Let$(X, d)$ be complete space $,$ a $T: X \rightarrow X$ mapping such as, $\dot{\text { ze }} T(X)=X .$
Prove that if exist $T$ with special conition (exist $r>1 $ such that $ d(T(x), T(y)) \geq r d(x, y)$for $x, y \in X), $ to $T$ has only one fixed point.
My attempt. We can conclude that $\frac{1}{a}$ such as $a \in [0,1)$ for Banach fixed point theorem? However, I am not sure if it is correct.
Not sure exactly what is meant by the phrase
$\dot{\text { ze }} T(X)=X, \tag 1$
but if we assume it means
$T(X) = X, \tag 2$
i.e., that $T$ is onto, that is, that
$\text{Range} (T) = X, \tag 3$
then we may argue as follows:
$T$ is also injective, since
$T(x) = T(y) \tag 4$
in concert with the hypothesis
$d(T(x), T(y)) \ge rd(x, y) \tag 5$
yields
$rd(x, y) \le 0, \tag 6$
which, since $r > 1$, implies
$d(x, y) \le 0; \tag 7$
but since $d$ is a metric on $X$ we have
$d(x, y) \ge 0; \tag 8$
now (7) and (8) together yield
$d(x, y) = 0, \tag 9$
whence
$x = y, \tag{10}$
and we conclude $T$ is injective. Since $T$ is both injective and surjective, it's inverse $T^{-1}$ is well defined, and thus the given condition
$d(T(x), T(y)) \ge rd(x, y) \tag{11}$
may be cast in the form
$d(x, y) \ge rd(T^{-1}(x), T^{-1}(y)), \tag{12}$
which is easy to see since
$x = T(T^{-1}(x), \; y = T(T^{-1}(y)); \tag{13}$
(12) is manifestly equivalent to
$r^{-1}d(x, y) \ge d(T^{-1}(x), T^{-1}(y)), \tag{14}$
which establishes $T^{-1}$ is a contraction, for
$r > 1 \Longrightarrow r^{-1} < 1; \tag{15}$
now $T^{-1}$ being a contraction mapping, the standard Banach fixed point theorem may be invoked to affirm the existence of a unique $z \in X$ such that
$T^{-1}(z) = z, \tag{16}$
and thus
$T(z) = T(T^{-1}(z)) = z, \tag{17}$
and $z$ is also a fixed point of $T$, which is clearly unique, since if in addition
$T(w) = w, \tag{18}$
we have
$w = T^{-1}(w), \tag{19}$
and $T^{-1}$ has a unique fixed point; thus
$w = z. \tag{20}$