By Kollar J., Lazarsfeld R., Morrison D. (eds.)

**Read or Download Algebraic Geometry Santa Cruz 1995, Part 2 PDF**

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**Additional resources for Algebraic Geometry Santa Cruz 1995, Part 2**

**Example text**

Definition 2. , Ili equivalently for any covector Qi E AnRn+N, we define the comass of 0 with respect to f, briefly the f -comass of 0 by n+N 11011f := supY(77) r7EA R For any vector C E AnRn+N we define the f -mass of := supWC) If 11 EAnI n+N by 110IIf<11 Proposition 2. We have 11 f > 0, f * (0) = 0 or +oo for all 0 E AnRn+N (ii) 11 4 11 = sup{q5(M(G))/ f (G) I G E M(N, n) }. (iii) the following claims are equivalent (a) II0IIf <<1 (b) ¢(M(G)) - f (G) < 0 V G E M(N, n) (2) I1 (c)0(C)-f(C)<0 V6 EA+ (drr) j* (0) = 0 (iv) F(b) = 116 11 f C d S E An,Rn+N Proof.

Here we consider the autonomous case which is simpler 1. Regular Variational Integrals 18 Theorem S. c.. c. l f (u) dµ < liminf f f (uk) dIL k--oo . 0 Proof. First we observe that f is the pointwise supremum of an increasing sequence of functions fj which are convex, Lipschitz, and piecewise linear. Actually each fj is the pointwise supremum of a finite number of affine functions fj(p) = max{2i(p) £i : 1R" -, IR affine, i = 1, ... , N(j)} , fj(p) > 0 Thus it suffices to prove the result for such fj that from now on we denote again by f.

Therefore Hahn-Banach's theorem yields Proposition 2. c.. Then f is never -oo if and only if there exists an affine map P : V -+ 1R such that $(x) < f (x) V x E V. Definition 4. c. , g < f } . c. c. function. Moreover I'C f is proper if f is not identically +oo and has an affine minorant. Also, Hahn-Banach's theorem yields Theorem 5. Suppose that TC f : V -+ lk has an of fine minorant. Then I'C f (x) = sup{2(x) I 1 : V -> R, 2 affine, P < f } sup{g(x) I g : V -+ 1R, g convex, g < f } . 1 Polyconvexity and Polyconvex Envelops 29 Notice that in the second equality of Theorem 5 the convex functions g take values in R, not in R U {+oo}.