I'm working through a problem very similar to Fixed point for expansion mapping., I started on a different path to the accepted answer and wasn't able to solve it that way. But it feels like I'm close. Is it possible to solve the problem this way and if so how?
The problem is: Given a Banach space $X$, a constant $K > 1$ and a mapping $T$ on $X$ satisfying \begin{equation} ||T(x) - T(y)|| \ge K ||x - y|| \qquad \forall\, x, y \, \in X \,, \tag{1} \end{equation}
Assuming $T(X) = X$, show that $T$ has a unique fixed point.
My attempt at a solution Let $(z^0_n)$ be any Cauchy sequence in $X$. $z^0_n \rightarrow z^0 \in X$ since $X$ is complete. Since $T(X) = X$, we have that for all $n = 1, 2, \ldots $ there exists $z^1_n \in X$ such that $z^0_n = T(z^1_n)$. Furthermore, $\exists z^1 \in X$ such that $z^0 = T(z^1)$.
\begin{equation} ||z^0_n - z^0|| = ||T(z^1_n) - T(z^1)|| \ge K ||z^1_n - z^1|| \end{equation}
Since $z^0_n \rightarrow z^0$ we get $z^1_n \rightarrow z^1$. We can continue inductively to show that these limits exist to form a sequence $(z^m)$. By iteratively applying (1) iteratively we get for any $n$
\begin{equation} ||z^{0} - z^{n}|| = ||T(z^1) - T(z^{n + 1})|| \ge K ||z^1 - z^{n + 1}|| \ge K^m ||z^m - z^{m+n}|| \\ \implies \frac{1}{K^m}||z^{0} - z^{n}|| \ge ||z^m - z^{m+n}|| \,. \tag{2} \end{equation}
Since $||z^0||$ and $||z^n||$ are finite, we get that $||z^m - z^{m+n}||$ can be made arbitrarily small with $m$ large enough. Therefore $(z^m)$ is a Cauchy sequence in $X$, and $z^m \rightarrow \tilde{z} \in X$. This also means that $T(z^m) \rightarrow \tilde{z}$.
From here I imagined that it would be easy to argue that $T(\tilde{z}) = \tilde{z}$ and prove that the fixed point exists, but I haven't been able to. Is it at all possible to continue from here to show the desired result?
Got it thanks to the hint by Daniel Fischer:
Take $w \in X$ such that $T(w) = \tilde{z}$. From the property of the mapping we get
\begin{equation} ||T(z^m) - \tilde{z}|| = ||T(z^m) - T(w)|| \ge K ||z^m - w|| \,. \end{equation}
Since LHS $||T(z^m) - \tilde{z}|| \rightarrow 0$ we get $z^m \rightarrow w$. By uniqueness of limits, $w = \tilde{z}$. Now from $w$'s definition
\begin{equation} \tilde{z} = T(\tilde{z})\, . \end{equation}
Finally, uniqueness is shown by taking any $v \in X$ such that $v = T(v).$ Then
\begin{equation} ||v - \tilde{z}|| = ||T(v) - T(\tilde{z})|| \ge K ||v - \tilde{z}|| \implies ||v - \tilde{z}|| = 0 \,, \end{equation}
so the fixed point $v = \tilde{z}$ is unique. $$\tag*{$\Box$}$$