According to Approach0, this question seems new.
There is a similar question, however in that question asked for a proof about a function, which satisfied the definition (see Problem), has a fix point. However a function, which has a fixed point, doesn't implies it is a contraction ($f(x)=x$).
Motivation: I am self-studying for my next semester math course. I did some past exam papers from that course (no solutions provided), then I met with this question, but I am not sure how to do.
Problem
Let $X$ be a complete metric space and $\phi: X \to X$ be continuous.If exists a function $f: X \to [0, \infty)$ such that$$d(x, \phi(x)) \leq f(x) - f(\phi(x)) \quad \text{for all } x \in X,$$Is $\phi$ necessarily a contraction?
My idea:
Definition of contraction: $\forall x,y\in X,\exists\alpha\in[0,1),d(\phi(x),\phi(y))\leq\alpha d(x,y)$.$$\begin{align}d(\phi(x),\phi(y))&\leq f(\phi(x))-f(\phi(y))=d(f(\phi(x)),f(\phi(y)))\end{align}$$However, I don't know to how connect $d(f(\phi(x)),f(\phi(y)))$ with $f(x)-f(\phi(x))$.
Thus, I think the statement is false, I tried $X=[0,\infty)$, $\phi(x)=x-1$ which is not a contraction and $f(x)=x^2$. $$d(x,\phi(x))=d(x,x-1)=1,f(x)-f(\phi(x))=x^2-(x-1)^2=2x-1$$However it is not true that $1\leq2x-1$ for $x\in[0,\infty)$.
Thus, my question is if the statement is true, how can I continue? If the statement is false, is there an way to construct such counterexamples and can anyone please explain the intuition behind. Thank!
