# An elementary question about metrics on the real plane [closed]

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 UstinovMay 31 '16 at 3:38

This question appears to be off-topic. The users who voted to close gave these specific reasons:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Neil Hoffman, Myshkin, Alexey Ustinov
• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Bill Johnson, Ilya Bogdanov
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• Isn't this equivalent to the line segment being the shortest path between two points for the p-metric ($p\geq1$) on the plane? – M.G. May 30 '16 at 21:20
• 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 $p = \infty$. – Todd Trimble May 30 '16 at 21:22
• 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 $d_1(x, z) + d_1(y, z) = 2 = d_1(x, z)$. – Todd Trimble May 30 '16 at 21:30
• Google "strictly convex space". – Bill Johnson May 30 '16 at 23:34