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If we have a smooth plane curve (Hausdorff dimension 1), we can thicken it by a small amount to get a 2D set (all points within distance $\epsilon$ to the curve).

What if we start with the graph of a scalar Wiener process, which has Hausdorff dimension 1.5? We can again thicken to get a 2D set, but in some sense this feels like overkill: we’re already halfway from 1 to 2.

Question: Is a natural way to “thicken less” to enlarge a Brownian motion graph into a Hausdorff dimension 2 set?

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    $\begingroup$ The "obvious" thickening is known as (not joking) the Wiener sausage. There are results as to how the volume scales as the thickness $\epsilon$ goes to zero, and some of those asymptotics might describe what happens if you thicken "infinitesimally". $\endgroup$ Commented Jun 14, 2020 at 19:27
  • $\begingroup$ @NateEldredge the picture on the left side of the wikipedia page doesn't with the naming... $\endgroup$ Commented Sep 1, 2020 at 19:09

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Let $Z$ be a set on the line of Hausdorff dimension $1/2$, e.g., the middle-half Cantor set or the zero set of another Brownian motion. Now let $W_t$ be a one dimensional Brownian motion, and consider the set $\Lambda:=\{(t+z,W_t): t \in [0,1], z \in Z\}$. This set will have Hausdorff dimension 2 (e.g. because the sum of two independent Brownian Zero sets has Hausdorff dimension 1 and then apply the Marstrand slicing theorem), though in the examples I mentioned its 2-dimensional measure will be 0.

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  • $\begingroup$ Nice! Though intuitively, it feels like this doesn’t quite correspond to the notion of a “thickening”, since it maps each point of a disconnected set. Not sure how to formalize that, however. $\endgroup$ Commented Aug 30, 2020 at 21:29
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    $\begingroup$ @GeoffreyIrving Every connected set with more than two points has Hausdorff dimension at least $1$, since it projects to an interval with dimension 1. This suggests that every similar idea with a connected set will be an overkill. $\endgroup$
    – Will Sawin
    Commented Aug 31, 2020 at 0:27

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