arXiv subjects
Publications and source records attributed to Bruce Hanson.
In this paper we give an example of a closed, strongly one-sided dense set which is not of uniform density type. We also show that there is a set of uniform density type which is not of strong uniform density type.
We characterize the subsets $E \subset \mathbb{R}$ for which there exists a continuous real valued function $f: \mathbb{R}\to\mathbb{R}$ such that lip $f$ is finite everywhere and Lip $f$ is infinite exactly on $E$.
We denote the local ``little" Lipschitz constant of a function $f: {{\mathbb R}}\to { {\mathbb R}}$ by $ {\mathrm{lip}}f$. In this paper we settle the following question: For which sets $E {\subset} { {\mathbb R}}$ is it possible to find a continuous function $f$ such that $ {\mathrm{lip}}f=\mathbf{1} _E$? In an earlier paper we introduced the concept of strongly one-sided dense sets. Our main result characterizes $ {\mathrm{lip}}1$ sets as countable unions of closed sets which are strongly one-sided dense. We also show that a stronger statement is not true i.e. there are strongly one-sided dense $F _\sigma$ sets which are not $ {\mathrm{lip}}1$.
Given a continuous function $f: {{\mathbb R}}\to {{\mathbb R}}$ we denote the so-called "big Lip" and "little lip" functions by $ {{\mathrm {Lip}}} f$ and $ {{\mathrm {lip}}} f$ respectively}. In this paper we are interested in the following question. Given a set $E {\subset} {{\mathbb R}}$ is it possible to find a continuous function $f$ such that $ {{\mathrm {lip}}} f=\mathbf{1}_E$ or $ {{\mathrm {Lip}}} f=\mathbf{1}_E$? For monotone continuous functions we provide the rather straightforward answer. For arbitrary continuous functions the answer is much more difficult to find. We introduce the concept of uniform density type (UDT) and show that if $E$ is $G_\delta$ and UDT then there exists a continuous function $f$ satisfying $ {{\mathrm {Lip}}} f =\mathbf{1}_E$, that is, $E$ is a $ {{\mathrm {Lip}}} 1$ set. In the other direction we show that every ${{\mathrm {Lip}}} 1$ set is $G_\delta$ and weakly dense. We also show that the converse of this statement is not true, namely that there exist weakly dense $G_{{\delta}}$ sets which are not $ {{\mathrm {Lip}}} 1$. We say that a set $E\subset \mathbb{R}$ is ${{\mathrm {lip}}} 1$ if there is a continuous function $f$ such that ${{\mathrm {lip}}} f=\mathbf{1}_E$. We introduce the concept of strongly one-sided density and show that every ${{\mathrm {lip}}} 1$ set is a strongly one-sided dense $F_\sigma$ set.
Suppose $\Lambda$ is a discrete infinite set of nonnegative real numbers. We say that $ {\Lambda}$ is type $1$ if the series $s(x)=\sum_{\lambda\in\Lambda}f(x+\lambda)$ satisfies a zero-one law. This means that for any non-negative measurable $f: {{\mathbb R}}\to [0,+ {\infty})$ either the convergence set $C(f, {\Lambda})=\{x: s(x)<+ {\infty} \}= {{\mathbb R}}$ modulo sets of Lebesgue zero, or its complement the divergence set $D(f, {\Lambda})=\{x: s(x)=+ {\infty} \}= {{\mathbb R}}$ modulo sets of measure zero. If $ {\Lambda}$ is not type $1$ we say that $ {\Lambda}$ is type 2. The exact characterization of type $1$ and type $2$ sets is not known. In this paper we continue our study of the properties of type $1$ and $2$ sets. We discuss sub and supersets of type $1$ and $2$ sets and we give a complete and simple characterization of a subclass of dyadic type $1$ sets. We discuss the existence of type $1$ sets containing infinitely many algebraically independent elements. Finally, we consider unions and Minkowski sums of type $1$ and $2$ sets.
Suppose $\Lambda$ is a discrete infinite set of nonnegative real numbers. We say that $ {\Lambda}$ is type $2$ if the series $s(x)=\sum_{\lambda\in\Lambda}f(x+\lambda)$ does not satisfy a zero-one law. This means that we can find a non-negative measurable "witness function" $f: {\mathbb R}\to [0,+ {\infty})$ such that both the convergence set $C(f, {\Lambda})=\{x: s(x)<+ {\infty} \}$ and its complement the divergence set $D(f, {\Lambda})=\{x: s(x)=+ {\infty} \}$ are of positive Lebesgue measure. If $ {\Lambda}$ is not type $2$ we say that $ {\Lambda}$ is type $1$. The main result of our paper answers a question raised by Z. Buczolich, J-P. Kahane, and D. Mauldin. By a random construction we show that one can always choose a witness function which is the characteristic function of a measurable set. We also consider the effect on the type of a set $ {\Lambda}$ if we randomly delete its elements. Motivated by results concerning weighted sums $\sum c_n f(nx)$ and the Khinchin conjecture, we also discuss some results about weighted sums $\sum_{n=1}^{\infty}c_n f(x+\lambda_n)$.
Given a function $f\colon \mathbb{R}\to \mathbb{R}$, the so-called "little lip" function $\operatorname{lip} f$ is defined as follows: \begin{equation*} \operatorname{lip} f(x)=\liminf_{r{\scriptscriptstyle \searrow} 0}\sup_{|x-y|\le r} \frac{|f(y)-f(x)|}{r}. \end{equation*} We show that if $f$ is continuous on $\mathbb{R}$, then the set where $\operatorname{lip} f$ is infinite is a countable union of a countable intersection of closed sets (that is an $F_{\sigma \delta}$ set). On the other hand, given a countable union of closed sets $E$, we construct a continuous function $f$ such that $\operatorname{lip} f$ is infinite exactly on $E$. A further result is that for the typical continuous function $f$ on the real line $\operatorname{lip} f$ vanishes almost everywhere.