Quantcast
Channel: Active questions tagged real-analysis - Mathematics Stack Exchange
Viewing all articles
Browse latest Browse all 9169

Prove/disprove: $d(x, \phi(x)) \leq f(x) - f(\phi(x))$ implies $\phi$ is a contraction

$
0
0

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!


Viewing all articles
Browse latest Browse all 9169

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>