Level sets of prevalent Hölder and Weierstrass functions

Typical continuous real functions

What does a “typical” real function on the unit interval \([0,1]\) look like? Is it smooth and predictable, like a straight line, or wild and chaotic, with sharp twists and turns? We examine Hölder functions and Weierstrass functions in this sense, focusing on their level sets—the sets of points where the function takes a specific value. These sets offer insights into the function’s behavior, revealing how “wild” or “tame” it is through the lens of dimension theory. We will explore two notions of “typicality” for functions: Baire typicality, which looks at what is common in a topological sense, and prevalence, which considers what takes up most of the “space” among all possible functions.

Behavior of Baire typical continuous functions

Intuitively, the graph of a smooth function looks like a gentle curve, while a chaotic one could resemble a jagged, space-filling scribble. To quantify this, we consider Hölder continuous functions, which are defined by how much they can “wiggle” over small distances. A function \(f \colon [0,1] \to \mathbb{R}\) is called \(\alpha\)-Hölder if there exists a constant \(C > 0\) such that

\[|f(x) - f(y)| \le C |x - y|^\alpha\]

for all \(x, y \in [0,1]\). Here, \(0<\alpha<1\) controls the function’s roughness: smaller \(\alpha\) means wilder behavior. For example, when \(\alpha\) is close to zero, the function can change rapidly, making its graph more complex. The upper Minkowski dimension of the graph of an \(\alpha\)-Hölder function is bounded by

\[\overline{\dim}_{\mathrm{M}}(\text{graph}(f)) \le 2 - \alpha.\]

This suggests that as \(\alpha\) gets smaller, the graph can become more intricate, approaching a “space-filling” curve in the plane in terms of its Haudorff dimension.

Recall that a set is nowhere dense if its closure has empty interior. A property is Baire typical if it is satisfied in a set whose complement is a countable union of nowhere dense sets. A set \(A\) is \(G_\delta\) if there exists a countable collection of open sets \(\{U_n\}_{n=1}^\infty\) such that

\[A = \bigcap_{n=1}^\infty U_n.\]

Here “\(G\)” stands for “Gebiet” which is German for “region” or “area”, and “\(\delta\)” indicates that there is a countable intersection.

Lemma 1. Let \(X\) be a complete metric space. If \(A \subseteq X\) is a dense \(G_\delta\) set, then its complement \(X \setminus A\) is a countable union of nowhere dense sets.

Proof. Let us assume that \(A \subseteq X\) is a dense \(G_\delta\) set. Then \(A\) intersects every non-empty open set in \(X\) and there exists a countable collection of open sets \(\{U_n\}_{n=1}^\infty\) such that \(A = \bigcap_{n=1}^\infty U_n\). The complement of \(A\) is a countable union of closed sets,

\[X \setminus A = \bigcup_{n=1}^\infty X \setminus U_n.\]

It remains to show each \(X \setminus U_n\) is nowhere dense, i.e., has empty interior. Suppose, for contradiction, that some \(X \setminus U_k\) has non-empty interior, so there exists a non-empty open set \(V \subseteq X \setminus U_k\). Then \(V \cap U_k = \emptyset\), and as \(A \subseteq U_k\), also \(V \cap A = \emptyset\). This contradicts the density of \(A\), as \(A\) fails to intersect the non-empty open set \(V\). Thus, each \(X \setminus U_n\) has empty interior and is nowhere dense. ■

The following result shows that the graph of a Baire typical continuous function has Hausdorff dimension one.

