Counterexamples to Banach Fixed Point (Banach's Contraction) Theorem with relaxed inequalities?

Question

Banach Fixed Point theorem states: Let $(X,d)$ be a complete metric space. Suppose that $f:X→X$ is a strong contraction, i.e. there exists $q ∈ [0, 1)$ such that

$d(f(x),f(y))$ $\le$ $q$ $d(x,y)$, then there is a unique point $x_0∈X$ s.t. $f(x_0)=x_0$

My questions are:

1- If we allow $q$ to be equal to $1$, does the theorem fail? Could someone provide an example?

2- If we substitute the strong contraction condition with the following condition: $d(f(x),f(y))$ $<$ $d(x,y)$, does the theorem fail? example?

Answer 1

2. Yes, it fails. Consider $f : [0, \infty) \to [0, \infty)$ given by

$$f(x) = \sqrt{1+x^2}.$$

Since

$$f'(\xi) = \frac{\xi}{\sqrt{1+\xi^2}} < 1 \text{ for each } \xi \in [0, \infty),$$

by the mean value theorem

$$|f(x) - f(y)| = f'(\xi) |x-y| < |x-y| \text{ for every } x, y \in [0, \infty).$$

But $f$ clearly has no fixed points.