Is $\overline {conv}(T(M))\subset T(\overline {conv}(M))$?

Question

Let $(E,\left \| . \right \| )$ be a Banach space $T:E\rightarrow E$ is a continuous and bounded mapping.

Let $x_0\in E$, we define a sequence $(x_n)_{n\in\mathbb N}$ as follows: $$x_{n+1}=T(x_n) \text{ for }n=0,1,... \:\:$$

Let $M_n=\overline {conv}(\{x_n,x_{n+1},...\})$ for each $n\in \mathbb N$.

-Did we have: $\overline {conv}(T(M_n))\subset T(\overline {conv}(M_n))$?

-If not do you have any counterexample in mind?

Thank you in advance.

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$\overline{conv}(A)$ is the closure of the convex hull of $A$ .

Answer 1

If $||T||<1$ then the sequence $x_n $ tends to $0$ since $$||x_n ||\leq ||T||^{n -1} ||x_1||$$ hence the set $M_n $ is compact and convex and therefore $\overline{\mbox{conv} }(M_n )=M_n .$ The set $T(M_n )=T(\overline{\mbox{conv} }(M_n )) $ is closed and convex since it is compact and convex therefore $$\overline{\mbox{conv} } T(M_n ) =T(M_n )=T(\overline{\mbox{conv} }(M_n )).$$ If $||T|\geq 1$ the we can cosider a operator $$T_1 =\frac{1}{||T|| +1 }T$$