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Let's assume X(s) is a fractal surface with Hausdorff dimension D. Now we take a nonlinear transformation f which transforms X(s) to f(X(s)). In this case, what will be the Hausdorff dimension of the transformed surface f(X(s))?

More clarification (asked by Theo Buehler) > Let's start with a simple example of one dimensional random walk. The path of the random walker becomes a fractal with the Hausdorff dimension 1.5. Let's call the path $X(t)$ at time $t$. Then we can think about the path $Y(t)=X(t)^3−2X(t)$. What will be the Hausdorff dimension of $Y(t)$?

Added (by anon) > Let's add the condition for $f$. $f$ is continuous, differentiable and bounded. In this case, will the Hausdorff dimension remain the same?

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Could you please make your hypotheses a bit more specific? I mean, what exactly is a fractal surface and more crucially what kind of non-linear transformation do you have in mind? (Maybe this is standard jargon in the theory of fractals having a special meaning but as I understand it, pretty much anything could happen.) –  t.b. Jul 7 '11 at 9:26
    
Let's start with a simple example of one dimensional random walk. The path of the random walker becomes a fractal with the Hausdorff dimension 1.5. Let's call the path $X(n)$ at step n. Then we can think about the path $Y(n) = X(n)^3-2 X(n)$. What will be the Hausdorff dimension of $Y(n)$? Does this clarify the question? –  Frederik Larsen Jul 7 '11 at 9:34
    
Thanks for the clarification, indeed it does! It would be a good idea to add this specific example and maybe some variants to your question because not everybody bothers to read the comments. Unfortunately, this is far too remote from my own field that I could say anything intelligent. –  t.b. Jul 7 '11 at 9:46
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You will need to say a bit more about this transformation! Otherwise, $f$ could be anything from a diffeomorphism (in which case $f(X)$ has the same dimension as $X$) down to a function that maps all of $X$ onto the unit circle or something, in which case $f(X)$ has dimension $1$. –  leftaroundabout Jul 7 '11 at 11:02
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What leftaroundabout said. But I wonder why this is in physics and not in mathematics? –  Raskolnikov Jul 7 '11 at 11:18

1 Answer 1

up vote 4 down vote accepted

If a map $f$ is Lipschitz ( that is, $|f(x)-f(y)| \le M |x-y|$ for some constant $M$ ), then is does not increase Hausdorff dimension: $\dim f(K) \le \dim K$. An example of a Lipschitz function is one that is differentiable with bounded derivative. Dimension does not increase, but it may decrease.

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Moreover, locally Lipschitz is enough (in a Lindelöf space). And of course if $f$ has an inverse function that is locally Lipschitz the dimension won't decrease. –  Robert Israel Jul 7 '11 at 16:17
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On the other hand, suppose $f: X \to Y$ is locally Lipschitz, all closed balls in $Y$ are compact, $X$ has Hausdorff dimension $d$, and $f^{-1}(\{y\})$ has Hausdorff dimension $s > 0$ for every $y \in f(X)$, then $f(X)$ has Hausdorff dimension at most $d - s$. –  Robert Israel Jul 7 '11 at 16:36
    
Thanks GEdgar and Robert Israel. Can you provide some more details how to prove it or can you give a reference for the proof? –  Frederik Larsen Jul 8 '11 at 7:29
    
I have found a reference. Lemma 6.1 in ``The Geometry of Fractal Sets'' by K. Falconer is the lemma I have been looking for. Thanks again to GEdgar and Robert Israel. –  Frederik Larsen Jul 8 '11 at 8:44

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