For a real-valued function $f$ continuous on a closed interval, the modulus of fractality $\nu(f, \varepsilon)$ is defined for every $\varepsilon > 0$ as the minimum number of squares with sides of length $\varepsilon$ parallel to the coordinate axes that can cover the graph of $f$. For a nonincreasing function $\mu: (0, +\infty) \to (0, +\infty)$, we consider the class $F^{\mu}$ of functions continuous on a closed interval and such that $\nu(f, \varepsilon) = O(\mu(\varepsilon))$. The relationship between the classes $F^{\mu_1}$ and $F^{\mu_2}$ is described for various $\mu_1$ and $\mu_2$. A connection is established between the classes $F^{\mu}$ and the classes of continuous functions of bounded variation $BV_{\Phi}[a, b] \cap C[a, b]$ for arbitrary convex functions $\Phi$. Namely, there is an inclusion
$$
BV_{\Phi}[a,b] \cap C[a,b] \subset F^{\frac{\Phi^{\,-1}(\varepsilon)}{\varepsilon^2}}.
$$
A counterexample is constructed showing that this inclusion cannot be improved. It is further shown that the equality of the classes $F^{\mu}$ and $BV_{\Phi}[a,b] \cap C[a,b]$ occurs only in the case
$$
BV[a, b] \cap C[a,b] = F^{1/\varepsilon},
$$
where $BV[a,b]$ are functions of classical bounded variation. For other cases, a counterexample is constructed showing that if $\mu(\varepsilon)$ grows faster than $\dfrac{1}{\varepsilon}$ as $\varepsilon \to +0$, then the class $F^{\mu} $ is not a subclass of any of the classes $BV_{\Phi}[a, b]$.
Keywords: fractal dimension, bounded variation
Received March 17, 2023
Revised October 20, 2023
Accepted October 23, 2023
Daniil Igorevich Masyutin, Krasovskii Institute of Mathematics and Mechanics of the Ural Branch of the Russian Academy of Sciences, Yekaterinburg, 620108 Russia, e-mail: newselin@mail.ru
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Cite this article as: D.I. Masyutin. On the connection between classes of functions of bounded variation and classes of functions with fractal graph. Trudy Instituta Matematiki i Mekhaniki UrO RAN, 2023, vol. 29, no. 4, pp. 155–168.
