# Direct Methods in the Calculus of Variations by Bernard Dacorogna

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By Bernard Dacorogna

This ebook is built for the examine of vectorial difficulties within the calculus of diversifications. the topic is a really lively one and nearly half the booklet includes new fabric. this can be a new version of the sooner e-book released in 1989 and it really is appropriate for graduate scholars. The e-book has been up-to-date with a few new fabric and examples additional. purposes are integrated.

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Example text

13, it is enough to show that any e ∈ co E can be written as N +1 e= i=1 λi xi , xi ∈ Eext , λ ∈ ΛN +1 . We consider two cases. Case 1. We first assume that e ∈ ∂ (co E) . 10 (ii) to get a ∈ RN , a = 0, and α ∈ R so that e; a = α ≤ x; a , for every x ∈ co E. Next define K := {x ∈ co E : x; a = α} and observe that it is a non empty (since e ∈ K) convex and compact set that lies in a subspace of dimension (N − 1) . We may therefore apply the induction hypothesis and write N e= i=1 λi xi , xi ∈ Kext , λ ∈ ΛN .

29) that |z1 − z|∞ , |z1 − x|∞ ≤ β ⇒ |f (z) − f (z1 )| ≤ 2αN 2N |z1 − z|∞ . 30) Now let z2 be such that |z2 − x|∞ ≤ β. Let u1 , u2 , · · · , uM ∈ [z1 , z2 ] (the segment in RN with endpoints z1 and z2 ) be such that u1 = z1 , u2 , · · · , uM = z2 and |um − um+1 |∞ ≤ β, m = 1, · · · , M − 1. Note that, since |z1 − x|∞ , |z2 − x|∞ ≤ β, then |um − x|∞ ≤ β, m = 1, · · · , M. 30), we immediately get |um − um+1 |∞ ≤ β ⇒ |f (um ) − f (um+1 )| ≤ 2αN 2N |um − um+1 |∞ . β Summing the above inequalities, we obtain |z1 − x|∞ , |z2 − x|∞ ≤ β ⇒ |f (z1 ) − f (z2 )| ≤ 2αN 2N |z1 − z2 |∞ β 51 Convex functions and whence the result.

Ii) If int E = ∅, then int E = E. (iii) ∂E = ∂E. 3 in Rockafellar [514]). ♦ Proof. We divide the proof into five steps. Step 1. We first show that if x ∈ int E and y ∈ E, then z := λx + (1 − λ) y ∈ int E, for every λ ∈ (0, 1] . 1) Since x ∈ int E, we can find ǫ > 0 such that Bǫ (x) := b ∈ RN : |b − x| < ǫ ⊂ E. 1) we show that Bλǫ (z) ⊂ E, for every λ ∈ (0, 1] . 3) So we choose a ∈ Bλǫ (z) and we let b := x + 1 1 1 (a − z) = a + (1 − )y. 2) we deduce that b ∈ E. 4) we obtain that a = λb + (1 − λ) y.