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Consider example $x^2 - 2 = 0$. I can rewrite so I get $x^2 + x - 2 = x$. If I define $\phi(x) = x^2 + x - 2$, I need to solve $\phi(x) = x$. $\phi$ is Lipschitz-continuous, since it's differentiable. On $[-\frac34,-\frac14]$ we have, using the mean value theorem for $a,b\in[-\frac34,-\frac14]$ $$ \phi(b) - \phi(a) = \phi'(\xi) (b-a).$$ Since $\phi'(\xi) = 2\xi + 1$ and $|\phi'(\xi)| < 1$ for $\xi\in[-\frac34,-\frac14]$, $\phi$ is a contraction and with the Banach theorem, there has to be a $\tilde x\in[-\frac34,-\frac14]$ with $\phi(\tilde x) = \tilde x$, hence $x^2 - 2 = 0$ has to have a solution on $[-\frac34,-\frac14]$, which it doesn't. Where is my mistake?

Martin Sleziak
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Stefan Hante
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    Can you state the version of the fixed point theorem you are using? If I follow the formulation of wikipedia, your argument only shows that $\phi$ is a contraction on $X=[-\frac{3}{4},-\frac{1}{4}]$, but $\phi(X)\not\subset X$ and thus this version cannot be applied. My guess is that you are not allowed to restrict to $[-\frac{3}{4},-\frac{1}{4}]$ since the image of this set is not a subset of itself. – Mathematician 42 Jul 03 '17 at 08:21
  • In fact, I'm fairly certain that you can only apply this theorem to subsets of your domain which are invariant under the function. Otherwise it would be very weird to expect fixed points. See also remark 3 on the wiki-page: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem Good question though, the argument almost convinced me :) – Mathematician 42 Jul 03 '17 at 08:25
  • Yes, I was using this version. I was completely blind for this premise. Sometimes that happens, I guess. – Stefan Hante Jul 03 '17 at 08:55

2 Answers2

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To apply Banach's theorem on $[-\frac34,-\frac14]$ we must have

$\phi([-\frac34,-\frac14]) \subseteq [-\frac34,-\frac14]$.

But this is not the case. Consider $\phi(-\frac12)$.

Fred
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For the Banach fixed point theorem to apply, you need that your to satisfy $\phi : X \rightarrow X$. In the example you provide, you chose the interval in such a way that this is not the case and the range of $\phi$ over $X$ is not contained in $X$.

KyleW
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