it is possible to not consider the condition q < 1 in the Banach Fixed Point Theorem (No contraction basically?)

Question

it is possible to not consider the condition q < 1 in the Banach Fixed Point Theorem (No contraction basically?) and still find a fixed point? Any particular example of a function?

f: R -> R

Answer 1

No. This can fail if we take a function which is Lipschitz continuous with constant $q=1$.

For instance, the iteration $x_{n+1}=f(x_n)$ with $f(x)=-x$ from any initial point $x_0\in\mathbb{R}\setminus\{0\}$ fails to converge to its unique fixed point of $0$.

Answer 2

Consider the function $ f : \mathbb{R} \to \mathbb{R} $ be given by $$ f(x) = \begin{cases} x + \frac{1}{x} & \mbox{ if } x \in [ 1, \infty) \\ 2 & \mbox{ if } x < 1 \end{cases} $$ Then we have $ | f(x) - f(y) | = | x- y | | 1 - \frac{1}{xy} | < | x-y | \ \forall \ x,y \in [ 1, \infty ) $.

But this function fails to have any fixed point.