Given the metric $d_p$ on the real plane, i.e.

$$ d_p(x,y) = d_p((x_1, y_1), (x_2, y_2)) = [|x_1 - x_2|^p+ |y_1 - y_2|^p]^{1/p} $$

for which values of $p$ ($\geq 1$) is it true that the following set is the usual line segment between points $x$ and $y$ :

$$ \{ z \in \mathbb{R}^2 \;|\; d_p(x,z) + d_p(z,y) = d_p(x,y)\}$$

more generally is there a simple meaningful characterization for a general metric $d$ for which the above is the usual line segment ?


closed as off-topic by Bill Johnson, Ilya Bogdanov, Neil Hoffman, Myshkin, Alexey Ustinov May 31 '16 at 3:38

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  • $\begingroup$ Isn't this equivalent to the line segment being the shortest path between two points for the p-metric ($p\geq1$) on the plane? $\endgroup$ – M.G. May 30 '16 at 21:20
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    $\begingroup$ Unless I've made a mistake, should be true if $1 < p < \infty$. Notice that the metric is invariant under translations, and so WLOG we may consider the case $y = 0$ (i.e. $y = (0, 0)$), where the line segment lies in this set by the simple calculation $d(tx, 0) = td(x, 0)$ for $t \in [0, 1]$ by homogeneity, where $d(tx, x) = (1-t)d(x, 0)$ again by translation. By strict convexity of the unit balls, I think the only cases where you could get more than the line segment are the extreme cases $p = 1$ and $p = \infty$. $\endgroup$ – Todd Trimble May 30 '16 at 21:22
  • $\begingroup$ So for example, in the $p = 1$ metric, the balls of radius $1$ about $y = (0, 0)$ and $x = (1, 1)$ don't just touch at $(1/2, 1/2)$; they intersect along the line segment connecting $(1, 0)$ to $(0, 1)$. For all those points $z$ on that line segment, you get the equation $d_1(x, z) + d_1(y, z) = 2 = d_1(x, z)$. $\endgroup$ – Todd Trimble May 30 '16 at 21:30
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    $\begingroup$ Google "strictly convex space". $\endgroup$ – Bill Johnson May 30 '16 at 23:34