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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: enter image description here

And here are some related definitions and theorems: enter image description here enter image description here

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!

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  • $\begingroup$ 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. $\endgroup$
    – VCG
    Sep 16 '16 at 1:49
  • $\begingroup$ 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 $x^*$ $\endgroup$
    – ask
    Sep 16 '16 at 3:07
  • $\begingroup$ And also we are in the non-linear case... $\endgroup$
    – ask
    Sep 16 '16 at 3:16
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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

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  • $\begingroup$ I believe you meant damped oscillations (reduced over time) rather than dumped $\endgroup$
    – Three Diag
    Oct 10 '16 at 10:23
  • $\begingroup$ @ThreeDiag Thanks for spotting this. Fixed. $\endgroup$ Oct 10 '16 at 23:14
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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 $\{t \in \mathbb{N} \mid x(t)<x^{*}\}$. By Acemoglu's inequality, $s>t$ implies $x(s)$ is strictly closer to $x^{*}$ than $x(t)$ is. Thus, since $x(t_{i})<x^{*}$ 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.

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  • $\begingroup$ 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. $\endgroup$
    – tdm
    Sep 21 '21 at 7:12
  • $\begingroup$ Thanks for the suggestion @tdm. I've updated the proof with more detail on that claim. $\endgroup$ Sep 21 '21 at 9:44

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