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I'm doing a practice problem which asks to prove that the sequence defined by $x_{n+1} = e^{-x_n}$ is convergent (or rather "study the convergence of $(x_n)$"). So I'd like to try and find some sufficient condition on $x_0$ for the sequence to converge.

I can see that $e^{-x}$ is $k$-lipschitzian with $k<1$ on $[a, \infty[$ for $a>0$. But the problem is that $e^{-x}$ does not map $[a, \infty[$ into itself. I started trying to find values of $a, b$ such that $[e^{-b}, e^{-a}]\subseteq[a, b]$, but then I wondered if maybe there was some simpler approach that I was missing.

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    $\begingroup$ Do you mean $x_{n+1}=e^{-x_n}$? $\endgroup$ – Wojowu Dec 27 '15 at 15:24
  • $\begingroup$ @Wojowu Yes. ${}$ $\endgroup$ – Jack M Dec 27 '15 at 15:25
  • $\begingroup$ $x\mapsto e^{-x}$ maps $[0,1]$ into itself. $\endgroup$ – PhoemueX Dec 27 '15 at 15:30
  • $\begingroup$ @PhoemueX But $e^{-x}$ is only $1$-lipschitzian on $[0, 1]$. $\endgroup$ – Jack M Dec 27 '15 at 15:31
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The solution is a corollary of another Banach fixed point theorem: see here Essentially, if we have contraction with constant $k=1$ and compactness (instead of completeness) the result will follow.

For any $x_0$ we have $x_1 \geq 0$ and so $x_2 \leq e^0 = 1$. A trivial induction shows that $x_n \in [0,1]$ for $n\geq 2$ As @Jack M has pointed out in the comments, $e^{-x}$ is indeed 1-lipschitz on $[0,1]$ so the conditions of the theorem in the link I provided are met: $K = [0,1]$. Thus, the result follows.

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  • $\begingroup$ So to show $x_n\to a$ (where $a$ is the unique fixed point), we use that $x_n$ has a convergent subsequence, show that the limit of this subsequence must be $a$ with the usual argument, then use the decreasing nature of $|x_n - a|$ to prove that the existence of a convergent subsequence implies the convergence of $(x_n)$ itself? $\endgroup$ – Jack M Dec 27 '15 at 16:48
  • $\begingroup$ Yes, and you can use that $a$ is the unique solution of $x = e^{-x}$ for the last part of the argument. $\endgroup$ – Milen Ivanov Dec 27 '15 at 16:58
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I like Milen's answer and upvoted it. Here's an approach that uses the standard contraction mapping theorem.

Let $f(x)=\exp(-x)$ and let $F(x)=f(f(x))$. We compute \begin{align} F'(x) &= e^{-x-e^{-x}} \\ F''(x) &= e^{-x-e^{-x}} \left(e^{-x}-1\right). \end{align} Note that $F''(x)<0$ for $x>0$. Thus $F'$ is decreasing on $[0,\infty)$ and $$F'(x)<F'(0)=\frac{1}{e}<1$$ for $x\in[0,\infty)$. It follows that $F$ is a contraction on $[0,\infty)$ and, thus, has a unique fixed point in $[0,\infty)$, say $x_0$. I claim that your recursively defined sequence converges to $x_0$.

Let $a$ be a real number with $0<a<x_0$ and let $b=f(a)$. Our observations on $F$ imply that $a<f(b)<x_0<f(a)<b$. In particular, $f$ maps $[a,b]$ into itself. Furthermore, it's easy to see that $|f'(x)|<|f'(a)|<1$ on $[a,b]$ so that $f$ is a contraction on $[a,b]$. Thus, iteration of $f$ must converge to $x_0$ when we start in $[a,b]$. Finally, every starting point in $\mathbb R$ must eventually land in some interval of the form $[a,f(a)]$.

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    $\begingroup$ Very nice trick, to consider $f(f(x))$ to get a Lipshitz constant $1/e$ $\endgroup$ – Milen Ivanov Dec 27 '15 at 17:00

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