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By Bruno Finzi

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26) uEHJ(O\E). )]v. 18) with ({3u, v) = fn uv. Moreover, it can be directly verified that the support cone K 0 of K, in this case, is K 0 = HJ(n \E), which is a linear space. 25). 3 (b) gives us the result in [104]. The case, where a is not necessarily symmetric, may also be treated by the methods of chapter 6. (b) We note that, ifO is an interior point of K, then, K 0 = V. In fact, in this case, K 0 is a cone that contains a neighborhood of 0. Thus, K 0 = V. J 0 0 However, the condition 0 E K (K is the interior part of K) is not necessary for K 0 = V, as shown in the next example.

18 2. , for all x E V, the mapping u f-t (Au, x), u EM, is continuous on M. (c) A is called bounded if it maps bounded sets of V into bounded sets of V*. Let j be a mapping from V to lR U {oo }. Definition ([14] and [65]) (a) D(j) = {u E V: j(u) < oo} is called the effective domain of j. , j ¢:. oo ). X,oo]) is open (respectively, weakly open) in V for all A E JR. It is well known ([14]) that j is lower semicontinuous (respectively, weakly lower semicontinuous) if and only if, r j(x):::; liminfj(y) (respectively j(x):::; liminfj(y)), y-x y~x for all x E V.

A) is symmetric and linear with respect to u. 36) uo(x) > 0, Vx E (0, a]. One has u 0 E KJ. In fact, consider D = CQ"(O, 1). Then, D is dense in HJ(O, 1), and, for ¢ E D, supp ¢ n (0, a] is a compact subset of (0, a]. 36), uo(x):::: inf supp n [O,a] uo = m > 0, Vx E supp¢ n (O,a]. tJ llr/>IIL""(0,1) = uo(x)- m > 0. 2 The case of a simple eigenvalue 53 For x E [O,a) \ supp¢, (uo + E)(x) = uo(x) :2:0. , uo + € E Ko, and, thus, uo E KJ. 36). 27). 33). 35). 3 In this example, we consider the following fourth-order variational inequality related to the equilibrium problem of beams (cf.

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