I am trying to find the fixed point of the function $g(x) = e^{-x}$. Wolfram Alpha tells me that this fixed point is approximately $x \approx 0,567$. However, if I apply the Banach fixed point theorem, I can prove that $g(x)$ has a fixed point in the interval $[2,\infty)$. I reasoned as follows:
Banach fixed point theorem:
Let $X$ be a Banach space, $D \subseteq X$ a closed interval and $T:D \rightarrow D$ a contraction, which means that $T$ is Lipschitz continuous with Lipschitz constant $L < 1$:
\begin{equation}
\Vert T(u) – T(v) \Vert_X \leq L \Vert u-v \Vert_X \text{ } \forall u,v \in D.
\end{equation}
Now $X = (\mathbb{R}, \Vert \cdot \Vert_1)$ is a Banach space and $D = [2,\infty)$ is a closed interval in $X$. Now, I show that $g(x)$ is a contraction: without loss of generality, assume $x < y$. Then
\begin{equation}
\Vert g_1(x) – g_1(y) \Vert_1 = |e^{-x} – e^{-y}| = e^{-x} – e^{-y} = e^{-x}(1-e^{x-y}).
\end{equation}
Since $e^a \geq 1+a \text{ } \forall a \in \mathbb{R}$, I obtain
\begin{equation}
\begin{split}
\Vert g_1(x) – g_1(y) \Vert \leq e^{-x}(1-(1+x-y)) = e^{-x}(y-x) \\
\leq e^{-2}(y-x) = e^{-2}|x-y| = e^{-2} \Vert x-y \Vert_1 = L \Vert x-y \Vert_1,
\end{split}
\end{equation}
where $L = e^{-2}$ < 1.
So the conditions of the Banach fixed point theorem are satisfied and $g(x)$ has a fixed point in the interval $[2, \infty)$. However… This is not true!
Can anyone tell me what is going wrong here?
Thank you very much in advance!
Best Answer
$e^{-x}$ does not map $[2,\infty)$ into itself.