Comment by danbruc

For point one the reason is of course that π(p) has a [global] minimum at 2. Actually showing that is not that easy because the involved integrals have no closed form solution but in principle it is not too hard. The circumference of the circles is 2 π(p) and equals four times the length of the quarter circle in the first quadrant which has all x and y positive and allows dropping the absolute values. The quarter circle is y(x) = (1 - x^p)^(1/p) with x in [0, 1]. Its length is the integral of ds over dx from 0 to 1 where ds is the arc length in the Lp norm ds = (dx^p + dy^p)^(1/p) which yields ds = (1 + (dy/dx)^p)^(1/p) dx. For dy/dx you insert the derivative of the quarter circle dy/dx = -x^(p - 1)(1 - x^p)^(1/p - 1) and finally you have to compute the derivative of the integral with respect to p and find the zeros to figure out where the extrema are. Well, technically you have to also look at the second and third derivative to confirm that it is a minimum and check the limiting behavior. The referenced paper works around the integral by modifying the function in a way such that it still agrees with the original function in some relevant points but yields a solvable integral and shows that using the modified function does not alter the result.