I'm studying contractions and I am trying to understand under which conditions a function is actually a contraction. I have understood that a continuous function $f: [a,b]\rightarrow\mathbb{R}$ is a contraction if:
i) $f$ is a selfmap, and
ii) given that it is differentiable on $(a,b)$, $|f'(x)|\leq\alpha$, $\forall x \in [a,b]$
Are these conditions right? If so, why $f(x)=arctan(x)+5$ should not be a contraction in the complete metric space $(\mathbb{R},d_1)$?
For a contraction, the (sufficient) condition you state would require $\alpha < 1$. Your example with $\arctan$ will only provide $\alpha =1$ (as $f^\prime(0)=1)$. I.e., it is in particular $1$-Lipschitz (as a differentiable function with derivative bounded by $c$ is $c$-Lipschitz).
However, a contraction requires by definition $\alpha$-Lipschitz with $\alpha < 1$. (And one can verify that your $f$ is not $\alpha$-Lipschitz for any $\alpha < 1$, as around $0$ we have $\lvert f(x) - f(0)\rvert = \lvert x\rvert + o(x)$.)