Suppose that $T: M \to M$ is a self map of a nonempty closed set $M$ in a complete metric Space $(X,d)$. Suppose further that $$d(Tx,Ty) \le k(a,b)d(x,y)$$ for all $x,y \in M$ with $0 \lt a \le d(x,y) \le b$ and arbitrary numbers $a,b$.Here $0 \le k(a,b) \lt 1$. Then show that $T$ has exactly one fixed point.
My question here is : Whether $k(a,b)$ depend on $x,y$?? From the question I conclude that they don't. If it doesn't then there is nothing to prove here for I can define a sequence $x_{n+1} =T(x_n)$ and proceed to show that it is Cauchy, hence convergent and the limit will turn out to be fixed point.
My try: My first claim is that given $ r\gt 0 $ there is a $x_r \in M$ such that $B[x_r,r] $ is invariant under $T$. Suppose I am able to justify this claim. Then I would choose $M_1=B[x_1,1]$ and again by choosing $r_2=\frac{1}{2}$ get a hold of $x_2$ such that $M_2=B[x_2,r_2] \subset M_1$. Proceeding this way I can get a decreasing sequence of balls and since $M$ is complete, $\cap_{n} M_n \ne \phi$. I expect that the point in the intersection to be the fixed point.
So I try to justify my claim. Suppose the claim doesn't hold. Then there exists $r \gt 0$, for all $x \in M$, $B[x,r]$ is not invariant under $T$. Consider $x_1 \in B[x,r]$. Then $d(x,x_1) \le r$ but $d(x, Tx_1) \gt r$. Then $d(x,Tx) \ge d(x,Tx_1)-d(Tx_1,Tx)$. If $d(x_1,x) \le \frac{r}{2}$, then $d(x,Tx) \ge \frac{r}{2}$. If $d(x_1,x) \gt \frac{r}{2}$, then $d(Tx_1,Tx) \le K(\frac{r}{2},r)d(x_1,x) \le rK(\frac{r}{2},r)$. Either way $d(Tx_1,Tx) \ge \min\{\frac{r}{2},r-rK(\frac{r}{2},r),\}=a \gt 0$.
I am unable to proceed from here.
Then I was referring to the following book "Geometrical Methods of Non-Linear Analysis" by Zabreiko & Krasnoselskii which does the proof for the above. It proceeds as I did and then does something which I don't understand. In the sixth line, the authors say that "for each $x_0 \in T$... " and then reaches the contradiction. I don't understand this step at all.
I think that I have to get hold of $ x \in M $ such that $ d (x, Tx) \lt a $. But I have no clue how to get hold of such $ x $. Even if I get a contradiction I am not able to show that the point in the intersection is the fixed point either.
Thanks for the help!!