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Hi, Please consider this object: Start with a realization of Brownian motion in 2D, which I'll denote by rho(t) where -infinity < t < +infinity. Next, lets smooth rho. There are various ways of doing this. To save space, I'll leave it to your imagination. But the point is that at short length scales the smoothed rho looks like a smooth continuous curve, but as you zoom out, it looks more and more like the original Brownian path. By the way, in the process of smoothing rho, I think the original time parameter is effectively lost.

So I claim two things.

1) You can find a circle, C, such that the smoothed rho never enters inside C.

2) Let us invert the smoothed rho about the circle C, so that it now lies entirely within C. I claim that everywhere inside C, the smoothed and inverted rho will have dimension 1, except at the center of C where it has a point fractal dimension of 2.

I'm actually a physicist, and don't have a great mathematical background, but I believe the above claims are valid. Would you agree with them? Is any of this is interesting, or has it been covered many times before?

Thanks, and I look forward to any comments, Chris

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    $\begingroup$ What is "point fractal dimension"? $\;$ $\endgroup$
    – user5810
    Feb 21, 2012 at 23:47
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    $\begingroup$ I don't think anyone will mind if you take the space to give a rigorous definition of what you mean by "smoothed rho." $\endgroup$
    – ShawnD
    Feb 21, 2012 at 23:49
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    $\begingroup$ To smooth it out, start with a time t_o and a point r_o = rho(t_o). Find the maximum time t_1 such that there exists a circle of radius R that contains rho(t) for all t in the interval t_o < t < t_1. So now we have a second point r_1 = rho(t_1) on rho. These two points are a distance R apart. Then repeat: Find the maximum time t_2 such that there exists a circle of Radius R that contains rho(t) for all t in the interval t_1 < t < t_2. Repeat for all positive integer i. We can "turn" this procedure "around" to negative integers. Then these points can be joined with a continuous spline. $\endgroup$
    – chris in physics
    Feb 22, 2012 at 0:24
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    $\begingroup$ Claim 1 seems wrong. With probability $1$, a random walk taking steps of size $1$ with independent and uniform angles will be dense in the plane. However, I don't think the uniformity of the random walk is important. $\endgroup$
    – Douglas Zare
    Feb 22, 2012 at 3:52
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    $\begingroup$ A curve of Hausdorff dimension 1 can definitely be dense in the plane. Even the set of points with rational coordinates, which has Hausdorff dimension 0, is dense in the plane $\endgroup$
    – Pablo Shmerkin
    Feb 22, 2012 at 8:24

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