$f$ is a real function and it is $\alpha$-Holder continuous with $\alpha>1$. Is $f$ constant?

Question

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.

Answer 1

Counterexample

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 $f(x)=f(y)$. $\quad\Box$

Answer 2

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 $\Omega$.

Answer 3

(I am using $f: X \to Y$ as this should also hold for arbitrary metric spaces $X$ and $Y$, the definition of $\alpha$-Hölder-continuous becomes “$\exists L < \infty$ with $d\big( f(x_0), f(x_1) \big) \leq L \cdot d( x_0, x_1 )^\alpha$ for all $x_0,x_1 \in X$”.)

We now need $X$ to have the property that any two points are connected by a $\beta$-Hölder-continuous path for $\beta > \frac{1}{\alpha}$. (Note that this is a slightly weaker condition than requiring the path to be Lipschitz, which in turn is equivalent to it having finite-length/being rectifiable)

Assume we have $\alpha > 1$, $f: X \to Y$ $\alpha$-Hölder-continuous, and two points $x_0 \neq x_1 \in X$ such that there exists some $\beta$-Hölder-continuous path $\gamma$ between them with $\beta > \frac{1}{\alpha}$. We will show that this implies $f(x_0) = f(x_1)$:

By our requirement on $\gamma$ there exists a constant $C < \infty$ such that $d\big( \gamma(t), \gamma(r) \big) \leq C \cdot d( x_0, x_1) \cdot |t-r|^\beta$ for all $t,r \in [0,1]$.

Let us define $g: [0,1] \to \mathbb{R}$ as $$ g(t) := \frac{ d\big( f(x_0), f(\gamma(t)) \big) }{ L \cdot d( x_0, x_1 )^\alpha }. $$

We now have, for any $t,r \in [0,1]$, $$\begin{align} | g(t) - g(r) | &= \frac{\left| d\big( f(x_0), f(\gamma(t)) \big) - d\big( f(x_0), f(\gamma(r)) \big) \right|}{ L \cdot d(x_0, x_1)^\alpha } \\ &\leq \frac{d\big( f(\gamma(t)), f(\gamma(r)) \big)}{ L \cdot d( x_0, x_1 )^\alpha } \\ &\leq \frac{ L \cdot d\big( \gamma(t), \gamma(r) \big)^\alpha }{ L \cdot d(x_0, x_1)^\alpha } \\ &\leq \frac{ L C^\alpha \cdot d(x_0, x_1)^\alpha \cdot |t-r|^{\alpha\beta} }{ L \cdot d(x_0, x_1)^\alpha } \\ &=C^\alpha \cdot |t-r|^{\alpha\beta} \end{align}$$ (here we first used the triangle-inequality and then twice Hölder-continuity)

We see that $g: [0,1] \to \mathbb{R}$ is $\alpha\beta$-Hölder-continuous, with $\alpha\beta > \frac{\alpha}{\alpha} = 1$, so by the known result it is constant on $(0,1)$, and therefore also constant on $[0,1]$.

Therefore we have $$ g(0) = \frac{ d\big( f(x_0), f(x_0) \big) }{ L \cdot d( x_0, x_1 )^\alpha } = 0 = g(1) = \frac{ d\big( f(x_0), f(x_1) \big) }{ L \cdot d( x_0, x_1 )^\alpha } $$ implying $$ 0 = d\big( f(x_0), f(x_1) \big) \quad\iff\quad f(x_0) = f(x_1) .$$