Carla Peri INTEGRAL GEOMETRY IN MINKOWSKI PLANE 1 ...

116Adding these inequalities we getLL 0 -2T1(S 0 +5)>0.Since an anticircle of radius r has area II r 2previous inequality becomes(16) rL-S-Ilr 2 >0,and perimeter 2IIr thewhere r satisfies the condition (15).From (16), we obtain, exactly as in the Euclidean case 4t, the following(17) L 2 - 4IIS > j [(L - 2llr,) 2 + (2Ur e - L) 2 ] ., LThis inequality shows that L 2 — 4115 = 0 only when r t - = r e = —- ,i.e. when K is an anticircle. Moreover we can statePROPOSITION. If a bounded convex set K is not an anticircle then the setof the convex sets congruent to K whose boundaries have at least fourpoints in common with the boundary of K has measure greater then zero.From (17) by using the inequality 2(x 2 + y 2 ) > (x +y) 2 , we canfinally deriveGeneralized Bonnesen inequality(18) L 2 -4ns>Ii 2 (r e -ri) 2(compare [4] for the Euclidean case).