Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $T(p) = p$?

share|cite|improve this question

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 $T(K) \subseteq K$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.