Is $f$ is a contraction, could be $F$ the empty set?

Question

Suposse we have a function $f: H \rightarrow H$ where $H$ is a hilbert space and $f$ is a contraction.

Is it possible that the set $F=\{x \in H : \exists \alpha \in (0,1) / x=\alpha f(x)\}$ is empty and $f$ have a fixed point?

Answer 1

Note that for any $\alpha \in (0, 1)$, the function $\alpha f$ is a contraction with constant $L \alpha$, where $L$ is the contraction constant of $f$. So according to Banach's fixed point theorem, for every $\alpha \in (0, 1)$ there is exactly one $x_\alpha \in H$ such that $x_\alpha = \alpha f(x_\alpha)$. So the set $F$ is never empty but consists of all the $x_\alpha$.

But $f$ has a fixed points according to Banach's fixed point theorem. Hence: No it is not possible.