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

Let $x_0\in E$ 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 $M$.

Motaka
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1 Answers1

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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 : $$T(U) = T([0,1]) = [1,2]$$

RcnSc
  • 136