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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!

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  • $\begingroup$ $(I-A)^{-1} = \sum_{n=0}^\infty A^n$ should allow you to prove $(1-A)$ is positive $\endgroup$ – Calvin Khor Sep 28 '18 at 5:12
  • $\begingroup$ Thanks @CalvinKhor . I tried to use the expression $(I- A)^{-1} = \sum^\infty_{n = 0} A^n$ to show the positivity of $(I-A)$, but this way did not lead to the desired result. Would you mind to explain this idea in a bit more detail please? How to utilize this expression to prove $(I-A)$ is positive? Many thanks :-) $\endgroup$ – Paradiesvogel Sep 28 '18 at 5:18
  • $\begingroup$ well $A^n x \ge \theta$ for every $n$, Im not familiar with ordered Banach spaces but it feels like the limit should go through maybe with some lemma like $$ a_i \ge \theta \implies \sum a_i \ge \theta,$$so that $(I-A)^{-1}$ is positive, and then you should be able to get $(I-A)$ positive with $a\ge\theta, ab\ge\theta \implies b\ge\theta$. Is this too naive? $\endgroup$ – Calvin Khor Sep 28 '18 at 5:24
  • $\begingroup$ Apparently a cone need not be closed. Is yours closed? math.stackexchange.com/questions/2065304/… $\endgroup$ – Calvin Khor Sep 28 '18 at 5:35
  • $\begingroup$ @CalvinKhor Thank you. I totally agree with you that the operator $(I-A)^{-1}$ is positive. But I did not see why the positivity of $(I-A)^{-1}$ could imply the positivity of $(I-A)$. Does such a fact that $a>\theta, ab>\theta \implies b>\theta$ apply to this case? Would you mind to explain this point please? Thanks so much! $\endgroup$ – Paradiesvogel Sep 28 '18 at 5:35
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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.

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  • $\begingroup$ Thank you very much @daw. It's really helpful :-) $\endgroup$ – Paradiesvogel Sep 28 '18 at 7:27

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