Theorem 2 (Mauldin-Williams (1986)). A Baire typical continuous function \(f \colon [0,1] \to \mathbb{R}\) satisfies:

\[\dim_{\mathrm{H}}(\mathrm{graph}(f)) = 1.\]

Proof. Recall that the vector space \(\mathcal{C}([0,1])\) of continuous real functions on \([0,1]\) equipped with the the supremum norm \(\Vert f \Vert_{\infty} = \sup_{x \in [0,1]} \vert f(x) \vert\) is a complete metric space. Therefore, by Lemma 1, it suffices to show that the set

\[A = \{ f \in \mathcal{C}([0,1]) : \dim_{\mathrm{H}}(\mathrm{graph}(f)) = 1 \}\]

is a dense \(G_\delta\) set in \(\mathcal{C}([0,1])\). Recall that, by the Stone-Weierstrass theorem, polynomials are dense in \(\mathcal{C}([0,1])\). The graph of a polynomial is a smooth algebraic curve, which has Hausdorff dimension one, and polynomials are thus contained in \(A\). Hence, the set \(A\) is dense in \(\mathcal{C}([0,1])\). To show that \(A\) is \(G_\delta\), define

\[A_{n,k} = \{ f \in \mathcal{C}([0,1]) : \mathcal{H}^{1+\frac{1}{n}}_\infty(\mathrm{graph}(f)) < \tfrac{1}{k} \}\]

for all \(n,k \in \mathbb{N}\). Note that if \(f \in \bigcap_{k=1}^\infty A_{n,k}\), then \(\mathcal{H}^{1+\frac{1}{n}}_\infty(\mathrm{graph}(f)) = 0\) and \(\dim_{\mathrm{H}}(\mathrm{graph}(f)) \leq 1+\frac{1}{n}\). Hence,

\[A = \bigcap_{n=1}^\infty \bigcap_{k=1}^\infty A_{n,k}.\]

It remains to show that each \(A_{n,k}\) is open. To that end, notice that if \(\varepsilon>0\) and \(f,g \in \mathcal{C}([0,1])\) are such that \(\|f-g\|_{\infty} < \varepsilon/\sqrt{2}\), then \(d_{\mathrm{H}}(\mathrm{graph}(f),\mathrm{graph}(g))<\varepsilon\), where \(d_{\mathrm{H}}\) is the Hausdorff distance. It is also easy to see that the Hausdorff content \(\mathcal{H}^{1+\frac{1}{n}}_\infty\) as a function defined on the space of compact sets of \(\mathbb{R}^2\) equipped with \(d_{\mathrm{H}}\) is upper semicontinuous. Therefore, the function \(f \mapsto \mathcal{H}^{1+\frac{1}{n}}_\infty(\mathrm{graph}(f))\) defined on \(\mathcal{C}([0,1])\) is upper semicontinuous and \(A_{n,k}\) is open. ■

Thus, a Baire typical continuous function behaves like a simple line in terms of its Hausdorff dimension, which measures the complexity of a set. This is surprising because, intuitively, we might expect a typical function to be wilder, especially for small \(\alpha\), where the dimension of an \(\alpha\)-Hölder function’s graph can approach \(2 - \alpha.\)

Level sets of prevalent Hölder continuous functions

A set of Hölder functions is prevalent if it dominates the function space in a measure-theoretic sense. Formally, a set in a Banach space is prevalent if there exists a \(d\)-dimensional probe space \(\mathcal{S}\) such that for any Hölder function \(f\), adding elements from \(\mathcal{S}\) keeps the function in the desired set for \(\mathcal{L}^d\)-almost all choices in \(\mathcal{S}\). Recall that \(\mathcal{S}\) is isometrically isomorphic to \(\mathbb{R}^d\) and hence, \(\mathcal{L}^d\) here is indeed the \(d\)-dimensional Lebesgue measure via this identification.

Theorem 3 (Clausel-Nicolay (2010)). A prevalent \(\alpha\)-Hölder function \(f \colon [0,1] \to \mathbb{R}\) satisfies:

\[\dim_{\mathrm{H}}(\text{graph}(f)) = 2 - \alpha.\]

This means that as \(\alpha\) approaches zero, a prevalent Hölder function’s graph becomes, in terms of its Hausdorff dimension, nearly space-filling, almost filling up a two-dimensional plane while remaining continuous—a remarkable balance of complexity and continuity.

