# Hölderian path connectedness

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

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