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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 $T(p) = p$?

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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

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