We seek a deeper understanding of how the function \(f\) varies across the unit interval. The level set \(f^{-1}(\{y\})\) consists of all points \(x\) where the function value \(f(x) = y\). Geometrically, this represents the intersection of the function’s graph with a horizontal line at height \(y\). The Hausdorff dimension of these level sets quantifies the function’s irregularity. For functions modeling random processes on \([0,1]\), it can also describe the process’s scaling properties. For smooth functions, level sets are typically discrete sets with a Hausdorff dimension of zero. In contrast, highly irregular or fractal-like functions may have level sets with positive Hausdorff dimension, indicating a more intricate structure. To estimate the Hausdorff dimension, Marstrand’s classical slicing theorem provides a valuable starting point.

Theorem 4 (Marstrand (1954)). A function \(f \colon [0,1] \to \mathbb{R}\) satisfies:

\[\dim_{\mathrm{H}}(f^{-1}(\{y\})) \le \dim_{\mathrm{H}}(\text{graph}(f)) - 1\]

for \(\mathcal{L}^1\)-almost all \(y \in \mathbb{R}\).

Recalling Theorem 3, the slicing theorem suggests that a prevalent \(\alpha\)-Hölder function’s level sets have Hausdorff dimension at most \(1-\alpha\). Can we make this bound hold for all \(y\) and when does equality hold? The following result tells us that for a prevalent Hölder function, the level sets are well-behaved, with their upper Minkowski dimension capped at \(1 - \alpha\) when \(\alpha\) is small, and many level sets achieve the maximum Hausdorff dimension of \(1 - \alpha\).

Theorem 5 (Anttila-Bárány-K (2025)). A prevalent \(\alpha\)-Hölder function \(f \colon [0,1] \to \mathbb{R}\) satisfies:
(1) \(\overline{\dim}_{\mathrm{M}}(f^{-1}(\{y\})) \le 1 - \alpha\) for all \(y \in \mathbb{R}\) if \(0 < \alpha < \frac{1}{2}\),
(2) \(\dim_{\mathrm{H}}(f^{-1}(\{y\})) = 1 - \alpha\) for all \(y\) in a set with positive Lebesgue measure if \(0 < \alpha < 1\).

Proof. To prove this, we use an Assouad embedding, which maps the interval \([0,1]\) into a higher-dimensional space \(\mathbb{R}^d\) using an \(\alpha\)-bi-Hölder map \(\Phi\). We choose a probe space \(\mathcal{S}\) to be the functions of the form \(x \mapsto \langle \Phi(x), \mathbf{t} \rangle\), where \(\mathbf{t} \in \mathbb{R}^d\). The goal is to show that the properties hold for \(f_{\mathbf{t}} = f + \langle \Phi, \mathbf{t} \rangle\) for \(\mathcal{L}^d\)-almost all \(\mathbf{t}\).

In the proof of (1), we analyze \(f_{\mathbf{t}}\) using Fourier analysis. Define the occupation measure \(\mu_{\mathbf{t}} = (f_{\mathbf{t}})_*\mathcal{L}^1\), which describes how the function distributes Lebesgue measure. The task is to show that

\[\int \mathcal{I}_s(\mu_{\mathbf{t}}) \, \mathrm{d}\mathbf{t} < \infty\]

over all bounded sets for \(s > 2\), where

\[\mathcal{I}_s(\mu_{\mathbf{t}}) = \int |\xi|^{s-1} |\hat{\mu}_{\mathbf{t}}(\xi)|^2 \, \mathrm{d}\xi\]

is the Sobolev energy of \(\mu_{\mathbf{t}}\). Having settled this, Cauchy–Schwarz inequality then gives us

\[\int_{|\xi|>1} |\hat{\mu}_{\mathbf{t}}| \, \mathrm{d}\xi \le \biggl( \int_{|\xi|>1} |\xi|^{s-1} |\hat{\mu}_{\mathbf{t}}(\xi)|^2 \, \mathrm{d}\xi \biggr)^{1/2} \biggl( \int_{|\xi|>1} |\xi|^{-(s-1)} \, \mathrm{d}\xi \biggr)^{1/2} < \infty\]

