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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.)2011-07-07
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    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?2011-07-07
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    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.2011-07-07
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    I would say that the graph of random walk has a dimension zero. Maybe you are talking about continuous Brownian motion? Then what will be the step $n$?2011-07-07
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    Perhaps one interpretation of the question is how the Hausdorff dimension of a fractal set of points changes under a nonlinear action on its elements, e.g. exponentiating all of the points of the Mandelbrot set. If that's the case then I think this is a very interesting question. I get the feeling that if the mapping is continuous, differentiable, and has bounded Jacobian determinant for all points in the set, the dimension won't change.2011-07-07
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    anon, yes, that's the precisely the question I have in mind. Also my feeling is that the Hausdorff dimension does not change. I have google whether there is any theorem related, but I was not able to find it.2011-07-07
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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 $1$.2011-07-07
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    What leftaroundabout said. But I wonder why this is in physics and not in mathematics?2011-07-07
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    A version of this question was kicked here from physics.SE. I've marked as duplicate and merged.2011-07-07

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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.2011-07-07
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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 $d - s$.2011-07-07
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    Thanks GEdgar and Robert Israel. Can you provide some more details how to prove it or can you give a reference for the proof?2011-07-08
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    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.2011-07-08