# Rigorous proof needed: Acemoglu (Intro Growth) Corollary $2.1.2$

I am reading Acemoglu's intro to modern economic growth. But I am having trouble understanding his proof to a theorem related with stability. Here is the theorem:

And here are some related definitions and theorems:

My question lies in his proof of corollary $2.1$. How did he show that the sequence $x(t)$ is monotone and bounded above by $x^*$? I think it is not enough to use that the derivative is less than $1$ and the distance to $x^*$ is shrinking to say $x(t)$ is monotone. Can someone give me some more details to understand the proof?

Thanks!

• Given that $|a|<1$ and the linear difference is affine, ratio being less than 1 (dividing the RHS over to the left) is enough to get monotone. Try some examples.
– VCG
Sep 16, 2016 at 1:49
• It only says the distance is decreasing, and it does not eliminate the case say $x(0)=1, x^*=3, x(1)=4, x(2)=2.5, x(3)=3.25...$, or even an extreme case that $x^*=1, x(2k-1)=\frac{1}{2}- \frac{1}{2k-1}, x(2k)=\frac{3}{2}+\frac{1}{2k}$, where $x(t)$ does not converge to $x^*$
Sep 16, 2016 at 3:07
• And also we are in the non-linear case...
Sep 16, 2016 at 3:16

The OP correctly identified a mistake here. Since the author claims monotonicity for a general function, let's disprove it for the simple linear case. Consider

$$x_{t+1} = g(x_t) = -0.5x_t$$

This satisfies all the requirements and conditions stated by the author, and also that the derivative is smaller than unity in absolute values $\forall x$. But it converges with "damped oscillations", and so it is not monotonic since, say,

$$x_0 = -3 < 0 = x^*$$ $$x_1 = 1.5$$ $$x_2 = -0.75$$ etc

• I believe you meant damped oscillations (reduced over time) rather than dumped Oct 10, 2016 at 10:23
• @ThreeDiag Thanks for spotting this. Fixed. Oct 10, 2016 at 23:14

Since the OP asked for a rigorous proof, here is one.

By Acemoglu's inequality in the first part of his proof, we can separate $$\{x(t)\}_{t=0}^{\infty}$$ into two subsequences, an increasing subsequence $$\{x(t_{i})\}_{t_{i}\in I}$$ bounded above by $$x^{*}$$ and a decreasing subsequence $$\{x(t_{j})\}_{t_{j} \notin I}$$ bounded below by $$x^{*}$$. Indeed, we define $$I$$ to be the set $$\{t \in \mathbb{N} \mid x(t). By Acemoglu's inequality, $$s>t$$ implies $$x(s)$$ is strictly closer to $$x^{*}$$ than $$x(t)$$ is. Thus, since $$x(t_{i}) for all $$t_{i} \in I$$, $$\{x(t_{i})\}_{t_{i}\in I}$$ is increasing. Similarly, $$\{x(t_{j})\}_{t_{j} \notin I}$$ is decreasing and bounded below by $$x^{*}$$

Assume that each of these subsequences is infinite. Then each converges to a limit, which we will denote by $$y$$ and $$z$$ respectively.

Note that it also follows from Acemoglu's inequality that $$y=x^{*}-\epsilon$$ and $$z=x^{*}+\epsilon$$ for some $$\epsilon \geq 0$$. Indeed, suppose that this is not the case (we will show this generates a contradiction). For convenience, assume $$\vert y -x^{*}\vert<\vert z-x^{*}\vert$$. Define $$\delta:=\vert z-x^{*}\vert-\vert y-x^{*}\vert>0.$$ Since $$\lim_{i \to \infty}x(t_{i})=y$$, we can pick $$t \in \mathbb{N}$$ such that $$\vert x(t)-y\vert<\frac{\delta}{2}.$$ The triangle inequality then tells us that $$\vert x(t)-x^{*}\vert <\vert y-x^{*}\vert+\frac{\delta}{2}.$$ Now pick some $$s \in \mathbb{N}$$ such that $$s >t$$ and $$x(s)>x^{*}$$ (such an $$s$$ exists by our assumption that both the constructed subsequences are infinite). Since $$x(s)>z>x^{*}$$, we have that $$\vert x(s)-x^{*}\vert>\vert z-x^{*}\vert.$$ Thus, $$\vert x(t)-x^{*}\vert <\vert y-x^{*}\vert+\frac{\delta}{2}<\vert z-x^{*}\vert <\vert x(s)-x^{*}\vert.$$ But since $$s>t$$, this contradicts Acemoglu's inequality.

Now we will consider the value of $$g(y)$$. Since, $$g(y)=g(\lim_{i \to \infty}x(t_{i}))=\lim_{i \to \infty}g(x(t_{i}))=\lim_{i \to \infty}x(t_{i}+1),$$ we must have that $$g(y)=y$$ or $$g(y)=z$$.

If $$g(y)=z$$, then we must have that $$y=z=x^{*}$$ since if $$y\neq x^{*}$$, we get a contradiction. Indeed, if $$y \neq x^{*}$$, then Acemoglu's inequality implies that $$g(y)$$ is strictly closer to $$x^{*}$$ than $$y$$ is; but $$y$$ and $$z$$ are equidistant from $$x^{*}$$. Thus $$g(y)=y$$ and similarly $$g(z)=z$$. (If only one of the subsequences is infinite, we get this immediately.) But this implies that $$y=z=x^{*}$$, as Acemoglu's inequality allows us to infer that if $$y \neq x^{*}$$, then $$g(y)$$ is strictly closer to $$x^{*}$$ than $$y$$ is.

• I'm not sure how you get to the claim that there are two monotone sequences whose limits are equally far from $x^\ast$. I think it would be good to be somwhat more specific where this comes from.
– tdm
Sep 21, 2021 at 7:12
• Thanks for the suggestion @tdm. I've updated the proof with more detail on that claim. Sep 21, 2021 at 9:44