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Let $E$ be a Banach space.

The definition of a cone in many in textbooks dealing with order in Banach space is:

A nonempty closed subset $K$ of $E$ is called a convex cone if:

  1. $K\neq \{0_E\}$.
  2. $K+\lambda K\subseteq K,\;\:\lambda \geq 0$.
  3. $K\cap -K=\{0_E\}$, that is if $x\in K$ and $-x\in K$, then $x=0_E$.

and $x \preceq y$ (in $E$) if and only if $y-x \in K$.

Now let we denote by $\mathcal{L}^{*}(E)$ the set of elements $A\in\mathcal{L}(E)$ satisfying the following conditions:

  1. $A K \subseteq K$ (that is $A$ is nondecreasing, i.e., $u \preceq v$ implies $A(u) \preceq A(v)$,
  2. For all $u \in K,\left\|A^{n} u\right\|_{E} \rightarrow 0$ as $n \rightarrow \infty$.

How do I construct an example of an operator in $\mathcal{L}^{*}(E)$?

I'm thinking about something like this, with the monotony condition.

almosteverywhere
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Motaka
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  • What do you mean by $K\cap -K={\theta}$, is $\theta$ the neutral element of the Banach space? And do you mean an operator for a general Banach space $E$ or an operator or an operator on a specific Banach space? – esteban Jan 29 '22 at 15:04
  • $E$ is a Banach space, it may be a specific one – Motaka Feb 01 '22 at 10:10

1 Answers1

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As @sein mentioned in his comment it depends on what you want. For a specific Banach space, consider the convex cone of nonnegative functions as a subset of $L^2(\mathbb{R})$. Define the operator $$ A f:= S_1 f \, , $$ where $S_1$ is the heat semigroup at $t\geq 1$, i.e. we have $S_1 f= \varphi_1* f$ with $\varphi_1$ the heat kernel at time $1$. Clearly, $A \in \mathcal{L}^*$, since $\varphi_1$ is nonnegative. Additionally, \begin{align} \lVert A^n f\rVert_{L^2(\mathbb{R})}= \lVert \varphi_n * f\rVert_{L^2(\mathbb{R})} \stackrel{n \to \infty}{\to} 0 \, , \end{align} where we simply use the fact that the heat equation converges to $0$ for all $L^2(\mathbb{R})$ initial data. I am sure there are even simpler examples than this.

Edit: On second thought here is a simpler example in the general setting; please correct me if I am making some stupid mistake. Set $$ A f= \lambda f \, , $$ for $\lambda \in (0,1)$. If $u \in K$, then $\lambda u = 0_E + \lambda u \in K$. The convergence to $0$ is straightforward.

almosteverywhere
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