397 research outputs found

    Restrictions of Brownian motion

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    Let {B(t) ⁣:0t1}\{ B(t) \colon 0\leq t\leq 1\} be a linear Brownian motion and let dim\dim denote the Hausdorff dimension. Let α>12\alpha>\frac12 and 1β21\leq \beta \leq 2. We prove that, almost surely, there exists no set A[0,1]A\subset[0,1] such that dimA>12\dim A>\frac12 and B ⁣:ARB\colon A\to\mathbb{R} is α\alpha-H\"older continuous. The proof is an application of Kaufman's dimension doubling theorem. As a corollary of the above theorem, we show that, almost surely, there exists no set A[0,1]A\subset[0,1] such that dimA>β2\dim A>\frac{\beta}{2} and B ⁣:ARB\colon A\to\mathbb{R} has finite β\beta-variation. The zero set of BB and a deterministic construction witness that the above theorems give the optimal dimensions.Comment: 6 page

    Continuous horizontally rigid functions of two variables are affine

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    Cain, Clark and Rose defined a function f\colon \RR^n \to \RR to be \emph{vertically rigid} if \graph(cf) is isometric to \graph (f) for every c0c \neq 0. It is \emph{horizontally rigid} if \graph(f(c \vec{x})) is isometric to \graph (f) for every c0c \neq 0 (see \cite{CCR}). In an earlier paper the authors of the present paper settled Jankovi\'c's conjecture by showing that a continuous function of one variable is vertically rigid if and only if it is of the form a+bxa+bx or a+bekxa+be^{kx} (a,b,k \in \RR). Later they proved that a continuous function of two variables is vertically rigid if and only if after a suitable rotation around the z-axis it is of the form a+bx+dya + bx + dy, a+s(y)ekxa + s(y)e^{kx} or a+bekx+dya + be^{kx} + dy (a,b,d,k \in \RR, k0k \neq 0, s : \RR \to \RR continuous). The problem remained open in higher dimensions. The characterization in the case of horizontal rigidity is surprisingly simpler. C. Richter proved that a continuous function of one variable is horizontally rigid if and only if it is of the form a+bxa+bx (a,b\in \RR). The goal of the present paper is to prove that a continuous function of two variables is horizontally rigid if and only if it is of the form a+bx+dya + bx + dy (a,b,d \in \RR). This problem also remains open in higher dimensions. The main new ingredient of the present paper is the use of functional equations

    Bruckner--Garg-type results with respect to Haar null sets in C[0,1]C[0,1]

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    A set AC[0,1]\mathcal{A}\subset C[0,1] is \emph{shy} or \emph{Haar null } (in the sense of Christensen) if there exists a Borel set BC[0,1]\mathcal{B}\subset C[0,1] and a Borel probability measure μ\mu on C[0,1]C[0,1] such that AB\mathcal{A}\subset \mathcal{B} and μ(B+f)=0\mu\left(\mathcal{B}+f\right) = 0 for all fC[0,1]f \in C[0,1]. The complement of a shy set is called a \emph{prevalent} set. We say that a set is \emph{Haar ambivalent} if it is neither shy nor prevalent. The main goal of the paper is to answer the following question: What can we say about the topological properties of the level sets of the prevalent/non-shy many fC[0,1]f\in C[0,1]? The classical Bruckner--Garg Theorem characterizes the level sets of the generic (in the sense of Baire category) fC[0,1]f\in C[0,1] from the topological point of view. We prove that the functions fC[0,1]f\in C[0,1] for which the same characterization holds form a Haar ambivalent set. In an earlier paper we proved that the functions fC[0,1]f\in C[0,1] for which positively many level sets with respect to the Lebesgue measure λ\lambda are singletons form a non-shy set in C[0,1]C[0,1]. The above result yields that this set is actually Haar ambivalent. Now we prove that the functions fC[0,1]f\in C[0,1] for which positively many level sets with respect to the occupation measure λf1\lambda\circ f^{-1} are not perfect form a Haar ambivalent set in C[0,1]C[0,1]. We show that for the prevalent fC[0,1]f\in C[0,1] for the generic yf([0,1])y\in f([0,1]) the level set f1(y)f^{-1}(y) is perfect. Finally, we answer a question of Darji and White by showing that the set of functions fC[0,1]f \in C[0,1] for which there exists a perfect Pf[0,1]P_f\subset [0,1] such that f(x)=f'(x) = \infty for all xPfx \in P_f is Haar ambivalent.Comment: 12 page
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