Fixed point of Lipschitz mapping Let X be a complete metric space containing the point $p_0$ and let r be a positive real number.  Define $K=\{p\ in\ X\ \big|\ d(p,p_0) \le r \}$.  Suppose that $T:K \to X$ is Lipschitz with Lipschitz constant c.  Suppose also that $cr + d(T(p_0),p_0) \le r$.  Prove that $T(K)\subseteq K$ and that $T:K \to K$ has a fixed point.
My thought process was to use Lipschitz assumption to show that $d(T(p),T(p_0)) \le cr$ and then use $cr + d(T(p_0),p_0) \le r$ to show get $d(T(p),T(p_0)) \le r$ proving that $T(K)\subseteq K$.  Here's where I'm stuck. I don't know if I should show that $0 \le c \lt 1$ and use Contraction Mapping Principle to prove exactly one fixed point or just simply show that there is a fixed point.  If I were to follow my second line of attack I would define $\{p_k\}$ as a Cauchy sequence converging to a point p in X such that $d(p,p_0) \le r$.  If I define $\{T(p_k)\}$ as a subsequence of $\{p_k\}$ can I show that $T(p) = p$?
 A: Assume first that $p_0$ is a fixed point for $T$. Then $d(T(p_0),p_0)=0$, thus $c \leq 1$. Then, if $p \in K$, $d(T(p),p_0) = d(T(p), T(p_0)) \leq c d(p, p_0) \leq d(p, p_0) \leq r$, which shows that $T(p) \in K$, thus $T(K) \subseteq K$.
Now, assume that $p_0$ is not a fixed point for $T$. Then $d(T(p_0),p_0) \gt 0$, thus $c \lt 1$ and $T$ is a contraction, and Banach's fixed point theorem shows that there exist a (unique) fixed point. Next, if $p \in K$, then by the triangle inequality $d(T(p), p_0) \leq d(T(p), T(p_0)) + d(T(p_0), p_0) \leq c d(p,p_0) + d(T(p_0), p_0) \leq cr + d(T(p_0), p_0) \leq r$ by hypothesis, which shows that $T(p) \in K$, and thus $T(K) \subseteq K$.
A: Its easy to show that T(p) is in K . Use the given inequality and Lipschitz condition and then triangle inequality to show d(T(p), p0) <= r .
So, we showed that T: K --> K  .. and that T is Lipschitz. Clearly K is closed. hence complete .. therefore we can directly apply Banach fixed point theorem to show the fixed point 
