2

1

I am reading Mandelbrot, and stubling upon his use of the limit ("almost a Hölder exponent") \lim_{\epsilon -> 0} log(f(x+\epsilon) - f(x))/log(\epsilon).

To simplify, lets assume that f is non-decreasing, and of course the limit of \epsilon is from the right. Calling this (me, not Mandelbrot!) the "log-derivative", we can calculate log-derivative x^\alpha at x=0 is \alpha (\alpha>0) For a differentiable function, such that the first derivative is positive, we can calculate (by Taylor) the log-derivative is 1, and if the function is almost a constant, such that many derivatives are zero, and k being the leasdt integer such that the kth derivative is positive, the log-derivative is k. And so on. The idea of course is that this will be defined, at least for some non-diferentiable function.

My question: Does this procedure have an official name? any references? It is difficult to google because log-derivative means something else ...

flag

1 Answer

1

Suppose $f(x) = \int_0^x d\mu(t)$ for some measure $\mu$. Then your log-derivative is (almost) equivalent to the quantity $\alpha(x) = \lim_{r \to 0} \frac{\log \mu(]x-r,x+r[)}{\log r}$. This is known as the local dimension of $\mu$ at $x$, and it has been much studied in geometric measure theory. One particular topic I know to be connected with Mandelbrot's work on random measures is called multifractal analysis, which refers to the study of the level sets of the local dimension.

link|flag
If $\alpha$ exists, it is the proper exponent $s$ to use in the limit $$ \lim_{r \to 0}\frac{\mu[x-r,x+r]}{r^s} . $$ This goes to $0$ or $\infty$ according as $s<\alpha$ or $s>\alpha$, and has a chance for a positive,finite limit only if $s=\alpha$. – Gerald Edgar Mar 27 2011 at 12:48

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.