$(X, ||. ||)$ is a Banach space. Let $T, S$ be two operators from $X$ to $X$. $T$ is the k-contraction with constant $k \in [0, 1)$ and fixed point $x \in X$. $S$ is the approximation to $T$ for $\epsilon > 0$: $$ || T(z) - S(z) || \leq \epsilon $$ for all $z \in X$. Fixed $x_0 = y_0 \in X$, denote $x_{m+1} = T(x_{m})$ and $y_{m+1} = S(y_{m})$ for $m \geq 0$.
I am trying to show the inequality relation: $$ ||x_{m} - y_{m} || \leq \epsilon \frac{1-k^m}{1-k} $$
I use the assumption of $T$ contraction mapping: $||x_{m+1} - x_{m}|| \leq k^{m} ||x_1 - x_0||$.
I rewrite the inside inequality and assume this holds for $m$ case, and try to prove for $m+1$ case:
$$ ||x_{m+1} - y_{m+1}|| = || x_{m+1} -x_{m} + x_{m} - y_{m} + y_{m} - y_{m+1}|| $$
$$ \leq ||x_{m+1} -x_{m} || + ||x_{m} - y_{m} || + ||y_{m} - y_{m+1}|| $$
The first term is bounded by $k^{m} ||x_{1} - x_{0}||$ the second term is bounded by the induction hypothesis. But how can I bound the third one? It is the approximation to $T$-contraction mapping.