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Let $X$ be a complete metric space and $\alpha \in (0,1)$. Suppose that for every $x,y \in X$ there exists $z \in X$ s.t. $$ d(x,z) \le \frac{1}{2^\alpha}d(x,y), \qquad d(y,z) \le \frac{1}{2^\alpha}d(x,y). $$

Then $X$ is Hölderian path-connected, i.e. for every $x,y \in X$ we can find a $\alpha$-Hölderian path $\gamma \colon [0,1] \to X$ such that $\gamma(0)=x$ and $\gamma(1)=y$.

Have you got any ideas on how to solve this problem? To be honest, I do not know how to start. I'm stumped for the hypothesis of completness... How can we use it?

I thank you in advance for any useful idea.

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Completeness is quite obviously a necessary condition, look at the rational number as a metric space. –  user20266 Jul 8 '12 at 10:46

1 Answer 1

up vote 3 down vote accepted

Some steps

  • Construct, inductively, for each integer $n$, $2^n+1$ points $z_{n,1},\dots,z_{n,2^n+1}$ such that $d(z_{n,k+1},z_{n,k})\leq 2^{-n\alpha}$ and $z_{n,1}=x$, $z_{n,2^n+1}=y$.
  • Define $f(k2^{-n})=z_{n,k}$. Now we shall extend it in a Hölder continuous map.
  • Let $t\in [0,1]$. We can find a sequence $\{t_j\}$ of the form $\{k_j2^{-n_j}\}$ which converges to $t$. Since $f$ satisfies a Hölder continuity condition, the sequence $f(t_k)$ is Cauchy in $X$.
  • Define $\gamma(t):=\lim_{k\to +\infty}f(t_k)$, checking that it's well-defined and $\alpha$-Hölderian.
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I've inserted an underscore in the first step, instead of $z{n,2^n+1}=y$ –  Siminore Jul 8 '12 at 13:06
    
@DavideGiraudo: great, exceptionally clear. Thank you very much. –  Romeo Jul 8 '12 at 18:49

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