... hA1 ∩U1 ,U1 = sup{u : u ∈ PSH(U1 ), u 1, u|A1∩U1 0} and its upper - semicontinuous regularization h∗ ∩U1 ,U1 Then h∗ ∩U1 ,U1 is A A plurisubharmonic on U1 Put F1 = {z1 ∈ U1 : hA1 ∩U1 ,U1 (z1 ... ∩U1 ,U1 (z1 )} A Some Results on the Relation Between 279 Theorem 7 .1 in [3] implies that F1 is pluripolar in U1 and h∗ ∩U1 ,U1 = hA1 ∩U1 ,U1 A on U1 \F1 Hence h∗ ∩U1 ,U1 (z1 ) = for z1 ∈ A1 ... × C and D2 × C respectively Setting E1 = {z1 ∈ D1 : u1 (z1 , w ) = −∞}, E2 = {z2 ∈ D2 : u2 (z2 , w ) = −∞} Then E1 and E2 are pluripolar sets in D1 and D2 respectively Let A1 = D1 \E1 0 and A2...
... to that in the BOP is 1.5 :1 in Russia, 2 :1 in Colombia, 2 :1 in India, and 3 :1 in Thailand—but reaches 11 :1 in Nigeria and 14 :1 in South Africa The relative sizes of urban and rural BOP health ... Africa, $54 in Asia, $56 in Eastern Europe, and $10 7 in Latin America Comparable numbers for health care are $ 15 4 in Africa, $13 1 in Asia, $ 15 2 in Eastern Europe, and $3 25 in Latin America and for ... Phone: ( 410 ) 51 6 -6 956 or (1- 800) 53 7 -54 87 Fax: ( 410 ) 51 6 -6998 Email: hfscustserv@press.jhu.edu U.K, Europe, Middle East, Africa, Asia Eurospan Group, c/o Turpin Distribution Phone: +44 (0) 17 67...
... a force Suppose A and B (Fig 1) are two bodies attached to tubes which can slide vertically up and down the rod E F, and that two balls C and D A A and B by rods hinged at and B, then the balls ... force between the velocities of A and B, and therefore acting their kinetic energy will change from time to time the kinetic energy lost by A and B will really have gone to increase the kinetic energy ... composed, and potential energy depending on the relative position of its parts The potential energy may be of various kinds; thus we may have potential energy due to gravity and potential energy...
... 206–208, 19 72 S C Bose and S K Laskar, “Fixed point theorems for certain class of mappings,” Journal of Mathematical and Physical Sciences, vol 19 , no 6, pp 50 3 50 9, 19 85 D Downing and W A Kirk, ... ∅, we have that ∞ B1 Am 2 .5 m Fix an arbitrary x1 ∈ S1 and define, by induction, a sequence {ym } such that {ym } ∈ Sm and the segment ym , ym does not meet Bm Given x ∈ B1 , there exists a ... x − xn − d 1 a/m m 1 ym a/ m 1 2.6 ym It is a routine to check that S is a continuous mapping from B1 to B1 Furthermore, S Am ⊂ ym , ym ⊂ Am for every m ≥ 4 Fixed Point Theory and Applications...