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Aug 11, 2016 at 14:32 comment added Mikhail Katz @fedja One possible modification of the argument if one does not want to approximate by polygons is the following. Consider the radius $r_t>0$ of a biggest ball that fits inside the curve $L_t$. There are may be several such balls. Then we consider the infimum of those $s_t>0$ for which there is a connected set in the complement of the $s_t$-neighborhood of $L_t$ and containing all of the maximal balls (with center at distance $r_t$ from $L_t$). Then we take a global minimum S of all the $s_t$ and work with this $S>0$ (positive by compactness). We now work with the cut locus truncated at S
Aug 11, 2016 at 13:34 comment added Mikhail Katz Perhaps we can continue via email?
Aug 11, 2016 at 13:33 comment added fedja If you can do a continuous inner approximation, I agree. But is this always possible?
Aug 11, 2016 at 13:03 comment added Mikhail Katz Let it gravitate if it wants. The entire deformation is compact therefore the path of midpoints is at distance from the boundary that is bounded away from zero.
Aug 11, 2016 at 12:57 comment added fedja And now comes the problem: when approximating, the midpoint tends to gravitate towards the narrow winding part, so the "deep inside" story is no longer sheer compactness. I'm not saying it cannot be done, just that some pesky details have to be taken care of :-)
Aug 11, 2016 at 12:55 history edited Mikhail Katz CC BY-SA 3.0
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Aug 11, 2016 at 12:55 comment added Mikhail Katz You are right, one has to approximate first. The rectifiability is for more regular curves, perhaps $C^1$?
Aug 11, 2016 at 12:51 comment added fedja The boundary of $0\le x\le 1/2$, $\sqrt{x}\sin(1/x)-x^2\le y\le \sqrt{x}\sin(1/x)+x^2$, say
Aug 11, 2016 at 12:48 comment added Mikhail Katz Can you describe your example in more detail?
Aug 11, 2016 at 12:47 comment added Mikhail Katz Anyway if you think this is of interest we can try to write this up. My coauthor Kanovei is an expert on Jordan curve theorem. Note that polygonal approximation is far from trivial, but it's been done.
Aug 11, 2016 at 12:47 comment added fedja I will still have a curve going ot the origin within the region and every such curve has infinite length.
Aug 11, 2016 at 12:46 comment added Mikhail Katz the "deep inside" is just a compactness argument.
Aug 11, 2016 at 12:45 comment added Mikhail Katz @fedja, when you take a tubular neighborhood of a curve and then look for the cut locus of the boundary of the tubular epsilon-neighborhood, you won't necessarily get the original curve back except special cases when you have control on curvature and epsilon sufficiently small.
Aug 11, 2016 at 12:41 comment added fedja The cut locus of such a tentacle will go along the curve all the way to the origin, so it will have infinite length. What am I missing? As to the polygons, you'll have to show that the midpoint of the cut locus of the approximation is not just inside but deep inside the region, which is, probably, true, but not quite obvious.
Aug 11, 2016 at 7:48 comment added Mikhail Katz @fedja I did not say the Jordan curve is rectifiable. rather the cut locus is rectifiable as proved in the literature around Marcel Berger. At any rate this is utterly irrelevant because for the purposes of this problem everything can be replaced by polygonal curves by compactness.
Aug 11, 2016 at 0:05 comment added fedja I believe the "tree" part but not the "rectifiable" part (imagine a narrow tentacle around $\sqrt{x}\sin(1/x)$) "Locally rectifiable" it is, but that's not enough to define the midpoint unless I misunderstand what the midpoint is.
Aug 10, 2016 at 10:33 history edited Mikhail Katz CC BY-SA 3.0
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Aug 10, 2016 at 10:22 history answered Mikhail Katz CC BY-SA 3.0