for \(\mathcal{L}^d\)-almost all \(\mathbf{t} \in \mathbb{R}^d\). The second integral converges precisely when \(s-1 > 1\), and the first is finite by assumption. The integral over \(\vert\xi\vert \leq 1\) is finite since \(\hat{\mu}_{\mathbf{t}}\) is bounded. It follows that \(\hat{\mu}_{\mathbf{t}} \in L^1(\mathbb{R})\) and hence, the Fourier inversion theorem implies that \(\mu_{\mathbf{t}}\) has a density \(g(x) = \frac{1}{2\pi} \int \hat{\mu}_{\mathbf{t}}(\xi) e^{i x \xi} \, \mathrm{d}\xi\). The inverse Fourier transform of an \(L^1\) function is continuous, so the density \(g\) is continuous. It is also bounded, because \(\vert g(x) \vert \leq \frac{1}{2\pi} \Vert\hat{\mu}_{\mathbf{t}}\Vert_{L^1} < \infty\). By the Radon-Nikodym theorem, we have \(\mu_{\mathbf{t}}(A) = \int_A g \,\mathrm{d}\mathcal{L}^1\) and hence,

\[\mu_{\mathbf{t}}(B(y,r')) \le Cr',\]

where \(C = \frac{1}{2\pi} \Vert\hat{\mu}_{\mathbf{t}}\Vert_{L^1}\). Let \(\{B(x_i,r)\}_{i=1}^{N_r}\) be a maximal \(r\)-packing of \(f^{-1}(\{y\})\). Note that \(y = f(x_i)\) and \(B(x_i,r) \cap B(x_j,r) = \emptyset\) whenever \(i \ne j\). Since \(|f(x_i)-f(z)| \le C|x_i-z|^\alpha \le Cr^\alpha\) for all \(z \in B(x_i,r)\), we have

\[\mu_{\mathbf{t}}(B(y,Cr^\alpha)) = \mathcal{L}^1(f^{-1}(B(y,Cr^\alpha))) \ge \sum_{i=1}^{N_r}\mathcal{L}^1(B(x_i,r))= 2N_rr.\]

Therefore,

\[\begin{align*} \overline{\dim}_{\mathrm{M}}(f^{-1}(\{y\})) &= \limsup_{r \downarrow 0} \frac{\log N_r}{\log r^{-1}} \\ &\le \limsup_{r \downarrow 0} \frac{\log \frac12 r^{-1}\mu_{\mathbf{t}}(B(y,Cr^\alpha))}{\log r^{-1}} \le \limsup_{r \downarrow 0} \frac{\log Cr^{\alpha-1}}{\log r^{-1}} \le 1-\alpha, \end{align*}\]

proving part (1).

To prove (2), the idea in Anttila-Bárány-K (2025) was to look at \(\theta\)-oblique occupation measures defined as \(\mu_{\mathbf{t}}^\theta = (\mathrm{proj}_\theta)_*\lambda_{\mathbf{t}}\), where \(\lambda_{\mathbf{t}}\) is the lift of the Lebesgue measure from the unit interval onto the graph of \(f_{\mathbf{t}}\) and \(\mathrm{proj}_\theta(x,y) = x\cos(\theta)+y\sin(\theta)\) is the orthogonal projection onto direction \(\theta\). The occupation measure \(\mu_{\mathbf{t}}\) is thus the \(\frac{\pi}{2}\)-oblique occupation measure \(\mu_{\mathbf{t}}^{\frac{\pi}{2}}\). The task is to show that \(\mu_{\mathbf{t}}^\theta \ll \mathcal{L}^1\) and, relying on the Marstrand’s slicing theorem,

\[\dim_{\mathrm{H}}(\mathrm{graph}(f_{\mathbf{t}}) \cap \mathrm{proj}_\theta^{-1}(\{y\})) \ge 1-\alpha\]

for \(\mathcal{L}^d\)-almost all \(\mathbf{t} \in \mathbb{R}^d\), \(\mathcal{L}^1\)-almost all \(\theta \in [0,2\pi]\), and \(\mu_{\mathbf{t}}^\theta\)-almost all \(y \in \mathrm{proj}_\theta(\mathrm{graph}(f_\mathbf{t}))\). There is a priori no reason that this observation holds for the slice with \(\theta = \frac{\pi}{2}\) corresponding to the level sets. However, by increasing the dimension of the probe space \(\mathcal{S}\), it is possible to adjust the slice direction by adding a linear function to the Hölder function. This finishes the proof. ■

