A question about fixed point theorem

Question

how to prove this theorem how to prove first if the sequence is there that is Cauchy...thanks

Answer 1

We proceed in analogy to Banach's fixed point theorem.

1. Existence.

Let $n$ be a natural number. $$ d(T^n x, T^{n+1} x) \le \lambda(d(T^{n-1}x, T^n x) + d(T^n x, T^{n+1} x)), $$ so $$ (1-\lambda) d(T^n x, T^{n+1} x) \le \lambda d(T^{n-1}x, T^n x) \Leftrightarrow d(T^n x, T^{n+1} x) \le \frac{\lambda}{1-\lambda} d(T^{n-1}x, T^n x) \le \left( \frac{\lambda}{1-\lambda} \right)^n d(x,Tx) $$ by induction. Define $\kappa := \frac{\lambda}{1-\lambda}$. By assumption, $$ 0 < \kappa < 1. $$ Hence, by the triangle inequality, whenever $m > n$, $$ d(T^m x, T^n x) \le \sum_{k=n+1}^m d(T^k x, T^{k+1} x) \le d(x,Tx) \sum_{k=n+1}^\infty \kappa^k \\ = \kappa^{n+1} d(x,Tx) \frac{1}{1 - \kappa}, $$ so that $T^n x$ is Cauchy, hence convergent to some $x^*$. This is also a fixed point since $$ d(T x^*, x^*) = d (T \lim_{n \to \infty} T^n x, x^*) = d (\lim_{n \to \infty} T^{n+1} x, x^*) = d(x^*, x^*) = 0, $$ applying the continuity of $d(\cdot, x^*)$ and $T$; the former is proved using the triangle inequality.

2. Uniqueness

Let $x, y$ be fixed points. Then $$ d(x,y) = \le \lambda(d(x,Tx) + d(y,Ty)) = 0, $$ so $x = y$.