Where is my mistake using the Banach theorem for $x^2 - 2 = 0$?

Question

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?

Answer 1

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

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

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

Answer 2

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 $X$.