I'm required to show that: $f(x) = e^x - 4x$ has a root in $(0,1)$ using the Banach Fixpoint theorem.

The fact that $f((0,1)) \neq (0,1)$ confuses me. How do I proceed without knowing that $f$ isn't a map of a set onto itself?

  • $\begingroup$ You may try some curve sketching and find a domain so that $f(D) \subseteq D$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 12 at 15:58
  • $\begingroup$ @GNUSupporter8964民主女神地下教會 How would I go about finding such a domain $D$? $\endgroup$ – user7802048 Feb 12 at 16:11
  • $\begingroup$ Sorry. I think my previous comment isn't correct. Perhaps we need to divide $f$ by a certain constant, so that $f$ is a contraction. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 12 at 16:17

You have to reformulate the requirement $f(x^\star) = e^{x^\star} -4x^\star = 0$ for a root such that the root $x^\star$ is a fixpoint of a function $g(x)$, that is $g(x^\star) = x^\star $. This is the case for $ g(x) = f(x)/4 +x = e^x/4$.

Now you have to show that $g(0,1) \subset (0,1)$ and $|g'(x)|<1$ on $x \in (0,1)$. Then, there is a fixpoint in $(0,1).$


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