Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be $\mathbb{R}^3$ with the sup norm $\|\cdot\|_{\infty}$. Let $Y=\{x\in X: \|x\|_{\infty}=1\}$. For $x,y\in Y$ define $d(x,y)$ to be the arc length of shortest paths on $Y$ joining $x,y$. (It is easy to see that generally there are more than one shortest path and these shortest paths must be a union of line segments on $Y$.) My question is: If $y\neq -x$, is the set $$Y\cap \{\lambda x+\mu y: \lambda\ge 0, \mu\ge 0\}$$ always a shortest path joining $x,y$?

share|improve this question
    
Sorry for the answer. I didn't see the sup norm. –  Beni Bogosel Nov 4 '11 at 18:49
2  
When you say "define $d(x,y)$ to be the arc length of shortest paths on $Y$ blablablabla", is it still with respect to the sup norm or with respect to the euclidean norm? –  Patrick Da Silva Nov 4 '11 at 18:55
    
With respect to the sup norm. –  TCL Nov 4 '11 at 20:18

1 Answer 1

up vote 5 down vote accepted

Somewhat surprisingly (for me), this is not the case.

Consider two points $p_x$ and $p_y$ on the faces $f_x$ and $f_y$ lying on the planes $x=1$ and $y=1$, respectively. On $f_x$, we can't travel in the $x$ direction, and on $f_y$ we can't travel in the $y$ direction, so the length of a path connecting the points is at least the sum of the distances of the points from the common edge. On the other hand, the length is also at least the difference in the $z$ coordinates of the two points. If we choose these two bounds to be the same, then there is exactly one shortest path connecting the points, namely the one that runs diagonally on each face in order to cover the same distance along $z$ as towards the edge.

The coordinates of the points in this case can be parametrized as

$$p_x=\pmatrix{1\\1\\z}+\pmatrix{0\\-a\\-a}\;,$$

$$p_y=\pmatrix{1\\1\\z}+\pmatrix{-b\\0\\b}\;,$$

where $z$ is the $z$ coordinate of the point where the shortest path crosses the edge, and $a,b\ge0$.

Now the path defined in the question lies in the plane

$$\lambda p_x+\mu p_y=\lambda\left[\pmatrix{1\\1\\z}+\pmatrix{0\\-a\\-a}\right]+\mu\left[\pmatrix{1\\1\\z}+\pmatrix{-b\\0\\b}\right]\;.$$

For this to be the shortest path, it would have to contain the point $(1,1,z)$. But setting this expression equal to $(1,1,z)$ yields $\lambda a=\mu b$ from the first two equations, and substituting that into the third leads to $\lambda z+\mu z=z$. Unless $z=0$, this implies $\lambda+\mu=1$, which contradicts the first two equations unless $a=0$ and $b=0$. Thus in general the shortest path does not lie in that plane.

share|improve this answer
    
Just to add the following as a record: Let $a=1/4,b=1/4,z=1/2$ in your example. Let $A=p_x=(1,3/4,1/4)$ , $B=p_y=(3/4,1,3/4)$ and $C=(1,1,1/2)$. The shortest path, which is the union of the line segments $AC$ and $CB$, has length $1/2$. While the path from $A$ to $B$ as given in the question is $AM\cup MB$, where $M=(1,1,4/7)$. This path has length $4/7$($>1/2$). –  TCL Nov 8 '11 at 16:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.