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!

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    $\begingroup$ The important line in the theorem is $T : D \to D$. This is the hypothesis you are missing. $\endgroup$ Mar 5, 2019 at 3:21

2 Answers 2


$e^{-x}$ does not map $[2,\infty)$ into itself.

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    $\begingroup$ The phrase "map into" is often used to denote injections (and "map onto" used to denote surjections), and this is injective. $\endgroup$ Mar 4, 2019 at 16:02
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    $\begingroup$ @Acccumulation - the phrase "map into" does not, and should not, mean "injection". "into" in standard English usage indicates inclusion, not uniqueness. Attaching such a specialized meaning to it in Mathematics simply invites confusion to readers. And while I obviously cannot say that no author has used such a thing, I personally am not familiar with any text where "map into" is given such a meaning. $\endgroup$ Mar 4, 2019 at 17:46
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    $\begingroup$ @Accumulation - So there is one person who is willing to make such an association in a forum post. However, I still say this is far from conventional. The texts I am familiar with do not use "into" to mean "injection", but simply when the function is not known to be surjective. $\endgroup$ Mar 4, 2019 at 18:03
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    $\begingroup$ 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. $\endgroup$
    – Teepeemm
    Mar 4, 2019 at 18:40
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    $\begingroup$ @Acccumulation I think I have uncommonly seen "into" and "onto" used to mean "injective" and "surjective", respectively. But in that sense, and including in the linked mathforum thread, I think they are always used like adjectives, which is not generally the case in English. Used as a preposition, I would have to assume the normal English meaning. $\endgroup$
    – aschepler
    Mar 5, 2019 at 3:09

$e^{-2}$ 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.

  • $\begingroup$ @FedericoPoloni You're right; the article on the BFPT says that L has to be in the interval [0,1), which I missed. $\endgroup$ Mar 4, 2019 at 21:23

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