Let $A \colon \mathbb{X} \to \mathbb{X}$ be a continuous linear operator on a real ordered Banach space $\mathbb{X}$ with the positive cone $\mathbb{K}$.
It is well know from the Neumann series that if the operator norm $\| A\|$ of $A$ is strictly less than $1$, i.e., $\| A\| <1$, then the liner operator equation $$ x = Ax +b \qquad (x \in \mathbb{X}, \,\, b \in \mathbb{K}),$$ has a unique solution, which is $x^*=(I-A)^{-1} b$.
I am wondering that if the operator $A$ is additionally assumed to be positive (i.e., $A x \geq \theta$ whenever $x \geq \theta$, where $\theta$ is the zero element of $\mathbb{X}$), then could we conclude the following result $$ x \geq x^* \implies x \geq Ax + b \,\,\,?$$
I was thinking to try to show that the linear operator $I-A$ is positive, then we obtain $ x \geq x^* \implies (I-A)x \geq (I-A)x^*$ (since the positivity of a linear operator is identical to the monotone increasing property), in which case $(I-A)x \geq b$ yields the desired result $x \geq Ax +b$. But I got stuck to showing the monotonicity of $I-A$. Besides, this might not be a good approach.
In fact, if we let $\mathbb{X} = \mathbb{R}$,, then the above linear operator equation becomes a linear equation with $A, b \in \mathbb{R}$, and it is clear that the conjecture $ x \geq x^* \implies x \geq Ax + b$ holds true.
Thus, I am curious that could we generalize such a result for an abstract Banach space?
Any suggestion or idea are most welcome! Thank you very much!
This is not true. Take $X=\mathbb R^3$, $K$ to be the ice-cream cone $$ K=\{x: \ x_3 \ge \sqrt{x_1^2 + x_2^2} \}. $$ Define $$ A=\pmatrix{0&0&0\\0&0&0\\0&0&0.5}. $$ Then for $x\in K$, $Ax\in K$. However, for every non-zero element of the boundary of $K$ it holds $(I-A)x\not\in K$, e.g., take $$ x=\pmatrix{0\\1\\1}, (I-A)x = \pmatrix{0\\1\\0.5}\not\in K. $$
Here is also a $2d$ example: $X=\mathbb R^2$, $K=\{x: \ x_1\ge0,x_2\ge0\}$, $$ A=\pmatrix{0.1&0.1\\0.1&0.1} ,\ x = \pmatrix{0\\1}, \ (I-A)x=\pmatrix{-0.1\\1.1}\not\in K. $$
These examples also show that the implication $$ x \ge (I-A)^{-1}b = x^* \ \Rightarrow\ (I-A)x \ge b $$ is not true in general: simply take $b=x^*=0$ in the above examples.