I want to understand the triangle inequality about contraction operator

Question

I read the Chow's paper, Multigrid algorithms and complexity results, https://dspace.mit.edu/handle/1721.1/14254

I have a question on page 42. Let me write the Lemma 2.4.3 on the page.

Lemma 2.4.3 Let $T_{1}$ be a $k$-stage contraction operator, with contraction factor $\alpha$, on a Banach space $(X, ||. ||)$. If $J_{1}\in X$ is the fixed point of $T_{1}$, then $$||J_{1}-J|| \le \frac{k}{1-\alpha} ||T_{1}J-J||, \forall J\in X$$.

Proof

Using the triangle inequality, we have $$||T_{1}^k J_{1}-J||\le \sum_{i=1}^k ||T_{1}^{k+1-i}J-T_{1}^{k-i}J||$$ ... The rest is omitted.

The inequality shows that, if $k=1$ , $ ||T_{1} J_{1}-J|| \le ||T_{1} J-J||$. I don't know how this inequality always becomes true without any constraints for $J$.

Answer 1

We know $T_1^k J_1 = J_1 \textbf{ }\forall k$.

We do $||T_1^k J_1 - J|| = ||(T_1^k J_1 - T_1^k J)+(T_1^k J - T_1^{k-1}J) + (T_1^{k-1}J - T_1^{k-2}J) + ...(T_1J - J)|| \leq $ (do triangle inequality on each bracketed expression) $\leq \sum_{i=1}^{k}||T_1^iJ - T_1^{i-1} J||+|| T_1^kJ_1 - T_1^k J)||$.

Can you see what to do next?