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The divider dimension of the graph of a function. (English) Zbl 0756.28004

Let \(f:[a,b]\to\mathbb{R}\) be a continuous function. For \(\varepsilon>0\) consider a maximal \(n=n(\varepsilon)\in\mathbb{N}\) such that for the sequence \((x_ i,f(x_ i))_{i=0,1,\dots,n}\) with \(x_ 0=a\) the Euclidean distance of \((x_ i,f(x_ i))\) and \((x_{i+1},f(x_{i+1}))\) is \(\varepsilon\) for all \(i=0,1,\dots,n-1\) and the distance of \((x_ i,f(x_ i))\) and \((z,f(z))\) is smaller than \(\varepsilon\) for \(z\) with \(x_ i<z<x_{i+1}\). The divider dimension of the graph of \(f\) is the number \(D=\lim_{\varepsilon\to 0}{\ln n\over -\ln\varepsilon}\) provided the limit exists. This dimension index was created in connection with the shape of coastlines. It is larger than Hausdorff dimension as well as box dimension, and it may be happen that it is larger than 2. The authors calculate it for \(0\leq x\leq 1\) for the function \(f(x)=\sum^ \infty_{i=1}\lambda_ i^{s-2}g(\lambda_ ix)\), where \(g\) is a “zig-zag” function of period 4 and \(1<s<2\), to be \(1/(2-s)\). The Hausdorff and box dimension is in this case are equal to \(s\). Finally, the average divider dimension of a one-dimensional Brownian motion on a fixed closed time interval is 2, and the authors conjecture that this must be true almost surely.

MSC:

28A80 Fractals
60J65 Brownian motion
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