Consider the complete metric space $(R^2,d_2) $ where $d_2$ is the Euclidean distance. I have to check whether the following operator $T:R^2 \rightarrow R^2$ is a contraction:
$ T(v) = (v_2,\frac{1}{2}v_1) $ for any $v=(v_1,v_2) \in R^2$
What this does is to flip the coordinates of a vector and cut by half the second.
This is the definition of a contraction: there exists some $\beta \in [0,1)$ such that
$ d(Tv,Tw) \leq \beta d(v,w)$ for all $v,w \in R^2$
First of all, I found that this mapping has a fixed point: $ v^\star = (0,0)$ but that's not the point of the exercise. I tried to apply the definition of contraction. Take two arbitrary vectors $v=(v_1,v_2)$, $w=(w_1,w_2)$, then $T(v)=(v_2,\frac{1}{2}v_1)$ and $T(w)=(w_2,\frac{1}{2}w_1)$. I have
$ d(Tv,Tw) = (v_2-w_2)^2+(\frac{1}{2}v_1-\frac{1}{2}w_1)^2 = (v_2-w_2)^2+\frac{1}{4}(v_1-w_1)^2 < (v_2-w_2)^2+(v_1-w_1)^2 = d(v,w)$
so I managed to show only that $ d(Tv,Tw) < d(v,w)$ which is the definition of weak contraction. And by the way this is not enough to invoke the contraction mapping theorem... So I'm a bit puzzled. Is this mapping a contraction or not? Any help would be really appreciated, thanks!
Your function $T$ is linear, so the condition is equivalent to showing $\|Tv\| \leq c\|v\|$ with $c < 1$ for all $v$. From here, we can see that $T$ cannot be a contraction: take for example $e_2$, which verifies $Te_2 = e_1$ and so $\|Te_2\| = \|e_1\|$.