Jordan plane curve such that $\frac{d(g(x),g(y))}{d(x,y)}\to0$?

Question

Write $g$ as the inverse of $f$.

Is there a continuous injective $f:S^1\to C\subset\mathbb{R}^2$ such that $$ \displaystyle\sup_{d(x,y)<r}\dfrac{d(g(x),g(y))}{d(x,y)}\to0 $$ as $r\to0$?

If you like, add more conditions - including that for each $x\in C$ there exists $r_x>0$ such that for any $\epsilon>0$ you can find $x_1,x_2\in D(x,\epsilon)$ such that $\angle (x_2 ,x,x_1)>r_x$.

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Motivation: somebody asked on StackExchange whether there is a non-constant function with derivative identically zero, on a set in Euclidean space that is path-connected but not necessarily open.

I gave a brief incomplete answer, that if the curve with the above specifications exists, then the person's example will be furnished by $g$.

The person, prompted by someone else's comment, also wondered if the uniqueness of how the derivative at a point is defined has any role - in other words, if $L\subset\mathbb{R}^2$ is a line and a differentiable function on $L\subset\mathbb{R}^2$ is defined, then of course the choice of how the derivative along the orthogonal direction is defined is arbitrary. That is why I suggested an additional condition above.

I have a hunch that this is a question on the level of Overflow. As a non-expert, I'm not sure though. (I got a copy of an interesting book about space-filling curves by Sagan. Unfortunately, the examples of Osgood curves that he provides don't fit this question. I have no more ideas.)

Answer 1

Sure. It is a standard fact that the von Koch snowflake $C$ can be parametrized by $$ f\colon S^1 \rightarrow C$$ where $$ C^{-1}|x-y|^{1/p} \leq |f(x)-f(y)| \leq C|x-y|^{1/p},$$ $p\in (1,2)$ is the Hausdorff dimension of the snowflake, and $C>0$ is some constant.

It follows that the inverse function $g\colon C \rightarrow S^1$ is Holder with constant $p>1$, from which your requirement follows.