If $X=\mathbb{R}^2$. For $x=(x_1,x_2)$, $y=(y_1,y_2)$ define


Prove or disprove $(X,d_{1/2})$ is a metric space.

Attempt at a solution: After multiplying it out, it seems to boil down to $2|x_1-y_1|^{1/2}|x_2-y_2|^{1/2}$ satisfying the triangle inequality. However, I can't seems to prove or disprove this one way or another.

Intution, however, is leading me towards the fact that it is in fact not a metric space.

  • $\begingroup$ It might be easy to prove one way or the other if the distance were defined in a readable way. $\endgroup$ – André Nicolas Sep 12 '12 at 4:17


try $x = (1,0), y = (0,1)$.

Another way of gaining an intuition for this is thinking about the unit ball defined by this "norm", i.e. the set of points whose distance to the origin is $1$. For various $p$, if the norm is $( \sum_{i=1}^2 |x_i - y_i|^p )^{1/p}$, the balls look like this (from Wikipedia):enter image description here

What do all the balls for legitimate norms have in common, and what does this have to do with the triangle inequality?


Try computing the triangle inequality on each pair from $(0,0), (0,d), (d,d)$.


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