I need to show that $d_∞(x, y) ≤ d_2(x, y) ≤ d_1(x, y) ≤ nd_∞(x, y)$

where $d_1=|x_1-0|+|y_1-0|$ and I'm setting $|x_1-0|+|y_1-0|<0$. Illustrating the $B(0,1)$ balls (centered at 0 with radius 1) gives me a square with 4 points of intersections $(1,0),(0,1),(0,-1),(-1,0)$.

I then add the metrics $d_2$=$\sqrt{|x_1-0|^2+|y_1-0|^2}$ and $d_∞=\max_ {1≤i≤n}|x_i − y_i|$

Here is what they look like:


Now this is completely counter intuitive since it's nested in the opposite order than I expected. Did I go wrong anywhere?

enter image description here

  • 3
    $\begingroup$ At the risk of being flagged, smaller distances must have bigger balls. You have plotted the level sets of the distance function. For example, if $d_2(x,y) \le 1$ you must have $d_\infty(x,y) \le 1$, hence the $2$-ball must be contained in the $\infty$-ball. $\endgroup$ – copper.hat Nov 2 '15 at 17:57
  • $\begingroup$ I see that it holds when I plug in points and evaluate the distances. Also I can see that from the Pythagoras that $d_2≤d_1$ but I don't know how to show $d_∞<d_2$ and $nd_∞>d_1$ $\endgroup$ – GRS Nov 2 '15 at 18:07
  • $\begingroup$ If $a_k$ are non negative you have $\max(a_1,...,a_n) \le a_1+...+a_n \le n \max(a_1,...,a_n)$. Also, $(\max(a_1,...,a_n))^2 = \max(a_1^2,...,a_n^2) \le a_1^2+...+a_n^2$. $\endgroup$ – copper.hat Nov 2 '15 at 18:14
  • $\begingroup$ I tried writing up the proof, but I got stuck on proving $d_2<d_1$. It seems to make sense in$R^2$ that this is Pythagoras, but how can I do this for $R^n$? $\endgroup$ – GRS Nov 2 '15 at 18:57
  • $\begingroup$ Show $d_2^2 \le d_1^2$. $(\sum_k |x_k|)^2 = \sum_i \sum_j |x_i||x_j| \ge \sum_i |x_i|^2$. $\endgroup$ – copper.hat Nov 2 '15 at 19:01

The balls have been drawn correctly. E,g, note that being in the smallest ball $d_1((x,y),(0,0)) \le 1$ implies (if the inequalities are correct, which they are) that both $d_\infty((x,y),(0,0)) \le 1$ and $d_2((x,y),(0,0)) \le 1$ as well. So indeed the $d_1$ ball is contained in the other two.

For $n=2$: $d_{1}((x,y),(u,v)) = |x-u| + |y-v| \le 2\max(|x-u|,|y-v|)$, as we estimate both terms with the max. For $n$ coordinates we get a factor $n$.

Also $\max(|x-u|,|y-v|)^2 \le |x-u|^2 + |y-v|^2$ as the max squared is one of the terms in the sum, and all terms are positive. Now take two square roots to see that $d_{\infty} \le d_2$.

  • $\begingroup$ I see the idea, but what about $nd_∞$? Can I simply use Pythagoras in this proof and say that 2 hypotenuses is always bigger than $d_1$ ? $\endgroup$ – GRS Nov 2 '15 at 18:11

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