Direct Methods in the Calculus of Variations (Applied by Bernard Dacorogna

By Bernard Dacorogna

This e-book is a brand new version of the authors past ebook entitled Direct equipment within the Calculus of diversifications, 1989. it's dedicated to the learn of vectorial difficulties within the calculus of diversifications. The e-book has been up-to-date considerably and a few extra examples were integrated. The publication will attraction researchers and graduate scholars in arithmetic and engineering.

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Example text

2) we deduce that b ∈ E. 4) we obtain that a = λb + (1 − λ) y. 3) is satisfied. This achieves the proof of Step1. Step 2. Let us now show that, in fact, the result of Step 1 holds even if y ∈ E. If λ = 1, nothing is to be proved; so assume that λ ∈ (0, 1) and ǫ > 0 is so that Bǫ (x) ⊂ int E. We set z := λx + (1 − λ) y. Let us prove that z ∈ int E. Since y ∈ E, we can find y ∈ E so that |y − y| < λǫ . 1−λ 34 Convex sets and convex functions Then set 1−λ 1 [z − (1 − λ) y] = x + (y − y) . 5) λ λ We therefore have x ∈ Bǫ (x) ⊂ E and hence x ∈ int E.

36) N +2 N +2 N +2 γi f (yi ) βi f (yi ) + λ = i=1 i=1 i=1 βi − βi f (yi ) N +2 ≤ βi f (yi ) . 35) and this concludes Step 3 and thus the theorem. 5, but before that we want to make more precise the connection between the convex hull of a set and the convex envelope of its indicator function. 36 Let E ⊂ RN and χE be the indicator function of E, namely χE (x) = Then 0 if x ∈ E +∞ if x ∈ / E. CχE = χco E . Moreover, if E F∞ := f : RN → R ∪ {+∞} : f |E ≤ 0 , F E := f : RN → R : f |E ≤ 0 , then co E = co E = E , x ∈ RN : f (x) ≤ 0, for every convex f ∈ F∞ x ∈ RN : f (x) ≤ 0, for every convex f ∈ F E .

28), we have immediately |z1 − z|∞ , |z1 − x|∞ ≤ βǫ ⇒ |f (z) − f (z1 )| ≤ ǫ. 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 |∞ .

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