By C.F. Dunkl; D.E. Ramirez
ISBN-10: 039027819X
ISBN-13: 9780390278197
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Sample text
Also, J4(xl, x2) _ -3xi and so (0,0) E 0(Z0). Setting q = (-11,0), then q and (0,0) belong to the same connected component of R2 \ 0(8D). Moreover, 0(x1, x2) = q if and only if (X 1, x2) _ (2', 0). O(x)) = -1; thus d(0, D, p) = -1. 3 Let a, b E R be such that a < b and set D = (a, b). Assume that 0 E C(D) and p ¢ {0(a), 0(b)}. Prove that d(0, D, p) E (-1,0, 11. 3. We first note that p ¢ 0(8D) and so d(0, D, p) is well defined. P(x) - a(a) (x - a) + 0(a). If 0(b) = O(a), then 0 is a constant and d(i,1, D, p) = 0.
2) Let V C D be an open set such that a E V, 4(a) = p and 4(x) 96 p for every x E V, x 96 a. 8 we have i(4,ai,P) = d(4,V,P) = sgn(J4(a)) Let Ai, ... , A be the eigenvalues of V4(a). \,,. where the complex eigenvalues occur in conjugate pairs a, a such that ar > 0, and so sgn(JJ(a)) = (-1)'. 3 The multiplication theorem Given two mappings 4 E Cl(D)N and 0 E Ci(4(D))N we want to compute d(t, o 4, D, p) in terms of d(i/), Di, p) and d(4, D, Di), where Di (i E N) are the connected components of RN \ 4(8D).
Define 0 : BN - SN-1 by 0(x) = H ( I X12' IXI2) , x E BN. It is clear that 0 is continuous at every x E BN\{0}. Using the uniform continuity of H and the fact that H(x, 0) = c for every x E BN, we deduce that ' is continuous at 0. Therefore, we find 0 E F such that the equation 0(x) = 0 has no solution in BN. Finally, we prove that (iii) As before, setting 0(x) . (i). Assume that cp is homotopic to a constant. _, x12) , x E BN, it is easy to verify that 0 is a continuous extension of 'p, homotopic to a constant.