# Contradiction with Banach Fixed Point Theorem

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 $$L < 1$$:

$$$$\Vert T(u) - T(v) \Vert_X \leq L \Vert u-v \Vert_X \text{ } \forall u,v \in D.$$$$

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

$$$$\Vert g_1(x) - g_1(y) \Vert_1 = |e^{-x} - e^{-y}| = e^{-x} - e^{-y} = e^{-x}(1-e^{x-y}).$$$$

Since $$e^a \geq 1+a \text{ } \forall a \in \mathbb{R}$$, I obtain

$$$$\begin{split} \Vert g_1(x) - g_1(y) \Vert \leq e^{-x}(1-(1+x-y)) = e^{-x}(y-x) \\ \leq e^{-2}(y-x) = e^{-2}|x-y| = e^{-2} \Vert x-y \Vert_1 = L \Vert x-y \Vert_1, \end{split}$$$$

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!

• The important line in the theorem is $T : D \to D$. This is the hypothesis you are missing. – Nate Eldredge Mar 5 at 3:21

$$e^{-x}$$ does not map $$[2,\infty)$$ into itself.
• Regardless, the main point of this answer is that the range of $e^{-x}$ is not in $[2,\infty)$. Whether "into" means "one-to-one" or not, $e^{-x}$ does not map $[2,\infty)$ into itself because the range is disjoint from the domain. – Teepeemm Mar 4 at 18:40
$$e^{-2}$$ is not in the interval $$[2,\infty)$$, so the image of the interval is not contained within the interval; i.e. $$g$$ does not map the interval to itself. In fact, the interval and its image are disjoint, so there can't possibly be a fixed point. One way of thinking of the Banach fixed-point theorem is that if you have an interval that is mapped to itself, then you can find a a sub-interval that is mapped to that sub-interval, and a sub-sub-interval of that sub-interval that is mapped to that sub-sub-interval, and so on, and the limit of that process is a single point that's mapped to itself. Once you get an interval that isn't mapped to itself, though, that interval doesn't necessarily have a fixed point, even though the space as a whole does. If you don't check that an interval is mapped to itself, you would find that the BFPT requires every interval to contain a fixed point, which is clearly absurd.