A question on contraction mapping theorem and fixed point iteration

Question

First of all, thank you for taking the time to read my post. Secondly, this is a question I got as a part of homework. However, the professor allows us to work in groups so I'm hoping that this is okay. First I will state the question, without actual numbers in the problem. I hope this will not affect the solution a lot. Also, I'm interested in the general methodology than the actual solution.

Question

Let $f(x) = e^{-x}.$

a) Show that the sequence $x_{n+1} = f(x_n)$ converges to a unique fixed point $z\in[a,b]$ for any initial value $x_0\in [a,b].$

b) Also show that the sequence $x_{n+1} = f(x_n)$ converges to the unique fixed point $z\in[a,b]$ for any initial value $x_0\in (-\infty,a)\cup(b,\infty).$

My attempt

So, I did part (a) using the contraction mapping theorem. I first proved that $f$ is a contraction on $[a,b]$ and then I showed that $f$ maps $[a,b]$ into itself. Then by the contraction mapping theorem, (a) follows directly.

For (b) however, the contraction mapping theorem couldn't be used because the interval is not a complete subset of the reals. I tried to show that for any initial point($x_0$), $x_1$ lies in $[a,b]$ so that the sequence would eventually lie within $[a,b]$ and then everything follows from part (a). Unfortunately, that didn't work either. Is there any way I can tackle the problem?

Thank you!

Answer 1

Hint: if you can show that there is some $n_0$ and a compact interval $I$ such that $x_{n_0} \in I$ no matter what $x_0$ is and $f(I) \subseteq I$, then you can just apply the contraction mapping principle to $\{ x_{n_0},x_{n_0+1},\dots \}$. In this case there is such an $n_0$ and it isn't that big.

Answer 2

Let $\phi(x) = x- f(x)$, note that $\phi'(x) = 1+e^{-x}\ge 1$. Note that $\phi$ is strictly increasing, $\lim_{x \to -\infty} \phi(x) = -\infty$ and $\lim_{x \to +\infty} \phi(x) = +\infty$. In particular, there is a unique point $x^*$ such that $\phi(x^*) = 0$. It is straightforward to check that this is the unique fixed point of $f$ and that $x^* >0$.

Note that for any $x_0$ we always have $x_n \ge 0$ for $n \ge 1$. Hence we can assume that $x_0 \ge 0$. Note that if $x_m=x^*$ for some $m$ then $x_n = x^*$ for $n \ge m$, so we can presume that $x_n \neq x^*$ for all $n$.

Note that $f'(x) \in [-1,0)$ for $x \ge 0$. The mean value theorem gives $x_{n+1}-x^* = f(x_{n})-f(x^*) = f'(\xi_{n}) (x_{n} -x^*)$ with $\xi_{n} $ in between $x^*$ and $x_{n}$. In particular, $|x_{n+1}-x^*| \le | x_n-x^*|$ and the sign changes with each iteration.

Hence $x_{n+2}-x^*$ and $ x_n-x^*$ have the same sign and the absolute values are non increasing, hence the sequence $x_2,x_4,...$ converges to some $x'$ and in a similar fashion $x_1,x_3,...$ converges to some $x''$.

Note that for $x \ge x^*$ we have $|f'(x)| \le |f(x^*)| < 1$. Since every second term is $\ge x^*$ we see that $|x_{n+2}-x^*| \le 1 \cdot |f'(x^*)|| x_n-x^*|$ and hence we must have $x='x''=x^*$, and so the sequence converges.

For Part a), we need $x^* \in [a,b]$ if we want the sequence to converge to some point in $[a,b]$.

Part b) is answered above.

Answer 3

Since $e^{-x}\gt0$, $x_n\gt0$ for $n\ge1$ and $x_n\lt1$ for $n\ge2$ and $x_n\gt\frac1e$ for $n\ge3$.

Then, for $n\ge4$ and some $\xi$ between $x_{n-1}$ and $x_n$ $$ \begin{align} |x_{n+1}-x_n| &=\left|\,e^{-x_n}-e^{-x_{n-1}}\right|\\[6pt] &=e^{-\xi}|x_{n-1}-x_n|\\[6 pt] &\le e^{-1/e}|x_{n-1}-x_n| \end{align} $$ Thus, the mapping is a contraction and has a unique limit point in $\left(\frac1e,1\right)$.