My task is to prove the following statement: if $T:V$—>$V$ is an $\alpha$-contraction and $V$ is a Banach space, then T has a globally attracting fixed point $\bar v$ in $V$, and for any initial point $v_0$, it holds that $$||T^n(v_0)-\bar v||\leq \alpha^n/(1-\alpha)||T^n(v_0)-v_0||. $$ I proved this result without induction when dealing with metric spaces, but now I’m trying induction.
Ultimately, the success of my proof comes down to demonstrating the following statement: $$||T^n(v_0)-v_0||\leq ||T^{n+1}(v_0)-v_0||.$$
Intuitively, this is obvious. The orbits around the initial point are going to get closer and closer to the fixed point, I.e. further and further from the initial point. However, I’m struggling to formalize this idea. Does anyone have a hint?
By the contraction property and $T\bar v=\bar v$, triangle inequality, one gets $$ \|T^nv_0 - T^n\bar v \|\le\alpha^n \|v_0-\bar v\| \le \alpha^n( \|v_0 - T^nv_0\| + \|T^nv_0-\bar v\|), $$ then $T^nv_0-\bar v$ to the left, $$ (1-\alpha^n)\|T^nv_0 - \bar v \|\le \alpha^n\|v_0 - T^nv_0\|, $$ which implies $$ \|T^nv_0 - \bar v \|\le \frac{\alpha^n}{1-\alpha^n}\|v_0 - T^nv_0\|. $$ Note that the desired inequality follows, since $\frac1{1-\alpha^n}< \frac1{1-\alpha}$.