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Let $T$ be a map from a compact metric space $X$ into itself satisfying $ d(Tx,Ty) < d(x,y)$ for all distinct $x,y$ in $X$. It is true that $T$ has a unique fixed point. Fix $x_0 \in X$ any point, and define the recursive sequence $\{ x_n \}$ by $Tx_0 = x_1$, and $T^{n} x_0 = x_n $. If this sequence converges, it is true that it converges to the unique fixed point in $X$. However, I am having trouble showing that the sequence does converge, i.e., is Cauchy. I can show the sequence is Cauchy if the mapping had been a contraction, but the technique there doesn't generalize well.

Edit: I came up with a simpler proof than from anything else I could find online, but it's not quite complete. Suppose for the sake of contradiction that $\{T^nx_0\}$ did not converge to the fixed point $\alpha$. Then $\exists \epsilon$ and a subseq $\{ T^{n(k)} x_0 \}$ s.t. for each $n(k)$, $d(T^{n(k)} x_0 , \alpha ) > \epsilon $. By compactness, $ \{ T^{n(k)} x_0 \}$ admits a convergent subseq. I want to say this subseq, say $\{ T^{n_k(m)} x_0\}_{m \in \mathbb{N}}$ and converging to $\beta$, must actually converge to $\alpha$ by continuity of $T$, but can can this be done? It's true that $T(T^{n_k(m))} x_0) \rightarrow T(\beta)$, but can I say $T^{n_k(m) + 1} x_0 \rightarrow \beta $, so that $\beta$ is actually the fixed point $\alpha$?

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I think this will help jstor.org/pss/2044417 –  Norbert Feb 13 '12 at 20:40
    
What's $T$? Just some random continuous map from $X$ to itself? Or is related to $f$ somehow? –  Jason DeVito Feb 13 '12 at 21:12
    
@JasonDeVito Sorry, $T$ is $f$! I edited the post. –  Peter Feb 13 '12 at 21:28
    
The article by S. Leader can be downloaded for free, ams.org/journals/proc/1982-086-01/S0002-9939-1982-0663887-2/… –  Will Jagy Feb 13 '12 at 21:44
    
Note that the fact you want to prove is Corollary 1 at the bottom of page 156, definitions all in a bunch on page 153. –  Will Jagy Feb 13 '12 at 22:45
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