Maybe this is a well know result, however, I could not find it. Before stating it, let me write here a well know result (at least for me)

Assume that $\Omega\subset\mathbb{R}^N$ is a open domain and $f:\Omega\to\mathbb{R}$. If there is constants $L>0$ and $\alpha>1$ such that $$|f(x)-f(y)|\leq L |x-y|^\alpha,\ \forall\ x,y\in\Omega$$ then, $f$ is constant in each connected componente of $\Omega$.

The above result can be proved, for example, by showing that $\nabla f=0$ and then we join points in the same connected component by a continuous curve.

Now my question is:

Assume that $\Omega\subset\mathbb{R}^N$ and $f:\Omega\to\mathbb{R}$. Suppose that there is constants $L>0$ and $\alpha>1$ such that $$|f(x)-f(y)|\leq L |x-y|^\alpha,\ \forall\ x,y\in\Omega$$ Can we conclude that $f$ is constant in each connected componente of $\Omega$?

Maybe it is necessary to add the hypothesis that each connected component of $\Omega$ is pathwise connected?

Remark: Note that in the question, $\Omega$ does not need to be a open set. It is now any set.

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    $\begingroup$ I think, the answer is negative for (at least some) fractal curves, however, I have to think about a clean proof. $\endgroup$ Nov 24, 2013 at 20:22

2 Answers 2



As usual, studiosus is right: the answer is negative. A natural parametrization of an arc of the von Koch snowflake gives a topological embedding $g:[0,1]\to \mathbb R^2$ such that $$|g(x)-g(y)|\ge C|x-y|^{p},\quad p=\frac{\log 3}{\log 4}$$ for all $x,y\in [0,1]$, with $C$ independent of $x,y$. The inverse $f=g^{-1}$ is a continuous map from a curve to $[0,1]$ which is Hölder continuous with exponent $1/p>1$.

More generally, for every $p\in (0,1)$ there is a topological embedding $g$ of $\mathbb R $ into a Euclidean space such that $$C |x-y|^p\le |g(x)-g(y)|\le C'|x-y|^{p}$$ for all $x,y\in\mathbb R $. This can be constructed directly, or obtained as a special case of Assouad's embedding theorem. The inverse of $g$ is Hölder continuous with exponent $1/p$ which can be arbitrarily large.

Positive result

To conclude that $f$ is constant, you need an additional geometric (not just topological) assumption on $\Omega$. It suffices to assume that $\Omega$ is quasiconvex (there is $C$ such that every two points $x,y\in \Omega$ can be joined by a curve of length at most $C|x-y|$).

Here is a weaker assumption: for every $x,y\in \Omega$ there is a connected set $E\subset \Omega$ that contains both $x$ and $y$ and has Hausdorff dimension less than $\alpha$.

Proof. Given $x,y\in\Omega$, take $E$ as above. The behavior of Hausdorff dimension under $\alpha$-Hölder maps is well known: $\operatorname{dim} f(E)\le \alpha \operatorname{dim} E$. Hence $\operatorname{dim} f(E)<1$. On the other hand, $f(E)$ is a connected subset of $\mathbb R$, i.e., either a point or an interval. Thus, $f(E)$ is a point, and $f(x)=f(y)$. $\quad\Box$


Claim: For $f$ as stated, and $p > 1$, assume that the region $\Omega$ is path connected by continuous paths of finite total variation. Then $f$ is constant on $\Omega$.

To see this, let $x$, $y$ be given, and choose a path $v(t) : [0,1]\rightarrow \Omega$ of finite arc length $l(v)$ that connects $x$ to $y$. Without loss of generality assume $l(v)\ne 0$ (otherwise $f(x)=f(y)$.) Let $\epsilon > 0$ be given. Find a partition $$ {0=t_{0} < t_{1} < \cdots t_{n}=1} $$ refined enough so that $|v(t_{j-1})-v(t_{j})| < (\epsilon/2l(v))^{1/(p-1)}$ for all $j$, which is possible because $x$ is continuous on $[0,1]$ and, hence, also uniformly continuous on $[0,1]$. Then $$ \begin{align} |f(x)-f(y)| & \le \sum_{j=1}^{n}|f(v(t_{j-1}))-f(v(t_{j}))| \\ & \le L\sum_{j=1}^{n}|v(t_{j-1})-v(t_{j})|^{p-1+1} \\ & \le \frac{\epsilon}{2l(v)}\sum_{j=1}^{n}|v(t_{j})-v(t_{j-1})|\le \frac{\epsilon}{2} < \epsilon. \end{align} $$ Because $\epsilon$ was arbitrary, then $f(x)=f(y)$. This is true for all points $y$ connected to $x$ by such paths, which is everything in this case. So $f$ is constant on $\Omega$.

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    $\begingroup$ Right, I don't know why I wrote "quasiconvex" in my answer, when "rectifiably connected" would suffice. (The Hausdorff dimension argument also covers the rectifiably connected case, since the image of a rectifiable curve has dimension 1.) $\endgroup$ Jan 7, 2014 at 2:28

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