# Convexity

• Feb 11th 2010, 04:20 PM
Richard
Convexity
Let $\displaystyle x$ and $\displaystyle y$ be distinct points in a Euclidean metric space. The set $\displaystyle \lbrace z \vert d(z,x) < d(z,y)\rbrace$ is convex. Does the same apply for any linear combination of the distances, i.e.: Is $\displaystyle \lbrace z \vert \lambda d(z,x) < (1 - \lambda) d(z,y)\rbrace$, where $\displaystyle 0 \leq \lambda \leq 1$, in any case convex?
• Feb 11th 2010, 04:29 PM
Drexel28
Quote:

Originally Posted by Richard
Let $\displaystyle x$ and $\displaystyle y$ be distinct points in a Euclidean metric space. The set $\displaystyle \lbrace z \vert d(z,x) < d(z,y)\rbrace$ is convex. Does the same apply for any linear combination of the distances, i.e.: Is $\displaystyle \lbrace z \vert \lambda d(z,x) < (1 - \lambda) d(z,y)\rbrace$, where $\displaystyle 0 \leq \lambda \leq 1$, in any case convex?

I just did this recently on another site :). What have you tried?
• Feb 11th 2010, 04:40 PM
Richard
Convexity
It's clear to me that for the simple case that $\displaystyle \lambda = 1 - \lambda = .5$, the set must be convex. To generate the said set, you just need to draw a line through the two points, half it, and draw a perpendicular line through the half. The halfplane on the side of $\displaystyle x$ is then the said set, which is convex. But I did not manage to generalise the proof.