Is there a two-dimensional unimodal function with fractal level sets

Question

Is there an open simply connected $U\subset\mathbb{R}^2$ and a continuous non-constant function $f: U\to \mathbb{R}$, such that for all $c\in \mathbb{R}$ both sets $$ f_{<c}~=~ f^{-1}\left( (-\infty,c)\right),~~~{\rm and}~~ f_{>c}~=~ f^{-1}\left( (c,\infty)\right)$$ are connected and all level sets $f_c=f^{-1}(c)$ are either empty, a one point set, or have Hausdorff dimension $\dim_H f_c >1$?

The question is related to Is there always a way up? because such a function might be a non-trivial example for a function with no upward paths. So far I considered three lines of argumentation:

1. It is possible to construct a continuous function on a ring where inner and outer boundaries are fractal curves by solving the Laplace equation with different constant potentials at the two boundary components. By nesting such rings it seems possible to construct a function with infinitely many fractal level sets, but in between all level sets are smooth.

2. Plate 187 in Mandelbrot (The fractal geometry of nature) shows a three dimensional structure where each cross-section is a Julia set. However this is not a function over an open subset $U\subset\mathbb{R}^2$ and also fractal landscapes as for example generated by Brownian motion do not fulfill the connectedness conditions imposed here, but it is maybe possible to fix this.

3. A dimensional argument indicates that such a function might not exist, because it decomposes $U$ into a one-dimensional family of sets $f_c$ with dimensions $>1$ such that in some sense $$2~=~ \dim_H U ~=~ 1 + \langle\dim_H f_c \rangle ~>2.$$ A possible starting point for this argument is Kirchheim (Trans. AMS, 347(5) 1995), but it seems not to apply directly.

Answer 1

I think this is essentially equivalent to Martin Hairer's comment but expressed a bit differently.

Let $F$ be any continuous, positive valued function defined on $[0,2\pi]$ with the property that $F(0)=F(2\pi)$ whose graph has Hausdorff dimension larger than $1$. A natural choice is this little variation of Weierstrass's classic nowhere differentiable function: $$ F(x) = 3 - \sum_{n=0}^{\infty} a^n \cos( b^n \, x), $$ for appropriate values of $a$ and $b$. Then, in polar coordinates, let $$ f(r,\theta) = 1-\frac{r}{F(\theta)}. $$ For $a=0.4$ and $b=5$, we get a function $f$ that varies from a maximum of 1 at $(0,0)$ down to 0 on the contour line with $r=F(\theta)$ that decreases linearly along each ray $\theta=$const. By choosing $U=\{(r,\theta):r< F(\theta)\}$ this provides a non-constant function where all non-trivial level sets have Hausdorff dimension $>1$.