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Extra resources for Introduction To The Calculus Of Variations
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I denotes the scalar product in Rn . (ii) The bidual of f is the function f ∗∗ : Rn → R ∪ {±∞} defined by f ∗∗ (x) = sup {hx; x∗ i − f ∗ (x∗ )} . 4. 53 (i) Let n = 1 and f (x) = |x| /p, where 1 < p < ∞. We then find 0 1 f ∗ (x∗ ) = 0 |x∗ |p p where p0 is, as usual, defined by 1/p + 1/p0 = 1. ¡ ¢2 (ii) Let n = 1 and f (x) = x2 − 1 . We then have f ∗∗ (x) = (iii) Let n = 1 and ⎧ ¡ ¢2 ⎨ x2 − 1 ⎩ f (x) = We immediately find that ∗ ⎧ ⎨ ⎩ ∗ 0 0 if |x| < 1 . if x ∈ (0, 1) +∞ otherwise. ∗ f (x ) = sup {xx } = x∈(0,1) if |x| ≥ 1 ⎧ ∗ ⎨ x ⎩ 0 if x∗ ≥ 0 if x∗ ≤ 0 f is often called the indicator function of (0, 1), and f ∗ the support function.
Before stating these results we need to define what kind of regularity will be assumed on the boundary of the domains Ω ⊂ Rn that we will consider. When Ω = (a, b) ⊂ R, there was no restriction. We will assume, for the sake of simplicity, that Ω ⊂ Rn is bounded. The following definition expresses in precise terms the intuitive notion of regular boundary (C ∞ , C k or Lipschitz). 40 (i) Let Ω ⊂ Rn be open and bounded. , n} ¡ ¢ ¡ ¢ H ∈ C k Q , H −1 ∈ C k U , H (Q+ ) = U ∩ Ω, H (Q0 ) = U ∩ ∂Ω with Q+ = {x ∈ Q : xn > 0} and Q0 = {x ∈ Q : xn = 0}.
This concludes the proof of the theorem when n = 1. 1 Let 1 ≤ p < ∞, R > 0 and BR = {x ∈ Rn : |x| < R}. Let for f ∈ C ∞ (0, +∞) and for x ∈ BR u (x) = f (|x|) . p (i) Show that u ∈ L (BR ) if and only if Z R rn−1 |f (r)|p dr < ∞ . 0 (ii) Assume that ¤ £ lim rn−1 |f (r)| = 0 . r→0 Prove that u ∈ W 1,p (BR ) if and only if u ∈ Lp (BR ) and Z R p rn−1 |f 0 (r)| dr < ∞ . 33. 2 Let AC ([a, b]) be the space of absolutely continuous functions on [a, b]. This means that a function u ∈ AC ([a, b]), if for every > 0 there exists δ > 0 so that for every disjoint union of intervals (ak , bk ) ⊂ (a, b) the following implication is true X X |bk − ak | < δ ⇒ |u (bk ) − u (ak )| < .