Level sets of prevalent Weierstrass functions

Weierstrass functions are a special class of Hölder functions, defined as

\[W_g^{\alpha,b}(x) = \sum_{k=0}^\infty b^{-\alpha k} g(b^k x),\]

where \(g\) is a Lipschitz function, \(b \ge 2\) is an integer, and \(0<\alpha<1\). These \(1\)-periodic functions are famous for being continuous but nowhere differentiable, making them a classic example of “wild” functions. In fact, the Weierstrass function is the first published example (1872) of a nowhere differentiable continuous function.

Theorem 6 (Ren-Shen (2021)). For any non-constant \(g\), all but finitely many \(0<\alpha<1\), and any integer \(b \ge 2\), an \(\alpha\)-Weierstrass function \(W_g^{\alpha,b} \colon [0,1] \to \mathbb{R}\) satisfies

\[\dim_{\mathrm{H}}(\text{graph}(W_g^{\alpha,b})) = 2 - \alpha.\]

This confirms that essentially all \(\alpha\)-Weierstrass functions, like prevalent \(\alpha\)-Hölder functions, have graphs that are, in terms of their Hausdorff dimension, nearly space-filling as \(\alpha\) approaches zero. We note that \(\alpha\)-Weierstrass functions are a strict subset of \(\alpha\)-Hölder functions. Indeed, we identify \(W_g^{\alpha,b}\) with the Lipschitz function \(g\) and define prevalence via this identification. The following theorem extends Theorem 5 to Weierstrass functions:

Theorem 7 (Buczolich-K-Maga (2025)). For any integer \(b \ge 2\), a prevalent \(\alpha\)-Weierstrass function \(W_g^{\alpha,b} \colon [0,1] \to \mathbb{R}\) satisfies:
(1) \(\overline{\dim}_{\mathrm{M}}((W_g^{\alpha,b})^{-1}(\{y\})) \le 1 - \alpha\) for all \(y \in \mathbb{R}\) if \(0 < \alpha < \frac{1}{2}\),
(2) \(\dim_{\mathrm{H}}((W_g^{\alpha,b})^{-1}(\{y\})) = 1 - \alpha\) for all \(y\) in a set with positive Lebesgue measure if \(0 < \alpha < 1\).

Proof. To prove (1), the task is to construct an \(\alpha\)-bi-Hölder Weierstrass embedding \(\Phi \colon [0,1] \to \mathbb{R}^d\) whose coordinate functions that are \(\alpha\)-Weierstrass functions, and proceed as in the proof of Theorem 5(1). However, in proving Theorem 5(2) the idea was first to analyze slices in almost every direction and then adjust the slice direction by adding a linear function to the Hölder function. In the Banach space of \(\alpha\)-Weierstrass functions, which consists of \(1\)-periodic functions, this adjustment is not feasible. Consequently, in addition to constructing an \(\alpha\)-bi-Hölder Weierstrass embedding, the task is to directly analyze the occupation measure of \(W_g^{\alpha,b} + \langle \Phi, \mathbf{t} \rangle\), showing it is absolutely continuous for \(\mathcal{L}^d\)-almost all \(\mathbf{t} \in \mathbb{R}^d\). ■

The method to prove Theorem 7(2) streamlines the proof of Theorem 5(2), showing that for prevalent \(\alpha\)-Hölder functions, the result can be established without relying on the “almost every rotation” argument. There also appears to be an interesting dichotomy: When \(\alpha\) is small, a prevalent Weierstrass function exhibits “wilder” behavior, yet its occupation measure’s density function may possess more favorable properties as we know it is continuous and bounded when \(0<\alpha<\frac12\).