Given $M := \mathbb R^2 \setminus B_r(0)$ I want to determine the function

$$\rho(x,y) = \inf \{ \int_a^b |\gamma'(t)|\,dt : \gamma : [a,b] \to M \text{ is a smooth curve connecting } x \text{ with }y \}$$

on $M$.

I understand that for $\overline{xy} := \{rx + (1-r)y : r\in [0,1]\}$ not intersecting $B_r(0)$ we have $\rho(x,y) = |x-y|$.

Otherwise I don't know how to procede, not even intuitively. When I think of connecting $x$ and $y$ then, I'm not sure whether to do this with two line segments or whether I should use something "curved".

Even if I could somehow prove that among "piecewise polynomial curves" only certain curves are "minimal", I wouldn't know why "piecewise polynomial curves" are the best ones to begin with.

What can I do?


1 Answer 1


In the case of intersecting; the shorter path is constructed with the tangents to the disk from each point and the arc between the points of tangency.

enter image description here

Consider moving $F$ on the boundary towards $C$ (but still on the boundary), clearly the new path passes by $F$ too but deviates from the previous straight line connecting $C$ and $F$, the shortest path. Now, move $F$ in the opposite sense that of course cannot be on the border. All the figure is displaced outwards making the path larger. From here the total distance can be calculated.

To prove it analytically doesn't seem difficult.

  • $\begingroup$ Thanks for answering. - Unfortunately I probably have to ask new question because I asked the wrong question: I wanted $B_r$ to refer to the closed ball... $\endgroup$
    – Jinx
    Nov 17, 2018 at 19:48
  • $\begingroup$ I think in this case there is no shortest path. $\endgroup$ Nov 17, 2018 at 19:54
  • $\begingroup$ So, would $\rho(x,y)$ be the length of the path in your picture, nonetheless? $\endgroup$
    – Jinx
    Nov 17, 2018 at 19:55
  • $\begingroup$ The length of the path in the drawing is the greater lower bound of the length of all possible paths. $\endgroup$ Nov 17, 2018 at 20:12
  • $\begingroup$ I'll take this as a 'yes' ;) $\endgroup$
    – Jinx
    Nov 17, 2018 at 21:07

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