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So the question states, Let $B = \{x = (x_1,x_2,x_3) \in \mathbb{R}^3: x_1^2 +x_2^2 +x_3^2 \leq 1 \}$ be the unit ball in $\mathbb{R}^3$. Compute the diameter of $B$ for each of the following metrics.

note: $diam(B) = \sup\{d(x,y): x,y \in B\}$

I know the diameter is 2 but I want to be able to do this in general for an arbitrary distance. I think seeing this one will help me do others. Here is what I have so far using euclidean distance.

Let $x,y \in B$ then \begin{eqnarray*} d(x,y) &=& ((x_1 - y_1)^2+(x_2 -y_2)^2 + (x_3-y_3)^2)^{1/2} \\ &=& ((x_1^2 +x_2^2+x_3^2)+(y_1^2+y_2^2+y_3^2) -2(x_1y_1+x_2y_2+x_3y_3))^{1/2} \\ &\leq& (1 + 1 - 2(x_1y_1+x_2y_2+x_3y_3))^{1/2} \\ &=& (2(1 -(x_1y_1+x_2y_2+x_3y_3))^{1/2} \end{eqnarray*}

I'm still yet to use the $\sup$ but I'm not sure how to move on from here. Do I need to use Lagrange multipliers or is there a better way to solve this?

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  • $\begingroup$ The term $(x_1y_1+x_2y_2+x_3y_3)$ is the inner product of two unit vectors. This will never have absolute value greater than $1$. How big a hint this is depends on whether you already have a proof of that fact. $\endgroup$ – David K Jan 31 '15 at 21:18
  • $\begingroup$ It is clear that this would be dot product of x and y but I don't see how that helps. $\endgroup$ – Ben Jan 31 '15 at 21:39
  • $\begingroup$ Would it be correct to assume that since we are looking for a diameter, x dot y = |x||y|cos(180) = -|x||y|=-1. $\endgroup$ – Ben Jan 31 '15 at 21:47
  • $\begingroup$ That's the idea. $\endgroup$ – David K Jan 31 '15 at 21:49
  • $\begingroup$ Awesome! Thank you for the hint! $\endgroup$ – Ben Jan 31 '15 at 21:59

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