is it possible to have a transformation $T: X \to X$such that there is contraction for T in $(X,d_1)$ but not in $(X,d_2)?$
I tried defining the function $T(x)=x/2$ and $d_1=|x-y|$ but I cannot seem to define a $d_2$ such that T doesn't contract? Is it even possible?
Conceptually I am confused if our existence of fixed point is dependent on the metric?
Consider $X:=[1,\infty)$ and $f:X \to X$, $f(x)=7\sqrt{x}$.
With respect to $d_1(x,y):=|x-y|$ the function $f$ is not a contraction, but with respect to the equivalent metric $d_2(x,y):=|\ln(x)-\ln(y)|$ we have $$ d_2(f(x),f(y))=|\ln(7\sqrt{x})-\ln(7\sqrt{y})|=\frac{1}{2}|\ln(x)-\ln(y)| =\frac{1}{2}d_2(x,y). $$ The existence of a fixed point is a property of the function.