# A conjecture about the positive continuous linear operator

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!

• $(I-A)^{-1} = \sum_{n=0}^\infty A^n$ should allow you to prove $(1-A)$ is positive – Calvin Khor Sep 28 '18 at 5:12
• 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 :-) – Paradiesvogel Sep 28 '18 at 5:18
• 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? – Calvin Khor Sep 28 '18 at 5:24
• Apparently a cone need not be closed. Is yours closed? math.stackexchange.com/questions/2065304/… – Calvin Khor Sep 28 '18 at 5:35
• @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! – Paradiesvogel Sep 28 '18 at 5:35

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.