = 1$, and $forall n>0, x_n = T(x_{n-1}) = T(1) = 1$, so :



      $$U=overline{conv}(x_0,x_1,...,x_n,...) = overline{conv}(0,1,...,1,...) = [0, 1]$$
      But :
      $$T(U) = T([0,1]) = [1,2]$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Take $E = mathbb{R}$, $T(x) = 1 + sin(pi x)$ and $x_0=0$.



        $x_1 = T(0) = 1$, and $forall n>0, x_n = T(x_{n-1}) = T(1) = 1$, so :



        $$U=overline{conv}(x_0,x_1,...,x_n,...) = overline{conv}(0,1,...,1,...) = [0, 1]$$
        But :
        $$T(U) = T([0,1]) = [1,2]$$






        share|cite|improve this answer











        $endgroup$



        Take $E = mathbb{R}$, $T(x) = 1 + sin(pi x)$ and $x_0=0$.



        $x_1 = T(0) = 1$, and $forall n>0, x_n = T(x_{n-1}) = T(1) = 1$, so :



        $$U=overline{conv}(x_0,x_1,...,x_n,...) = overline{conv}(0,1,...,1,...) = [0, 1]$$
        But :
        $$T(U) = T([0,1]) = [1,2]$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 19 '18 at 12:47

























        answered Dec 19 '18 at 12:41









        RcnScRcnSc

        1265




        1265






























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